A+B,氵题一道,84种解法!大佬羡慕

千里之行始于足下梦开始的地方 \huge千里之行始于足下梦开始的地方 千里之行始于足下梦开始的地方
A + B p r o b l e m C + + 史上最详解,氵题一道, 84 种解法! \huge A+B problem C++史上最详解,氵题一道,84种解法! A+BproblemC++史上最详解,氵题一道,84种解法!


(来自大佬们的评价:)(附加多种语法的代码

来自https://www.acwing.com/solution/content/69403/的题解

为了不误导初学者

先放一波正常代码:

/*
             A          K         K         CCCCCCCCCCCC SSSSSSSSSSS  PPPPPPPPPPPPP
            A A         K       K          C             S            P           P
           A   A        K     K           C              S            P           P
          A     A       K   K            C               S            P           P
         A       A      K K              C               SSSSSSSSSSS  PPPPPPPPPPPPP
        AAAAAAAAAAA     K   K            C                         S  P
       A           A    K     K           C                        S  P
      A             A   K       K          C                       S  P
     A               A  K         K         CCCCCCCCCCC  SSSSSSSSSSS  P


             A          K         K         IIIIIIIIIII    OOOOOOOOOOO    IIIIIIIIIII
            A A         K       K                I         O         O         I
           A   A        K     K                  I         O         O         I
          A     A       K   K                    I         O         O         I
         A       A      K K                      I         O         O         I
        AAAAAAAAAAA     K   K                    I         O         O         I
       A           A    K     K                  I         O         O         I
      A             A   K       K                I         O         O         I
     A               A  K         K         IIIIIIIIIII    OOOOOOOOOOO    IIIIIIIIIII
*/
#include 
using namespace std;
int a, b;                   //定义两个全局整型变量 a 和 b 
int main() {
    scanf("%d%d", &a, &b);  //读入 a 和  b,当然用 cin 也没毛病
    printf("%d\n", a + b);  //输出 a +b,当然用 cout 也没毛病
    return 0;
} 

这个引用表示强调……

首先声明此题非常困(jian)难(dan),连国(ru)家(men)队(cai)员(niao)都做不(de)出,我想了1年(miao)才(jiu)把这题想出来了。

此帖强烈建议收藏!!!

													 祝
												点赞的盆友们!
								身体健康万事如意周赛AK题解爆赞期末考试三科满分!

前方高能!!!警告!!!

本文长度恐怖!请各位大佬蒟蒻们小心! \huge本文长度恐怖!请各位大佬蒟蒻们小心! 本文长度恐怖!请各位大佬蒟蒻们小心!

建议慢慢食用…

其实有四五个算法是查的但是我也懵了……

算法一、DFS一号

#include 
using namespace std;
int n = 2, a[5], s;
int dfs(int x, int sum) {
    if (x > n) return sum;
    int i = dfs(x + 1, sum);
    int j = dfs(x + 1, sum + a[x]);
    if (i == s) return i;
    if (j == s) return j;
    return -1;
}
int main() {
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]), s += a[i];
    cout << dfs(1, 0) << endl;
    return 0;
}

算法二、DFS二号

#include 
using namespace std;
int a, b;
int dfs(int x) {
    if (x <= 5) return x;
    return dfs(x / 2) + dfs(x - x / 2);
} 
int main() {
    scanf("%d%d", &a, &b);
    printf("%d\n", dfs(a) + dfs(b));
    return 0;
}

算法三、BFS

#include 
using namespace std;
int n = 2, a[5], s;
queue q;
void bfs() {
    q.push(0);
    int c = 0;
    while (q.size()) {
        c++;
        int f = q.front(); q.pop();
        if (f == s) {printf("%d\n", f); exit(0);}
        q.push(f + a[c]);
        q.push(f);
    }
}
int main() {
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]), s += a[i];
    bfs();
    return 0;
}

算法四、直接算咯 (这叫算法?!

#include 
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
    return 0;
} 

算法五、二分

#include 
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    int l = 0, r = 200000000;
    while (l < r) {
        int mid = l + r >> 1;
        if (mid == a + b) {printf("%d\n", mid); return 0;}
        if (mid <  a + b) l = mid + 1;
        if (mid >  a + b) r = mid - 1;
    }
    cout << l << endl;
    return 0;
}

算法六、稍微有点暴力的枚举
但是还是1892ms
卡过了欸

#include 
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    for (int i = 0;i <= 200000000; i++) if (a + b == i) {printf("%d\n", i); break;}
    return 0;
} 

算法七、最短路之dijkstra
思路:定义节点1到节点2路径长度为a,节点2到节点3路径长度为b
则答案为节点1到节点3的最短路(也就是a+b)

#include 
using namespace std;
int w[5][5], d[5], v[5];
int n = 3;
void dijkstra() {
    memset(d, 0x3f, sizeof d);
    memset(v, 0, sizeof v);
    d[1] = 0;
    for (int i = 1;i < n; i++) {
        int x = 0;
        for (int j = 1;j <= n; j++)
            if (!v[j] && (x == 0 || d[j] < d[x])) x = j;
        v[x] = 1;
        for (int y = 1;y <= n; y++)
            d[y] = min(d[y], d[x] + w[x][y]);
    }
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    memset(w, 0x3f, sizeof w);
    w[1][2] = a; w[2][3] = b;
    dijkstra();
    printf("%d\n", d[3]);
    return 0;
}

算法八、最短路之SPFA
思路同上

#include 
using namespace std;
int a, b, n = 3;
int w[5][5], d[5], v[5];
queue q;
void spfa() {
    memset(d, 0x3f, sizeof d);
    memset(v, 0, sizeof v);
    d[1] = 0, v[1] = 1;
    q.push(1);
    while (q.size()) {
        int x = q.front(); q.pop();
        v[x] = 0;
        for (int i = 1;i <= n; i++) {
//          if (w[x][i] == 0x3f) continue;
            if (d[i] > d[x] + w[x][i]) {
                d[i] = d[x] + w[x][i];
                if (!v[i]) q.push(i), v[i] = 1;
            }
        }
    }
}
int main() {
    scanf("%d%d", &a, &b);
    memset(w, 0x3f, sizeof w);
    w[1][2] = a; w[2][3] = b;
    spfa();
    printf("%d\n", d[3]);
    return 0;
}

算法九、最短路之Floyd
思路同上

#include 
using namespace std;
int d[5][5], n = 3;
int main() {
    int a, b; scanf("%d%d", &a, &b);
    memset(d, 0x3f, sizeof d);
    d[1][2] = a; d[2][3] = b;
    for (int k = 1;k <= n; k++)
        for (int i = 1;i <= n; i++)
            for (int j = 1;j <= n; j++)
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
    printf("%d\n", d[1][3]);
    return 0;
}

算法十、高精

#include
using namespace std;
string a0, b0;
int a[1005], b[1005];
int main(){
    cin >> a0 >> b0;
    int l1 = a0.size(), l2 = b0.size();
    for (int i = 0;i < l1; i++) a[l1 - i] = a0[i] - 48;
    for (int i = 0;i < l2; i++) b[l2 - i] = b0[i] - 48;
    l1 = max(l1, l2);
    for (int i = 1;i <= l1; i++) {
        a[i] += b[i];
        if (a[i] > 9) a[i + 1] += 1, a[i] %= 10;
    }
    if (a[max(l1, l2) + 1] > 0) l1++;
    for (int i = l1;i >= 1; i--) printf("%d", a[i]);
    return 0;
}

算法十一、最小生成树之kruskal

思路其实和最短路的一样,只是改成用最小生成树的方法求罢了

#include 
using namespace std;
struct rec {
    int x, y, z;
} edge[5];

int fa[5], m = 2, ans = 0;

int get(int x) {
    if (x == fa[x]) return x;
    return fa[x] = get(fa[x]);
}
int cmp(rec a, rec b) { return a.z < b.z; }

int main() {
    int a, b; scanf("%d%d", &a, &b);
    edge[1] = (rec){1, 2, a};
    edge[2] = (rec){2, 3, b};
    for (int i = 1;i <= m + 1; i++) fa[i] = i;
    sort(edge + 1, edge + 1 + m, cmp);
    for (int i = 1;i <= m; i++) {
        int x = get(edge[i].x);
        int y = get(edge[i].y);
        if (x == y) continue;
        fa[x] = y;
        ans += edge[i].z;
    }
    printf("%d\n", ans);
    return 0;
}

算法十二、最小生成树之prim
思路同上

#include 
using namespace std;
int w[5][5], d[5], n = 3, ans, v[5];

void prim() {
    memset(d, 0x3f, sizeof d);
    memset(v, 0, sizeof v);
    d[1] = 0;
    for (int i = 1;i < n; i++) {
        int x = 0;
        for (int j = 1;j <= n; j++)
            if (!v[j] && (x == 0 || d[j] < d[x])) x = j;
        v[x] = 1;
        for (int y = 1;y <= n; y++)
            if (!v[y]) d[y] = min(d[y], w[x][y]);
    }
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    memset(w, 0x3f, sizeof w);
    w[1][2] = a; w[2][3] = b;
    prim();
    int ans = 0;
    for (int i = 2;i <= n; i++) ans += d[i];
    printf("%d\n", ans);
    return 0;
}

算法十三、前缀和

#include 
using namespace std;
int a[5], s[5];
int main() {
    for (int i = 1;i <= 2; i++) scanf("%d", &a[i]), s[i] += a[i] + s[i - 1];
    printf("%d\n", s[2]);
    return 0;
}

算法十四、后缀和

#include 
using namespace std;
int a[5], s[5];
int main() {
    for (int i = 2;i >= 1; i--) scanf("%d", &a[i]), s[i] += a[i] + s[i + 1];
    printf("%d\n", s[1]);
    return 0;
}

算法十五、位运算

#include 
using namespace std;
int add(int a, int b) {
    if (b == 0) return a;
    return add(a ^ b, (a & b) << 1);
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    printf("%d\n", add(a, b));
    return 0;
}

算法十六、树的直径——BFS
emmm……思路继续和最短路的一样。。。

#include 
using namespace std;
const int maxn = 1e5 + 10;

int head[maxn * 2],edge[maxn * 2],Next[maxn * 2],ver[maxn * 2];
int vis[maxn], dist[maxn];
int n = 3, p, q, d;
int tot = 0;
int maxd = 0;

void add(int u,int v,int w) {
    ver[ ++ tot] = v,edge[tot] = w;
    Next[tot] = head[u],head[u] = tot;
}

int BFS(int u) {
    queueQ;
    while(!Q.empty()) Q.pop();
    memset(vis, 0, sizeof vis);
    memset(dist, 0, sizeof dist);  
    Q.push(u);
    int x, max_num = 0;
    while(!Q.empty()) {
        x = Q.front();
        Q.pop();
        vis[x] = 1;
        for(int i = head[x]; i ; i = Next[i]) {
            int y = ver[i];
            if(vis[y]) continue;
            vis[y] = 1;
            dist[y] = dist[x] + edge[i];
            if(dist[y] > maxd) {
                maxd = dist[y];
                max_num = y;
            }
            Q.push(y);
        }
    }
    return max_num;
}
int main(void) {
    int a, b; scanf("%d%d", &a, &b);
    add(1, 2, a); add(2, 1, a);
    add(2, 3, b); add(3, 2, b);
    BFS(BFS(1));
    int ans = 0;
    for (int i = 1;i <= n; i++) ans = max(ans, dist[i]);
    printf("%d\n", ans);
    return 0;
}

算法十七、树的直径——DFS
思路同上

#include
#define maxn 100000
using namespace std;
struct node {
    int u, v, w, nex;
} edge[2 * maxn + 10];
int n = 3, d[maxn + 10], head[maxn + 10], f_num, cnt = 0, ans;
inline void add(int x,int y,int z) {
    edge[++cnt].u = x;
    edge[cnt].v = y;
    edge[cnt].w = z;
    edge[cnt].nex = head[x];
    head[x] = cnt;
}
inline void dfs(int x, int fa) {
    if(ans < d[x]) {
        ans = d[x];
        f_num = x;
    }
    for (int i = head[x]; i != -1; i = edge[i].nex) {
        int j = edge[i].v;
        if (j == fa)continue;
        d[j] = d[x] + edge[i].w;    
        dfs(j, x);
    }
}
int main() {
    memset(head, -1, sizeof(head));
    int a, b; scanf("%d%d", &a, &b);
    add(1, 2, a); add(2, 1, a);
    add(2, 3, b); add(3, 2, b);
    dfs(1, 0);
    ans = 0;
    d[f_num] = 0;
    dfs(f_num, 0);
    for (int i = 1;i <= n; i++) ans = max(ans, d[i]);
    printf("%d", ans);
    return 0;
}

算法十八、树的直径——树形DP
还是一样咯

#include 
using namespace std;
int f[5], n = 3, cnt, h[5], ans, dis[5];
struct edge {
    int to, next, vi;
} e[5];
void add(int u, int v, int w) {
    e[cnt].to= v;
    e[cnt].vi = w;
    e[cnt].next = h[u];
    h[u] = cnt++;
}
void dp(int u, int fa) {
    for (int i = h[u]; ~i; i = e[i].next) {
        int v = e[i].to;
        if (v == fa) continue;
        dp(v, u);
        ans = max(ans, dis[v] + dis[u] + e[i].vi);
        dis[u] = max(dis[u], dis[v] + e[i].vi);
    }
}
int main() {
    memset(h, -1, sizeof h);
    int a, b; scanf("%d%d", &a, &b);
    add(1, 2, a); add(2, 1, a);
    add(2, 3, b); add(3, 2, b);
    dp(1, 0);
    printf("%d\n", ans);
    return 0;
}

算法十九、网络流
从别的大佬那边搞过来的,但是一点都看不懂┭┮﹏┭┮

#include
using namespace std;
#define set(x) Set(x)
#define REP(i,j,k) for (int i=(j),_end_=(k);i<=_end_;++i)
#define DREP(i,j,k) for (int i=(j),_start_=(k);i>=_start_;--i)
#define mp make_pair
#define x first
#define y second
#define pb push_back
template inline bool chkmin(T &a,const T &b){ return a > b ? a = b, 1 : 0; }
template inline bool chkmax(T &a,const T &b){ return a < b ? a = b, 1 : 0; }
typedef long long LL;
typedef pair node;
const int dmax = 1010, oo = 0x3f3f3f3f;
int n, m;
int a[dmax][dmax] , ans;
int d[dmax], e[dmax];
priority_queue  q;
inline bool operator >(node a,node b){ return a.y>b.y; }
bool p[dmax];
void Set(int x){ p[x] = 1; }
void unset(int x){ p[x] = 0; }
bool check(int x){ return x != 1 && x != n && !p[x] && e[x] > 0; }
void preflow(){
    e[1] = oo;
    d[1] = n - 1;
    q.push(mp(1, n - 1));
    set(1);
    while (!q.empty()) {
        bool flag = 1;
        int k = q.top().x;
        q.pop(), unset(k);
        DREP(i, n, 1)
        if ((d[k] == d[i] + 1 || k == 1) && a[k][i] > 0){
            flag = 0;
            int t = min(a[k][i], e[k]);
            e[k] -= t;
            a[k][i] -= t;
            e[i] += t;
            a[i][k] += t;
            if (check(i)) {
                q.push(mp(i, d[i]));
                set(i);
            }
            if (e[k] == 0) break;
        }
        if (flag) {
            d[k] = oo;
            REP(i, 1, n)
            if (a[k][i] > 0) chkmin(d[k], d[i] + 1);
        }
        if (check(k)) {
            q.push(mp(k, d[k]));
            set(k);
        }
    }
    ans = e[n];
}
int main() {
    n = 2, m = 2;
    int x, y;
    scanf("%d%d", &x, &y);
    a[1][2] += x + y;
    preflow();
    printf("%d\n", ans);
    return 0;
}

算法二十、线段树
转化为区间求和问题

#include 
#define l(x) tree[x].l
#define r(x) tree[x].r
#define sum(x) tree[x].sum
#define add(x) tree[x].add
using namespace std;
struct SegmentTree {
    int l, r; //区间左右端点 
    long long sum, add; //sum 区间和  add 延迟标记 
} tree[400010];
int a[100010], n = 1, m = 2;
void build (int p, int l, int r) {
    l(p) = l, r(p) = r;
    if(l == r) {sum(p) = a[l]; return;}
    int mid = l + r >> 1;
    build(p * 2, l, mid);
    build(p * 2 + 1, mid + 1, r);
    sum(p) = sum(p * 2) + sum(p * 2 + 1);
}
void spread(int p) {
    if(add(p)) { //节点p有标记 
        sum(p * 2) += add(p) * (r(p * 2) - l(p * 2) + 1); //更新左子节点信息 
        sum(p * 2 + 1) += add(p) * (r(p * 2 + 1) - l(p * 2 + 1) + 1); //更新右子节点
        add(p * 2) += add(p); //给左子节点打延迟标记 
        add(p * 2 + 1) += add(p); //给右子节点打延迟标记 
        add(p) = 0; //清除p的标记 
    }
}
void change(int p, int l, int r, int d) {
    if(l <= l(p) && r >= r(p)) { //完全覆盖 
        sum(p) += (long long) d * (r(p) - l(p) + 1); //更新节点信息 
        add(p) += d; //给节点打延迟标记 
        return;
    }
    spread(p); //下传延迟标记 
    int mid = l(p) + r(p) >> 1;
    if(l <= mid) change(p * 2, l, r, d);
    if(r > mid) change(p * 2 + 1, l, r, d);
    sum(p) = sum(p * 2) + sum(p * 2 + 1);
}
long long ask(int p, int l, int r) {
    if(l <= l(p) && r >= r(p)) return sum(p);
    spread(p);
    int mid = l(p) + r(p) >> 1;
    long long val = 0;
    if(l <= mid) val += ask(p * 2, l, r);
    if(r > mid) val += ask(p * 2 + 1, l, r);
    return val;
}
int main() {
    a[1] = 0;
    build(1, 1, n);
    while(m--) { 
        int d = 0;
        scanf("%d", &d);
        change(1, 1, 1, d);
    }
    printf("%lld\n", ask(1, 1, 1));
    return 0;
}

算法二十一、树状数组
思路一样,区间求和

#include
using namespace std;
const int SIZE = 100010;
int a[SIZE], n = 1, m = 2;
long long c[2][SIZE], sum[SIZE];

long long ask(int k, int x) {
    long long ans = 0;
    for(; x ; x -= x & -x) ans += c[k][x];
    return ans;
}

void add(int k,int x,int y) {
    for(; x <= n; x += x & -x) c[k][x] += y;
}

int main() {
    a[1] = 0;
    while(m--) {
        int d = 0;
        scanf("%d", &d);
        add(0, 1, d);
        add(0, 2, -d);
        add(1, 1, d);
        add(1, 2, -2 * d);
    }
    long long ans = sum[1] + 2 * ask(0, 1) - ask(1, 1);
    ans -= sum[0] + 1 * ask(0, 0) - ask(1, 0);
    printf("%lld\n", ans);
    return 0;
}

算法二十二、分块
思路一样,区间求和

#include
using namespace std;
long long a[50000010], sum[50000010], add[50000010];
int L[50000010], R[50000010];
int pos[50000010];
int n = 1, m = 2, t;

void change(int l, int r, long long d) {
    int p = pos[l], q = pos[r];
    if (p == q) {
        for (int i = l; i <= r; i++) a[i] += d;
        sum[p] += d*(r - l + 1);
    }
    else {
        for (int i = p + 1; i <= q - 1; i++) add[i] += d;
        for (int i = l; i <= R[p]; i++) a[i] += d;
        sum[p] += d*(R[p] - l + 1);
        for (int i = L[q]; i <= r; i++) a[i] += d;
        sum[q] += d*(r - L[q] + 1);
    }
}

long long ask(int l, int r) {
    int p = pos[l], q = pos[r];
    long long ans = 0;
    if (p == q) {
        for (int i = l; i <= r; i++) ans += a[i];
        ans += add[p] * (r - l + 1);
    }
    else {
        for (int i = p + 1; i <= q - 1; i++)
            ans += sum[i] + add[i] * (R[i] - L[i] + 1);
        for (int i = l; i <= R[p]; i++) ans += a[i];
        ans += add[p] * (R[p] - l + 1);
        for (int i = L[q]; i <= r; i++) ans += a[i];
        ans += add[q] * (r - L[q] + 1);
    }
    return ans;
}

int main() {
    a[1] = 0;
    t = sqrt(n*1.0);
    for (int i = 1; i <= t; i++) {
        L[i] = (i - 1)*sqrt(n*1.0) + 1;
        R[i] = i*sqrt(n*1.0);
    }
    if (R[t] < n) t++, L[t] = R[t - 1] + 1, R[t] = n;
    for (int i = 1; i <= t; i++)
        for (int j = L[i]; j <= R[i]; j++) {
            pos[j] = i;
            sum[i] += a[j];
        }
    while (m--) {
        int d;
        scanf("%d", &d);
        change(1, 1, d);
    }
    printf("%lld\n", ask(1, 1));
}

算法二十三、LCT
来自洛谷

#include
using namespace std;
struct node
{
    int data,rev,sum;
    node *son[2],*pre;
    bool judge();
    bool isroot();
    void pushdown();
    void update();
    void setson(node *child,int lr);
}lct[233];
int top,a,b;
node *getnew(int x)
{
    node *now=lct+ ++top;
    now->data=x;
    now->pre=now->son[1]=now->son[0]=lct;
    now->sum=0;
    now->rev=0;
    return now;
}
bool node::judge()
{
    return pre->son[1]==this;
}
bool node::isroot()
{
    if(pre==lct)return true;
    return !(pre->son[1]==this||pre->son[0]==this);
}
void node::pushdown()
{
    if(this==lct||!rev)return;
    swap(son[0],son[1]);
    son[0]->rev^=1;
    son[1]->rev^=1;
    rev=0;
}
void node::update()
{
    sum=son[1]->sum+son[0]->sum+data;
}
void node::setson(node *child,int lr)
{
    this->pushdown();
    child->pre=this;
    son[lr]=child;
    this->update();
}
void rotate(node *now)
{
    node *father=now->pre,*grandfa=father->pre;
    if(!father->isroot()) grandfa->pushdown();
    father->pushdown();
    now->pushdown();
    int lr=now->judge();
    father->setson(now->son[lr^1],lr);
    if(father->isroot()) now->pre=grandfa;
    else grandfa->setson(now,father->judge());
    now->setson(father,lr^1);
    father->update();
    now->update();
    if(grandfa!=lct) grandfa->update();
}
void splay(node *now)
{
    if(now->isroot())return;
    for(; !now->isroot(); rotate(now))
    if(!now->pre->isroot())
    now->judge()==now->pre->judge()?rotate(now->pre):rotate(now);
}
node *access(node *now)
{
    node *last=lct;
    for(; now!=lct; last=now,now=now->pre) {
        splay(now);
        now->setson(last,1);
    }
    return last;
}
void changeroot(node *now)
{
    access(now)->rev^=1;
    splay(now);
}
void connect(node *x,node *y)
{
    changeroot(x);
    x->pre=y;
    access(x);
}
void cut(node *x,node *y)
{
    changeroot(x);
    access(y);
    splay(x);
    x->pushdown();
    x->son[1]=y->pre=lct;
    x->update();
}
int query(node *x,node *y)
{
    changeroot(x);
    node *now=access(y);
    return now->sum;
}
int main()
{
    scanf("%d%d",&a,&b);
    node *A=getnew(a);
    node *B=getnew(b);
    connect(A,B);
    cut(A,B);
    connect(A,B);
    printf("%d",query(A,B));
    return 0;
}

算法二十四、Splay
来自洛谷

#include 
#define ll long long
#define N 100000
using namespace std;
int sz[N], rev[N], tag[N], sum[N], ch[N][2], fa[N], val[N];
int n, m, rt, x;
void push_up(int x){
    sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1;
    sum[x] = sum[ch[x][1]] + sum[ch[x][0]] + val[x];
}
void push_down(int x){
    if(rev[x]){
        swap(ch[x][0], ch[x][1]);
        if(ch[x][1]) rev[ch[x][1]] ^= 1;
        if(ch[x][0]) rev[ch[x][0]] ^= 1;
        rev[x] = 0;
    }
    if(tag[x]){
        if(ch[x][1]) tag[ch[x][1]] += tag[x], sum[ch[x][1]] += tag[x];
        if(ch[x][0]) tag[ch[x][0]] += tag[x], sum[ch[x][0]] += tag[x];
        tag[x] = 0;
    }
}
void rotate(int x, int &k){
    int y = fa[x], z = fa[fa[x]];
    int kind = ch[y][1] == x;
    if(y == k) k = x;
    else ch[z][ch[z][1]==y] = x;
    fa[x] = z; fa[y] = x; fa[ch[x][!kind]] = y;
    ch[y][kind] = ch[x][!kind]; ch[x][!kind] = y;
    push_up(y); push_up(x);
}
void splay(int x, int &k){
    while(x != k){
        int y = fa[x], z = fa[fa[x]];
        if(y != k) if(ch[y][1] == x ^ ch[z][1] == y) rotate(x, k);
        else rotate(y, k);
        rotate(x, k);
    }
}
int kth(int x, int k){
    push_down(x);
    int r = sz[ch[x][0]]+1;
    if(k == r) return x;
    if(k < r) return kth(ch[x][0], k);
    else return kth(ch[x][1], k-r);
}
void split(int l, int r){
    int x = kth(rt, l), y = kth(rt, r+2);
    splay(x, rt); splay(y, ch[rt][1]);
}
void rever(int l, int r){
    split(l, r);
    rev[ch[ch[rt][1]][0]] ^= 1;
}
void add(int l, int r, int v){
    split(l, r);
    tag[ch[ch[rt][1]][0]] += v;
    val[ch[ch[rt][1]][0]] += v;
    push_up(ch[ch[rt][1]][0]);
}
int build(int l, int r, int f){
    if(l > r) return 0;
    if(l == r){
        fa[l] = f;
        sz[l] = 1;
        return l;
    }
    int mid = l + r >> 1;
    ch[mid][0] = build(l, mid-1, mid);
    ch[mid][1] = build(mid+1, r, mid);
    fa[mid] = f;
    push_up(mid);
    return mid;
}
int asksum(int l, int r){
    split(l, r);
    return sum[ch[ch[rt][1]][0]];
}
int main(){
    //总共两个数
    n = 2;
    rt = build(1, n+2, 0);//建树
    for(int i = 1; i <= n; i++){
        scanf("%d", &x);
        add(i, i, x);//区间加
    }
    rever(1, n);//区间翻转
    printf("%d\n", asksum(1, n));//区间求和
    return 0;
}

算法二十五、LCA
来自洛谷

#include                                                  //头文件
#define NI 2                                                          
//从来不喜欢算log所以一般用常数 不知道算不算坏习惯 因为3个节点 所以log3(当然以2为底)上取整得2
struct edge
{
    int to,next,data;                                              //分别表示边的终点,下一条边的编号和边的权值
}e[30];                                                                     //邻接表,点少边少开30是为了浪啊
int v[10],d[10],lca[10][NI+1],f[10][NI+1],tot=0;      //数组开到10依然为了浪
//数组还解释嘛,v表示第一条边在邻接表中的编号,d是深度,lca[x][i]表示x向上跳2^i的节点,f[x][i]表示x向上跳2^i的距离和
void build(int x,int y,int z)                                      //建边
{
    e[++tot].to=y; e[tot].data=z; e[tot].next=v[x]; v[x]=tot;
    e[++tot].to=x; e[tot].data=z; e[tot].next=v[y]; v[y]=tot;
}
void dfs(int x)                                                        //递归建树
{
    for(int i=1;i<=NI;i++)                                   //懒,所以常数懒得优化
        f[x][i]=f[x][i-1]+f[lca[x][i-1]][i-1],
        lca[x][i]=lca[lca[x][i-1]][i-1];                   //建树的同时进行预处理
    for(int i=v[x];i;i=e[i].next)                              //遍历每个连接的点
    {
        int y=e[i].to;
        if(lca[x][0]==y) continue;
        lca[y][0]=x;                                       //小技巧:lca[x][0]即为x的父亲~~(向上跳2^0=1不就是父节点嘛)
        f[y][0]=e[i].data;
        d[y]=d[x]+1;
        dfs(y);                                            //再以这个节点为根建子树【这里真的用得到嘛??】
    }
}
int ask(int x,int y)                                             //询问,也是关键
{                                                                        
    if(d[x]=0;i--)                                  //不知道能不能正着循环,好像倒着优,反正记得倒着就好了
        if(lca[x][i]!=lca[y][i])                            //如果x跳2^i和y跳2^j没跳到一起就让他们跳
            ans+=f[x][i]+f[y][i],
            x=lca[x][i],y=lca[y][i];
    return ans+f[x][0]+f[y][0];                           //跳到LCA上去(每步跳的时候都要更新信息,而且要在跳之前更新信息哦~)
}
int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    build(1,2,a);
    build(1,3,b);                                                       //分别建1 2、1 3之间的边
    dfs(1);                                                                //以1为根建树
    printf("%d",ask(2,3));                                         //求解2 3到它们的LCA的距离和并输出
}
算法二十六、字典树
来自洛谷

#include
#include
#include
#include
using namespace std;
struct node{
    int str[26];
    int sum;
}s[1000];
char str1[100];
int t=0,tot=0,ss=0;
bool f1;
void built()
{
    t=0;
    for(int i=0;i

嘿嘿洛谷的没有啦~

算法二十七、Bellman-Ford
思路和别的最短路解法一样~

#include 
using namespace std;
int dis[50], u[50], v[50], w[50], n, m;
void bellman(int start) {
    for (int i = 1;i <= n; i++) dis[i] = 0x3f3f3f3f;
    dis[start] = 0;
    for (int i = 1;i < n; i++)
        for (int j = 1;j <= m; j++)
            if (dis[v[j]] > dis[u[j]] + w[j]) dis[v[j]] = dis[u[j]] + w[j];
}
int main() {
    n = 3; m = 2;
    for (int i = 1;i <= m; i++) cin  >> w[i], u[i] = i, v[i] = i + 1;
    bellman(1);
    printf("%d\n", dis[3]);
    return 0;
}

算法二十八、可耻的打表

#include 
using namespace std;
int a, b; int main() { 
    scanf("%d%d", &a, &b);
    if (a == 3 && b == 4) printf("7");
    if (a == 45 && b == 55) printf("100");
    if (a == 123 && b == 321) printf("444");
    if (a == 91086199 && b == 18700332) printf("109786531");
    if (a == 42267194 && b == 60645282) printf("102912476");
    if (a == 69274392 && b == 10635835) printf("79910227");
    if (a == 5710219 && b == 85140568) printf("90850787");
    if (a == 75601477 && b == 24005804) printf("99607281");
    if (a == 70597795 && b == 90383234) printf("160981029");
    if (a == 82574652 && b == 22252146) printf("104826798");
    return 0;           //hh,这个len没加上return 0,还是我加的……
}

算法二十九、SPFA求最短路之SLF优化
呃呃呃就是加了个SLF优化而已

#include
using namespace  std;
const int maxn = 100000 + 10;
const int INF = 0x7FFFFFFF;

int pre[maxn], dis[maxn], path[maxn];
bool vis[maxn];
int head[maxn], n, m;

int tot, cnt;
struct node {
    int v, w, next;
} E[2 * maxn];
void add(int u, int v, int w) {
    E[tot].v = v;
    E[tot].w = w;
    E[tot].next = head[u];
    head[u] = tot++;
}
void init() {
    tot = 0;
    memset(vis, 0, sizeof vis);
    memset(head, -1, sizeof head);
}
void spfa(int st) {
    for (int i = 1;i <= n; i++) vis[i] = false, dis[i] = INF;
    int now, next;
    dis[st] = 0; vis[st] = 1;
    deque q;
    q.push_back(st);
    pre[st] = -1;
    while(!q.empty()) {
        now = q.front();
        q.pop_front();
        vis[now] = 0;
        for (int i = head[now]; i != -1;i = E[i].next) {
            next = E[i].v;
            if(dis[next] > dis[now] + E[i].w) {
                dis[next] = dis[now] + E[i].w;
                pre[next] = now;
                if(!vis[next]) {
                        vis[next] = 1; 
                        if (q.empty() || dis[next] > dis[q.front()]) q.push_back(next);
                        else q.push_front(next);
                }
            }
        }
    }
}
void print(int x) {
    if(pre[x] == -1) return;
    print(pre[x]);
    printf("%d ", x);
}
int main() {
    init();
    n = 3; m = 2;
    int w;
    for (int i = 1;i <= m; i++) {scanf("%d", &w); add(i, i + 1, w);}
    spfa(1);
    if(dis[n] == INF) puts("-1");
    else printf("%d\n", dis[n]);
    return 0;
}

算法三十、SPFA之LLL优化

#include
#define MAXN 10010
#define MAXM 500010
#define MAX 2147483647
using namespace std;
int n, m, t, c = 1;
int head[MAXN], path[MAXN];
bool vis[MAXN];
struct node {
    int next, to, w;
}a[MAXM << 1];
inline int relax (int u, int v, int w) {
    if (path[v] > path[u] + w) {
        path[v] = path[u] + w;
        return 1;
    }
    return 0;
}
inline void add(int u, int v, int w) {
    a[c].to = v;
    a[c].w = w;
    a[c].next = head[u];
    head[u] = c++;
}
void spfa() {
    int u, v, num = 0;
    long long x = 0;
    list q;
    for (int i = 1;i <= n; i++){path[i] = MAX; vis[i] = 0;}
    path[1] = 0;
    vis[1] = 1;
    q.push_back(1);
    num++;
    while (!q.empty()) {
        u = q.front();
        q.pop_front();
        num--; x -= path[u];
        while (num && path[u] > x / num){
            q.push_back(u);
            u = q.front();
            q.pop_front();
        }
        vis[u] = 0;
        for (int i = head[u]; i ; i = a[i].next) {
            v = a[i].to;
            if (relax(u, v, a[i].w) && !vis[v]) {
                vis[v] = 1;
                if(!q.empty() && path[v] < path[q.front()]) q.push_front(v);
                else q.push_back(v);
                num++; x += path[v];
            }
        }
    }
}
int main() {
    n = 3; m = 2;
    for (int i = 1;i <= m; i++) {
        int w;
        scanf("%d", &w);
        add(i, i + 1, w);
    }
    spfa();
    printf("%d\n", path[n]);
    return 0;
}

算法三十一、SPFA之SLF+LLL优化算法

#include 
using namespace std;
const int INF = 1 << 30;
const int gg = 200000 + 11;
int head[gg], dis[gg], n, m, cnt;
bool vis[gg];
int sum, tot;
struct node{
    int net, to, w;
} a[gg];

inline void add(int i, int j, int w) {
    a[++cnt].to = j;
    a[cnt].net = head[i];
    a[cnt].w = w;
    head[i] = cnt;
}

inline void spfa(int s) {
    deque q;
    for (int i = 1;i <= n; i++) dis[i] = INF;
    dis[s] = 0; vis[s] = 1;    
    q.push_back(s);
    tot = 1;
    while(!q.empty()) {
        int u = q.front();
        q.pop_front();
        vis[u] = false;
        tot--;
        sum -= dis[u];
        for (int i = head[u]; ~i ; i = a[i].net) {
            int v = a[i].to;
            if (dis[v] > dis[u] + a[i].w) {
                dis[v] = dis[u] + a[i].w;
                if(!vis[v]) {
                    vis[v] = 1;
                    if (q.empty() || dis[v] > dis[q.front()] || dis[v] * tot <= sum) q.push_back(v);
                    tot++;
                    sum += dis[v];
                }
            }
        }
    }
}

int main() {
    memset(head, -1, sizeof head);
    n = 3; m = 2;
    for (int i = 1;i <= m; i++) {
        int w = 0;
        scanf("%d", &w);
        add(i, i + 1, w);
    }
    spfa(1);
    if (dis[n] == INF)  puts("-1");
    else printf("%d\n", dis[n]);
    return 0;
}

算法三十二、只用一个变量跑A+B
把一个long long拆成两个int
指针啊!!!

#include
using namespace std;
long long a;
int main() {
    scanf("%d%d", (int*)(&a), (int*)(&a+1));
    printf("%d\n", *((int*)&a) + *((int*)(&a+1)));
    return 0;
}

算法三十三、矩阵乘法

#include
using namespace std;
int a, b;
int x[2][2] = {
    {0, 1},
    {1, 1}
};
void mo(int f[]) {
    int ans[2] = {0};
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++) ans[i] += f[j] * x[i][j];
    for(int i = 0; i < 2; i++) f[i] = ans[i];
}
int main() {
    cin >> a >> b;
    int f[3] = {a, b};
    mo(f);
    cout << f[1];
    return 0;
}

算法三十四、STL+dijkstra

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int N=405;
struct Edge {
    int v,w;
};
vector edge[N*N];
int n;
int dis[N*N];
bool vis[N*N];
struct cmp {
    bool operator()(int a,int b) {
        return dis[a]>dis[b];
    }
};
int Dijkstra(int start,int end)
{
    priority_queue,cmp> dijQue;
    memset(dis,-1,sizeof(dis));
    memset(vis,0,sizeof(vis));
    dijQue.push(start);
    dis[start]=0;
    while(!dijQue.empty()) {
        int u=dijQue.top();
        dijQue.pop();
        vis[u]=0;
        if(u==end)
            break;
        for(int i=0; idis[u]+edge[u][i].w) {
                dis[v]=dis[u]+edge[u][i].w;
                if(!vis[v]) {
                    vis[v]=true;
                    dijQue.push(v);
                }
            }
        }
    }
    return dis[end];
}
int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    Edge Qpush;

    Qpush.v=1;
    Qpush.w=a;
    edge[0].push_back(Qpush);

    Qpush.v=2;
    Qpush.w=b;
    edge[1].push_back(Qpush);

    printf("%d",Dijkstra(0,2));
    return 0;
}

算法三十五、数学表达式

#include 
using namespace std;
long long a, b;
int main() {
    cin >> a >> b;
    cout << a - b + (a * 2) - (a - b) - a + (a + (b - a)) << endl;
    return 0;
}

算法三十六、define大法

#include 
#define ___ int
#define $$$ main
#define _$_$_ return
#define _ cin
#define $ cout
#define __ using
#define $$ namespace
#define o_o std
__ $$ o_o;
___ $$$(){
    ___ _$o$_,$o_o$;
    _ >> _$o$_ >> $o_o$;
    $ << _$o$_ + $o_o$;
    _$_$_ 0;
}

算法三十七、压位高精度加法

奇怪的知识又增加了!

#include 
using namespace std;
const int mod = 100000000;
vector add(vector &A, vector &B) {
    vector C;
    int t = 0;
    for (int i = 0; i < A.size() || i < B.size(); i++) {
        if (i < A.size()) t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % mod);
        t /= mod;
    }
    if (t) C.push_back(t);
    return C;
}
int main() {
    string a, b; cin >> a >> b;
    vector A, B, C;
    for (int i = a.size() - 1, s = 0, j = 0, t = 1; i >= 0; i--) {
        s += (a[i] - '0') * t;
        j++; t *= 10;
        if (j == 8 || i == 0) A.push_back(s), s = 0, j = 0, t = 1;
    }
    for (int i = b.size() - 1, s = 0, j = 0, t = 1; i >= 0; i--) {
        s += (b[i] - '0') * t;
        j++; t *= 10;
        if (j == 8 || i == 0) B.push_back(s), s = 0, j = 0, t = 1;
    }
    C = add(A, B);
    cout << C.back();
    for (int i = C.size() - 2; i >= 0; i--) printf("%08d", C[i]);
    return 0;
}

算法三十八、加一大堆东东……

虽然好像没优化多少
#pragma GCC diagnostic error "-std=c++11"
#pragma GCC target("avx")
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#include 
using namespace std;
int main() {
    int a, b; scanf("%d%d", &a, &b);
    printf("%d", a + b);
    return 0;
}

算法三十九、暴力枚举优化版
和算法六区别“不大”但是时间优化了300ms……
时间:1567ms

就是在 min(2×a,2×b)
到 max(2×a,2×b)
之间枚举,效率高了“亿”点点

#include 
using namespace std;
int main() {
    int a, b; scanf("%d%d", &a, &b);
    for (int i = min(2 * a, 2 * b);i <= max(2 * a, 2 * b); i++)
        if (a + b == i) {
            printf("%d", i); //注意要输出i,不然如果输出a+b循环就没意义了……
            return 0;
        }
}

算法四十、矩阵DP
就是和方格取数之类的这些同样的思维~

#include 
using namespace std;
int a[110][110], n = 2;
int main() {
    for (int i = 1;i <= n; i++)
        for (int j = 1;j <= n; j++) scanf("%d", &a[i][j]);
    for (int i = 1;i <= n; i++)
        for (int j = 1;j <= n; j++) 
            if (max(a[i - 1][j], a[i][j - 1]) > -1) a[i][j] += max(a[i - 1][j], a[i][j - 1]);
    printf("%d\n", a[n][n]);
    return 0;
}

算法四十一、拖延时间大法

yyds!永远的拖延时间!

3176 ms天哪!

#include 
//STL 通用算法
#include 
//STL 位集容器
#include 
//字符处理
#include 
//定义错误码
#include 
//浮点数处理
#include 
//对应各种运算符的宏
#include 
//定义各种数据类型最值的常量
#include 
//定义本地化函数
#include 
//定义数学函数
#include 
//复数类
#include 
//信号机制支持
#include 
//异常处理支持
#include 
//不定参数列表支持
#include 
//常用常量
#include 
//定义输入/输出函数
#include 
//定义杂项函数及内存分配函数
#include 
//字符串处理
#include 
//定义关于时间的函数
#include 
//宽字符处理及输入/输出
#include 
//宽字符分类
#include 
//STL 双端队列容器
#include 
//异常处理类
#include 
//文件输入/输出
#include 
//STL 定义运算函数(代替运算符)
#include 
//定义各种数据类型最值常量
#include 
//STL 线性列表容器
#include 
//本地化特定信息
#include 
//STL 映射容器
#include 
//STL通过分配器进行的内存分配
#include 
//动态内存分配
#include 
//STL常用的数字操作
#include 
//参数化输入/输出
#include 
//基本输入/输出支持
#include 
//输入/输出系统使用的前置声明
#include 
//数据流输入/输出
#include 
//基本输入流
#include 
//STL迭代器
#include 
//基本输出流
#include 
//STL 队列容器
#include 
//STL 集合容器
#include 
//基于字符串的流
#include 
//STL 堆栈容器
#include 
//标准异常类
#include 
//底层输入/输出支持
#include 
//字符串类
#include 
//运行期间类型信息
#include 
//STL 通用模板类
#include 
//对包含值的数组的操作
#include 
//STL 动态数组容器


//头文件拖延编译时间(虽然不能拖延运行时间,但能拖一点编译时间也很不错了hh) 
using namespace std;
int main(){
    int a; int b; //不用int a, b;,拖延运行时间
    cin >> a >> b; //cin拖延运行时间
    int ans = 1 * 10000 / 10 / 10 / 10 / 10 * 5 * 2 / 10 - 1; //ans表达式拖延编译和运行时间
    for (int i = 1;i <= a; i++) ans += 5, ans -= 4; //拖延时间 
    for (int i = 1;i <= b; i++) ans += 5, ans -= 4; //拖延时间 
    ans = ans - ans + ans + ans - ans; //表达式拖延时间
    cout << ans << endl; //cout和多输出回车拖延时间 
    return 0;
}

算法四十二、极限卡点

卡到了9970ms……

#include 
using namespace std;
int st = clock();
int main() {
    int a, b; scanf("%d%d", &a, &b);
    while (clock() - st < 995000) {}
    printf("%d", a + b);
    return 0;
}

算法四十三、快读快写

#include
using namespace std;
int read() {
    int s = 0, f = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-') f = -1;
        ch = getchar();
    }
    while(isdigit(ch)) {  
        s = s * 10 + ch - '0';
        ch = getchar();
    }
    return s * f;
}
void write(int x) {
    if(x < 0) {
        putchar('-'); 
        x = -x;
    }
    if(x > 9) write(x / 10);
    putchar(x % 10 + '0');
    return;
}
int main() {
    int a, b; a = read(); b = read();
    write(a + b);
    return 0;
}

算法四十四、终极大杀器快读快写

#include
using namespace std;
static char buf[100000], *pa = buf, *pd = buf;
#define gc pa == pd && (pd = (pa = buf) + fread(buf, 1, 100000, stdin), pa == pd) ? EOF : *pa++
inline int read() {
    register int x(0); register char c(gc);
    while (c < '0' || c > '9') c = gc;
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = gc;
    return x;
}
void write(int x) {
    if(x < 0) {
        putchar('-'); 
        x = -x;
    }
    if(x > 9) write(x / 10);
    putchar(x % 10 + '0');
    return;
}
int main() {
    int a, b; a = read(); b = read();
    write(a + b);
    return 0;
}

算法四十五、sort大大大大大大大大大法

sort yyds!
#include 
using namespace std;
const int MAXN = 1e8 + 10;
int n, a[MAXN];
int main(){
    n = 2;
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]);
    sort(a + 1, a + 1 + n);
    int ans = 0;
    for (int i = 1;i <= n; i++) ans += a[i]; printf("%d", ans);
    return 0;
}

算法四十六、冒泡排序
E……

#include 
using namespace std;
const int MAXN = 1e8 + 10;
int a[MAXN], n;
int main(){
    n = 2;
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]);
    for (int i = n;i > 1; i--)
        for(int j = 1;j < i; j++)
            if(a[j] > a[j + 1]) swap(a[j], a[j + 1]);
    int ans = 0;
    for (int i = 1;i <= n; i++) ans += a[i]; printf("%d", ans);
    return 0;
}

算法四十七、选择排序
………………

#include 
using namespace std;
const int MAXN = 1e8 + 10;
int a[MAXN], n;
int main(){
    n = 2;
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1;i < n; i++) {
        int w = i, Min = a[i];
        for (int j = i;j <= n; j++) if(Min > a[j]) w = j, Min = a[j]; //寻找最小数和它的位置
        swap(a[i], a[w]);
    }
    int ans = 0;
    for (int i = 1;i <= n; i++) ans += a[i]; printf("%d", ans);
    return 0;
}

算法四十八、插入排序

#include 
using namespace std;
const int MAXN = 1e8 + 10;
int a[MAXN], n;
int main(){
    n = 2;
    for (int i = 1;i <= n; i++) {
        scanf("%d", &a[i]); int x = i - 1;
        while(a[x] > a[x + 1] && x > 0) swap(a[x], a[x + 1]), x--;
    }
    int ans = 0;
    for (int i = 1;i <= n; i++) ans += a[i]; printf("%d", ans);
    return 0;
}

算法四十九、希尔排序

azzzzzzzzzzzzzzzzzz

#include 
using namespace std;
const int MAXN = 1e8 + 10;
int n, a[MAXN];
int main(){
    n = 2;
    for (int i = 0;i < n; i++) scanf("%d", &a[i]);
    for (int step = n / 2; step > 0; step /= 2)
        for (int i = 0;i < step; i++)
            for (int j = i + step;j < n; j += step)
                if(a[j] < a[j - step]) {
                    int temp = a[j];
                    int k = j - step;
                    while (k >= 0 && temp < a[k]) {
                        swap(a[k + step], a[k]);
                        k -= step;
                    }
                }
    int ans = 0;
    for (int i = 0;i < n; i++) ans += a[i]; printf("%d ", ans);
    return 0;
}

算法五十、归并排序

#include  
using namespace std;
const int MAXN = 1e8 + 10;
int n, a[MAXN], T[MAXN];
void Mergesort(int l, int r) {
    if (l == r) return; //区间内只有一个数,返回
    int mid = l + r >> 1; //相当于(l + r) / 2
    Mergesort(l, mid); //递归左半部分
    Mergesort(mid + 1, r); //递归右半部分
    int i = l, j = mid + 1, k = l;
    while (i <= mid && j <= r) //合并
        if (a[i] <= a[j]) T[k++] = a[i++];
        else T[k++] = a[j++];
    while (i <= mid) T[k++] = a[i++];
    while (j <= r) T[k++] = a[j++];
    for (int q = l; q <= r; q++) a[q] = T[q]; //转移
}
int main() {
    n = 2;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    Mergesort(1, n);
    int ans = 0;
    for (int i = 1; i <= n; i++) ans += a[i]; printf("%d", ans);
    return 0;
}

算法五十一、快速排序

#include  
using namespace std;
const int MAXN = 1e8 + 10;
int n, a[MAXN];
void quickSort(int l, int r) {
    if (l >= r) return;
    int i = l, j = r, base = a[l]; //base取最左边的数为基准数
    while(i < j) {
        while (a[j] >= base && i < j) j--;
        while (a[i] <= base && i < j) i++;
        if (i < j) swap(a[i], a[j]);
    }
    a[l] = a[i]; a[i] = base; //基准数归位
    quickSort (l, i - 1); //递归左边
    quickSort (i + 1, r); //递归右边
}
int main() {
    n = 2;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    quickSort(1, n);
    int ans = 0;
    for (int i = 1; i <= n; i++) ans += a[i]; printf("%d", ans);
    return 0;
}

算法五十二、堆排序

#include  
using namespace std;
int n;
const int MAXN = 1e8 + 10;
int h[MAXN], s;
void down(int u) {
    int t = u;  // t记录最小值
    if (2 * u <= s && h[2 * u] < h[t]) t = 2 * u; // 左儿子
    if (2 * u + 1 <= s && h[2 * u + 1] < h[t]) t = 2 * u + 1; // 右儿子
    if (t != u) { //需要调整
        swap(h[t], h[u]);
        down(t); //递归
    }
}
int main() {
    n = 2;
    for (int i = 1; i <= n; i ++) scanf("%d", &h[i]);
    s = n;
    for (int i = n / 2; i >= 1; i--) down(i); //初始化堆j
    int ans = 0;
    while (n--) {
        ans += h[1];
        h[1] = h[s]; s--;
        down(1); 
    }
    printf("%d", ans);
    return 0;
}

算法五十三、计数排序

#include  
using namespace std;
const int MAXN = 1e8 + 10;
long long n, cnt[MAXN];
int main() {
    n = 2; int x = 0, Max = -0x3f3f3f, Min = 0x3f3f3f; //初始化最大值和最小值
    for (int i = 1; i <= n; i ++) {
        scanf("%d", &x); cnt[x]++; //统计
        Max = max(Max, x); Min = min(Min, x); //更新最大值和最小值
    }
    int ans = 0;
    for (int i = Min;i <= Max; i++)
        while(cnt[i]) cnt[i]--, ans += i;
    printf("%d", ans);
    return 0;
}

算法五十四、桶排序

#include 
using namespace std;
const int MAXN = 1e8 + 10;
int n, Min = MAXN, Max = 0, sum[MAXN], ans;
bool f[45];
vector bucket[45];//定义桶,这里定义40个桶
void insertsort(int s) {
    for (int i = 0;i < bucket[s].size(); i++)
        for (int j = i;j >= 1; j--) if(bucket[s][j - 1] > bucket[s][j]) swap(bucket[s][j], bucket[s][j - 1]);//这里是从小到大排序
    for (int i = 0;i < bucket[s].size(); i++) ans += bucket[s][i];
}
void bucketsort() {
    for (int i = 1;i <= n; i++)
        bucket[int((sum[i] - Min) / ((Max - Min + 1) / 40.0))].push_back(sum[i]), f[int((sum[i] - Min) / ((Max - Min + 1) / 40.0))] = 1;//运用最大最小值来合理分配桶
    for (int i = 0;i <= 40; i++) if(f[i]) insertsort(i); //如果当前桶有数值,则对桶内的数进行排序(这里用选择排序)
}
int main() {
    n = 2;
    for (int i = 1;i <= n; i++) {
        scanf("%d", &sum[i]);
        Min = min(Min, sum[i]), Max = max(Max, sum[i]); //为了能够合理利用空间,确保第一个桶和最后一个桶都有数,所以存储最大最小值
    }
    bucketsort(); printf("%d", ans);
    return 0; 
}

算法五十五、基数排序

#include 
using namespace std;
int maxbit(int data[], int n) {
    int d = 1, p = 10; //d保存最大的位数 
    for (int i = 0;i < n; i++) while(data[i] >= p) p *= 10, d++;
    return d;
}
void radixsort(int data[], int n) { //基数排序 
    int d = maxbit(data, n);
    int tmp[n];
    int cnt[15], i, j, k, radix = 1;
    for (i = 1;i <= d; i++) { //进行d次排序
        memset(cnt, 0, sizeof(cnt)); //清空计数器
        for (j = 0;j < n; j++) {
            k = (data[j] / radix) % 10;
            cnt[k]++;
        }
        for (j = 1;j < 10; j++) cnt[j] += cnt[j - 1];
        for (j = n - 1;j >= 0; j--) {
            k = (data[j] / radix) % 10;
            tmp[cnt[k] - 1] = data[j];
            cnt[k]--;
        }
        for (j = 0;j < n; j++) data[j] = tmp[j];
        radix *= 10;
    }
}
const int MAXN = 1e8 + 10;
int n, a[MAXN];
int main(){
    n = 2;
    for (int i = 0;i < n; i++) scanf("%d", &a[i]);
    radixsort(a, n);
    int ans = 0;
    for (int i = 0;i < n; i++) ans +=  a[i]; printf("%d", ans);
}

算法五十六、鸡尾酒排序

#include 
using namespace std;
const int MAXN = 1e8 + 10;
int n, a[MAXN];
int main() {
    n = 2;
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]);
    int cnt = 0;
    while (1) {
        int f = 0; cnt++;
        if(cnt & 1)
            for (int i = 1;i < n; i++) if(a[i] > a[i + 1]) swap(a[i], a[i + 1]), f = 1;
        else
            for (int i = n;i > 1; i--) if(a[i] < a[i - 1]) swap(a[i], a[i - 1]), f = 1;
        if(!f) break;
    }
    int ans = 0;
    for (int i = 1;i <= n; i++) ans += a[i]; printf("%d", ans);
    return 0;
}

算法五十七、二叉排序树排序

(怎么还是排序hh)

#include
#define LL long long
#define INF 0x7FFFFFFF
using namespace std;
const int N = 1e8 + 10;
int n, idx, rt, ans;
int a[N], t[N];
int ch[N][2];
void insert(int &x, int val) {
    if (!x) {
        x = ++ idx;
        t[x] = val;
        return;
    }
    else {
        if(val < t[x]) insert(ch[x][0], val);
        else insert(ch[x][1], val);
    }
}
void dfs(int x) { //中序遍历二叉排序树
    if(!x) return;
    dfs(ch[x][0]);
    ans += t[x];
    dfs(ch[x][1]);
}
int main() {
    n = 2;
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1;i <= n; i++) insert(rt, a[i]);
    dfs(rt); printf("%d", ans);
    return 0;
}

算法五十八、侏儒排序

#include 
using namespace std;
const int MAXN = 1e8 + 10;
int n, a[MAXN];
int main() {
    n = 2;
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]);
    int s = 1;
    while(s <= n) {
        if(a[s - 1] <= a[s] || s == 0) s++;
        else swap(a[s], a[s - 1]), s--;
    }
    int ans = 0;
    for (int i = 1;i <= n; i++) ans += a[i]; printf("%d", ans);
    return 0;
} 

算法五十九、猴子排序
(虽然在排序上理论会TLE……但是A+B还是能AC的……)
(排序终于结束了……hh)

#include 
using namespace std;
const int MAXN = 1e8 + 10;
int n, a[MAXN];
int check() {
    for (int i = 1;i < n; i++) if(a[i] > a[i + 1]) return 0;
    return 1;
}
int main() {
    n = 2;
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]);
    while (1) {
        random_shuffle(a + 1, a + 1 + n);   //随机打乱数组的系统函数 
        if(check()) break;
    }
    int ans = 0;
    for (int i = 1;i <= n; i++) ans += a[i]; printf("%d", ans);
    return 0;
}

算法六十、快速幂

#include
using namespace std;
int qmi(int m, int k, int p) {
    int res = 1 % p, t = m;
    while (k) {
        if (k & 1) res = res * t % p;
        t = t * t % p;
        k >>= 1;
    }
    return res;
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    printf("%d", qmi(a, 1, 100000010) + qmi(b, 1, 100000010));
    return 0;
}

算法六十一、差分

#include 
using namespace std;
int n = 2, m = 5, b[10];
int main() {
    for (int i = 1;i <= n; i++) {
        int x; scanf("%d", &x);
        b[1] += x; b[m + 1] -= x;
    }
    int ans = 0, x = 0;
    for (int i = 1;i <= m; i++) {
        x += b[i]; ans = max(ans, x);
    }
    printf("%d", ans);
    return 0;
}

算法六十二、模拟人工计算
当然这注释是和人工计算最接近的hh

#include 
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b); //人眼看见数据
    if (a == 0) {printf("%d", b); return 0;}  //大脑瞬间“打表”被老师发现了,血量减少50……
    if (b == 0) {printf("%d", a); return 0;}  //大脑瞬间“打表”被老师发现了,血量减少50……
    int f1 = 0, f2 = 0; //大脑申请了两个空间……(还好没炸掉)
    if (a < 0) f1 = 1;  //大脑正在判断,请勿打扰……
    if (b < 0) f2 = 1;  //大脑正在判断,请勿打扰……
    a = abs(a); b = abs(b);  //哇!大脑使用了去掉负号的大法!!!
    int ans = 0;   //大脑申请了一个空间
    if (f1) ans -= a;  //大脑正在艰难地判断着……
    //大脑指挥手拿起笔在草稿纸上划来划去……
    //大脑感到烦躁
    //眼睛看到老师转了一下身子,立刻反馈给大脑
    //大脑指挥手在计算器键盘上写下了算式……
    //眼睛看到答案,反馈给大脑,大脑立刻指挥手关掉了计算器……
    //眼睛看到老师转回来了
    else ans += a;  //大脑正在艰难地判断着……
    if (f2) ans -= b;//大脑正在艰难地判断着……
    //大脑指挥手拿起笔在草稿纸上划来划去……
    //大脑感到烦躁
    //眼睛看到老师转了一下身子,立刻反馈给大脑
    //大脑指挥手在计算器键盘上写下了算式……
    //眼睛看到答案,反馈给大脑,大脑立刻指挥手关掉了计算器……
    //眼睛看到老师转回来了
    else ans += b;//大脑正在艰难地判断着……
    //眼睛观察到老师正在身后冷冷地看着……
    //大脑感到紧张
    //大脑让身体不由自主地颤抖起来
    //大脑差点死机
    //大脑复活
    //立刻写下答案
    printf("%d", ans);
    //大脑又死机了……
    //耳朵听到老师在叫自己起来
    //大脑指挥身体起来了
    //开始下一题……(下一个数据)
    return 0;
}

算法六十三、二进制

额……好像有点不太正常

#include
using namespace std;
int a, b, s, s1, i, na, nb;
int main() {
    scanf("%d%d", &a, &b);
    if (a <= 0) na = 1, a = abs(a);
    while(a) {
        if ((a & 1) != 0) s += pow(2, (a & 1) * i);
        a >>= 1; i++;
    }
    i = 0;
    if (na == 1) s = abs(s);
    if (b <= 0) nb = 1, b = abs(b);
    while(b) {
        if ((b & 1) != 0) s1 += pow(2, (b & 1) * i);
        b >>= 1; i++;
    }
    if (nb == 1) s1 = abs(s1);
    printf("%d", s + s1);
    return 0;
}

算法六十四、ST表
区间最小值加区间最大值 = A+B

#include 
using namespace std;
int n, a[10], st1[10][17], st2[10][17];
void ST_work1() {
    for (int i = 1; i <= n; i++) st1[i][0] = a[i];
    int t = log(n) / log(2) + 1;
    for (int j = 1; j < t; j++) {
        for (int i = 1; i <= n - (1 << j) + 1; i++)
            st1[i][j] = max(st1[i][j - 1], st1[i + (1 << (j - 1))][j - 1]);
    }
}
int ST_query1(int l, int r) {
    int k = log(r - l + 1) / log(2);
    return max(st1[l][k], st1[r - (1 << k) + 1][k]);
}
void ST_work2() {
    for (int i = 1; i <= n; i++) st1[i][0] = a[i];
    int t = log(n) / log(2) + 1;
    for (int j = 1; j < t; j++) {
        for (int i = 1; i <= n - (1 << j) + 1; i++)
            st1[i][j] = min(st1[i][j - 1], st1[i + (1 << (j - 1))][j - 1]);
    }
}
int ST_query2(int l, int r) {
    int k = log(r - l + 1) / log(2);
    return min(st1[l][k], st1[r - (1 << k) + 1][k]);
}
int main() {
    n = 2;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    ST_work1(); int ans1 = ST_query1(1, 2);
    ST_work2(); int ans2 = ST_query2(1, 2);
    printf("%d", ans1 + ans2);
    return 0;
}

算法六十五、01背包

#include 
using namespace std;
int c[1010], w[1010], dp[1010], n, m;
int main() {
    n = 2; m = 2;  //2个物体,背包容积2
    for (int i = 1;i <= n; i++) c[1] = 1, scanf("%d", &w[i]);  //设体积为1,读入价值
    for (int i = 1;i <= n; i++)
        for (int j = m; j >= c[i]; j--)
            dp[j] = max(dp[j], dp[j - c[i]] + w[i]);
    printf ("%d\n", dp[m]);
    return 0;
}

算法六十六、完全背包

#include 
using namespace std;
int c[1010], w[1010], dp[1010], n, m;
int main() {
    n = 2; m = 3;
    for (int i = 1;i <= n; i++) scanf("%d", &w[i]);
    for (int i = 1;i <= n; i++)
        for (int j = c[i]; j <= m; j++)
            dp[j] = max(dp[j], dp[j - c[i]] + w[i]);
    printf ("%d\n", dp[m]);
    return 0;
}

算法六十七、多重背包之暴力拆分法

#include 
using namespace std;
int c[110], w[110], s[110], dp[1010], n, m;
int main() {
    n = 2; m = 2;
    for (int i = 1;i <= n; i++) scanf("%d", &w[i]), c[i] = s[i] = 1;
    for (int i = 1;i <= n; i++)
        for (int x = 1;x <= s[i]; x++)
            for (int j = m; j >= c[i]; j--)
                dp[j] = max(dp[j], dp[j - c[i]] + w[i]);
    printf ("%d\n", dp[m]);
    return 0;
}

算法六十八、多重背包之二进制拆分

#include 
using namespace std;
int n, m, v[10010], w[10010], f[2010];
int main() {
    n = 2; m = 2;
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        int a = 1, b, s = 1, k = 1; scanf("%d", &b);
        while (k <= s) {
            cnt++;
            v[cnt] = a * k; w[cnt] = b * k;
            s -= k; k <<= 1;
        }
        if (s > 0) {
            cnt++;
            v[cnt] = a * s;
            w[cnt] = b * s;
        }
    }
    n = cnt;
    for (int i = 1; i <= n; i ++ )
        for (int j = m; j >= v[i]; j -- )
            f[j] = max(f[j], f[j - v[i]] + w[i]);
    printf("%d\n", f[m]);
    return 0;
}

算法六十九、二维费用背包问题

#include 
using namespace std;
int N, V, M;
int v[1010], m[1010], w[1010], f[1010][1010];
int main () {
    N = V = M = 2;
    for (int i = 1; i <= N; i++) scanf("%d", &w[i]), v[i] = m[i] = 1;
    for (int i = 1; i <= N; i++)
        for (int j = V; j >= v[i]; j--)
            for (int k = M; k >= m[i]; k--)
                f[j][k] = max(f[j - v[i]][k - m[i]] + w[i], f[j][k]);
    printf("%d\n", f[V][M]);
    return 0;
}

算法七十、分组背包问题

#include 
using namespace std;
int f[110][110], v[110][110], w[110][110], s[110];
int n, m, k;
int main() {
    n = m = 2;
    for (int i = 1; i <= n; i++) {
        s[i] = 1;
        for (int j = 0; j < s[i]; j++) scanf("%d", &w[i][j]), v[i][j] = 1;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            f[i][j] = f[i - 1][j];
            for (int k = 0; k < s[i]; k++)
                if (j >= v[i][k]) f[i][j] = max(f[i][j], f[i - 1][j - v[i][k]] + w[i][k]);
        }
    }
    printf("%d", f[n][m]);
    return 0;
}

算法七十一、有依赖的背包问题

#include 
using namespace std;
int n, m, v[110], w[110];
int h[110], e[110], ne[110], idx;
int f[110][110];

void add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }
void dfs(int u) {
    for (int i = h[u]; ~i; i = ne[i]) {  // 循环物品组
        int son = e[i];
        dfs(e[i]);
        // 分组背包
        for (int j = m - v[u]; j >= 0; j -- )  // 循环体积
            for (int k = 0; k <= j; k ++ )  // 循环决策
                f[u][j] = max(f[u][j], f[u][j - k] + f[son][k]);
    }
    // 将物品u加进去
    for (int i = m; i >= v[u]; i -- ) f[u][i] = f[u][i - v[u]] + w[u];
    for (int i = 0; i < v[u]; i ++ ) f[u][i] = 0;
}
int main() {
    n = m = 2;
    memset(h, -1, sizeof h);
    int root;
    for (int i = 1; i <= n; i++) {
        int p; v[i] = 1; scanf("%d", &w[i]);
        if (i == 1) p = -1; else p = 1; //A+B中的特判
        if (p == -1) root = i;
        else add(p, i);
    }
    dfs(root);
    printf("%d", f[root][m]);
    return 0;
}

算法七十二、多重背包队列优化

#include 
using namespace std;
int n, m;
int f[20010], g[20010], q[20010];
int main() {
    n = m = 2;
    for (int i = 1; i <= n; i ++ ) {
        int v = 1, w, s = 1; scanf("%d", &w);
        memcpy(g, f, sizeof f);
        for (int j = 0; j < v; j ++ ) {
            int hh = 0, tt = -1;
            for (int k = j; k <= m; k += v) {
                if (hh <= tt && q[hh] < k - s * v) hh ++ ; //剔除超出长度元素
                if (hh <= tt) f[k] = max(f[k], g[q[hh]] + (k - q[hh]) / v * w); //更新当前答案
                while (hh <= tt && g[q[tt]] - (q[tt] - j) / v * w <= g[k] - (k - j) / v * w) tt -- ;
                //维持单调性
                //这里也可以这样写,更易理解
                //while (hh <= tt && g[q[tt]] <= g[k] - (k - q[tt]) / v * w) tt -- ;
                q[++tt] = k;
            } 
        }
    }
    printf("%d", f[m]);
    return 0;
}

算法七十三、istream

还是大佬们厉害,我没看懂……

#include 
#include 
#include 
#include 
#include
using namespace std;
int main()
{
    istream_iterator in(cin), eof;
    cout << accumulate(in, eof ,0) << endl;
    return 0;
}

算法七十四、进制转换
我本来想要十进制转二进制一个算法,十进制转三进制一个算法……然后发现甚至可以到十进制转108
进制……
而且它们都属于进制转换,所以我决定直接写一个通用的进制转换程序。
然后随机进制转换,在那个进制下加法,再转化为十进制存储下来,如果所有的答案都一样,则输出这个答案。

我是不是很无聊

#include 
using namespace std;
int A[30], B[30], a, b;
int check(int mod) {
    int x = a, y = b, ta = 0, tb = 0;
    while (x) {
        A[++ta] = x % mod;
        x /= mod;
    }
    while (y) {
        B[++tb] = y % mod;
        y /= mod;
    }
    for (int i = 1; i <= max(ta, tb); i++) {
        A[i] += B[i];
        if (A[i] >= mod) A[i + 1] += A[i] / mod, A[i] %= mod;
    }
    if (A[ta + 1]) ta++;
    int Ans = 0;
    for (int i = ta; i > 0; i--) Ans = Ans * mod + A[i];
    return Ans;
}
int main() {
    scanf("%d%d", &a, &b);
    int ans[100010];
    for (int i = 1; i <= 100000; i++) {
        srand((unsigned)time(NULL));
        int o = (int) rand() % 1000000 + 2;  //取随机进制
        ans[i] = check(o);
    }
    bool f = 1;
    for (int i = 2; i <= 100000; i++) if (ans[i] != ans[i - 1]) { f = 0; break; }
    if (f) printf("%d\n", ans[1]);
    else puts("老子不干了!WA就WA吧!一行WA上西天!");  //誓死不AC(逃)
    return 0;
}

算法七十五、指针算法

#include 
using namespace std;
int main() {
    int a, b;
    cin>>a>>b;
    int *c = &a, *d = &b, ans;
    ans = *c + *d;
    cout << ans << endl;
}

算法七十六、vector

#include 
using namespace std;
vector v;
int main() {
    int x = 0;
    while (cin>>x) v.push_back(x);
    int ans = 0;
    for (int i = 0; i < v.size(); i++) ans += v[i];
    cout << ans;
    return 0;
}

算法七十七、queue

#include 
using namespace std;
queue q;
int main() {
    int x;
    while (cin>>x) q.push(x);
    int ans = 0;
    while (q.size()) ans += q.front(), q.pop();
    cout << ans;
    return 0;
}

算法七十八、deque

#include 
using namespace std;
deque a, b;
int main() {
    int x;
    while (cin>>x) a.push_front(x);
    int ans = 0;
    while (a.size()) ans += a.back(), b.push_back(a.back()), a.pop_back();
    int res = 0;
    while (b.size()) res += b.front(), b.pop_front();
    if (ans == res) cout << (ans + res) / 2;
    return 0;
}

算法七十九、list

#include 
using namespace std;
list a, b;
int main() {
    int x;
    while (cin>>x) a.push_front(x);
    int ans = 0;
    while (a.size()) b.push_back(a.back()), ans += a.back(), a.pop_back();
    int res = 0;
    while (b.size()) res += b.front(), b.pop_front();
    if (ans == res) cout << (ans + res) / 2;
    return 0;
}

算法八十、map

#include 
using namespace std;
map m;
int main() {
    int x = 0;
    while (cin>>x) m[x] = "老子不会老子WA行了吧???";
    int ans = 0;
    for (auto iter = m.begin(); iter != m.end(); ++iter) {
        ans += iter->first;
    }
    cout << ans;
    return 0;
}

算法八十一、set

#include 
using namespace std;
set s;
int main() {
    int x = 0;
    while (cin>>x) s.insert(x);
    int ans = 0;
    for (auto iter = s.begin(); iter != s.end(); ++iter) {
        ans += *iter; 
    }
    cout << ans;
    return 0;
}

算法八十二、pair

#include 
using namespace std;
int main() {
    pair a[110];
    int x = 0, c = 0, t = 1;
    while (cin>>x) {
        if (c == 0) a[t].first = x, c = 1;
        else a[t].second = x, t++, c = 0;
    }
    int ans = 0;
    for (int i = 1; i <= 100; i++) {
        ans += a[i].first + a[i].second;
    }
    cout << ans;
    return 0;
}

算法八十三、stack

#include 
using namespace std;
stack s;
int main() {
    int x = 0;
    while (cin>>x) s.push(x);
    int ans = 0;
    while (s.size()) ans += s.top(), s.pop();
    cout << ans;
    return 0;
}

算法八十四、priority_queue

#include 
using namespace std;
priority_queue q;
int main() {
    int x = 0;
    while (cin>>x) q.push(x);
    int ans = 0;
    while (q.size()) ans += q.top(), q.pop();
    cout << ans;
    return 0;
}

懵了没? \huge懵了没? 懵了没?
搞了那么久,我们来个正常的吧……

满足某人的需求

观众们:你是认真的吗?

C代码

#include 

int main()
{
    int a,b;
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
    return 0;
}

C++代码

#include 
#include 

using namespace std;

int main()
{
    int a,b;
    cin >> a >> b;
    cout << a+b << endl;
    return 0;
}

Pascal代码

var a, b: longint;
begin
    readln(a,b);
    writeln(a+b);
end.

Python代码

import sys

for line in sys.stdin:
    print sum(map(int, line.split()))
Python2代码

s = raw_input().split()
print int(s[0]) + int(s[1])
Python3代码

print(sum(map(int, input().split())))

Java代码

import java.io.*;
import java.util.*;
public class Main {
    public static void main(String args[]) throws Exception {
        Scanner cin=new Scanner(System.in);
        int a = cin.nextInt(), b = cin.nextInt();
        System.out.println(a+b);
    }
}

JavaScript代码(Node.js)

const fs = require('fs')
const data = fs.readFileSync('/dev/stdin')
const result = data.toString('ascii').trim().split(' ').map(x => parseInt(x)).reduce((a, b) => a + b, 0)
console.log(result)
process.exit() // 请注意必须在出口点处加入此行

Ruby代码

a, b = gets.split.map(&:to_i)
print a+b

PHP代码

Rust代码

use std::io;

fn main(){
    let mut input=String::new();
    io::stdin().read_line(&mut input).unwrap();
    let mut s=input.trim().split(' ');

    let a:i32=s.next().unwrap()
               .parse().unwrap();
    let b:i32=s.next().unwrap()
               .parse().unwrap();
    println!("{}",a+b);
}

Go代码

package main

import "fmt"

func main() {
    var a, b int
    fmt.Scanf("%d%d", &a, &b)
    fmt.Println(a+b)
}

C# Mono代码

using System;

public class APlusB{
    private static void Main(){
        string[] input = Console.ReadLine().Split(' ');
        Console.WriteLine(int.Parse(input[0]) + int.Parse(input[1]));
    }
}

Visual Basic Mono代码

Imports System

Module APlusB
    Sub Main()
        Dim ins As String() = Console.ReadLine().Split(New Char(){" "c})
        Console.WriteLine(Int(ins(0))+Int(ins(1)))
    End Sub
End Module

Kotlin代码

fun main(args: Array) {
    val (a, b) = readLine()!!.split(' ').map(String::toInt)
    println(a + b)
}

Haskell代码

main = do
    [a, b] <- (map read . words) `fmap` getLine
    print (a+b)

Scala代码

object Main extends App {
    println(scala.io.StdIn.readLine().split(" ").map(_.toInt).sum)
}

Perl代码

my $in = ;
chomp $in;
$in = [split /[\s,]+/, $in];
my $c = $in->[0] + $in->[1];
print "$c\n";

FoxPro代码

use
replace all c with a+b

怎么样?这个还算正常吧……hh

你不打赏你对得起我打这个吗?

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