HBU Medium problem set
可能没很多人看，后面没写思路，有问题单独问我吧
目录
7-1 区间翻转
7-2 次大值的和
7-3 买东西
7-4 双重子串
7-5 对决
7-6 最大公约数
7-7 填数字
7-8 放小球
7-9 给树染色
7-10 摆放小球
7-11 区间的个数
7-12 选零食
7-13 小球之间的距离
7-14 1还是2
7-15 青春猪头之我没学过C语言
7-16 打怪兽
7-17 背上书包去旅行
7-18 向前走
7-19 热水器
7-20 走方格
7-21 朋友圈
7-22 走方格
7-23 和与积
7-24 增加硬币
7-25 日落大道
7-26 ICPC保定站
7-27 简单的异或
7-28 选我啊！
7-29 豪华酒店
7-30 摆放小球
7-31 最大公约数的和
7-32 一道简单题
7-33 一个无限大的集合
7-1 区间翻转
思路：
贪心，我们每次操作，除最后是这种“RRRRRL” 或 “LRRRR” 操作后只能使满意度+1的特殊情况外，总能使满意度+2.
代码：
#include
using namespace std;
typedef pair PII;
typedef long long LL;
const int N = 200010, M = N * 2;
int n, m, r1, r2;
string s;
int main() {
cin >> n >> m;
getchar();
cin >> s;
int res = 0;
for (int i = 0; i < n - 1; i++) {
if (s[i] == s[i + 1])res++;
}
res = min(res + m * 2, n - 1);
cout << res;
return 0;
} 7-2 次大值的和
思路：
计算每一个值的对结果的贡献。即它作为次大值出现是什么情况。
即左边出现一个比他大的值&&右边没有比他大的值，或者对称情况
这里要用单调栈来算每个数左边距离其最近的比它大的数，右边同理，最后将这些情况的结果计算出来，详见代码。
代码：
#include
using namespace std;
const int N = 1000010;
int n;
int a[N],b[N];
int l_low[N], l_up[N], r_low[N], r_up[N];
long long res;
int stk1[N], stk2[N], tt;
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
b[a[i]] = i;
}
for (int i = 1; i <= n; i++) {
while (tt && a[stk1[tt]] < a[i])tt--;
if (!tt) {
l_low[i] = 1;
l_up[i] = -1;
}
else {
l_low[i] = stk1[tt] + 1;
l_up[i] = stk1[tt];
}
stk1[++tt] = i;
}
tt = 0;
for (int i = n; i >= 1; i--) {
while (tt && a[stk2[tt]] < a[i])tt--;
//cout << tt << " " << stk2[tt] << endl;
if (!tt) {
r_low[i] = n;
r_up[i] = -1;
}
else {
r_low[i] = stk2[tt] - 1;
r_up[i] = stk2[tt];
}
stk2[++tt] = i;
}
for (int i = 1; i <= n; i++) {
int llow = i - l_low[i];
int rlow = r_low[i] - i;
int lup, rup;
if (l_up[i] == -1) {
lup = 0;
}
else {
lup = 1;
while (--l_up[i] && a[i] > a[l_up[i]])lup++;
}
if (r_up[i] == -1) {
rup = 0;
}
else {
rup = 1;
while (++r_up[i] != n + 1 && a[i] > a[r_up[i]])rup++;
}
//cout << llow << " " << lup << " " << rlow << " " << rup << endl;
res += ((long long)(llow + 1) * rup + (long long)(rlow + 1) * lup) * a[i];
}
cout << res;
return 0;
} 7-3 买东西
思路：
贪心：每次用一张券使得最贵的物品价格除以2
代码：
#include
using namespace std;
typedef pair PII;
typedef long long LL;
const int N = 100010, M = N * 2, mod = 1e9 + 7;
int n, m;
priority_queue q;
int main() {
cin >> n >> m;
for (int i = 0; i < n; i++) {
int x;
cin >> x;
q.push(x);
}
while (m--) {
int x = q.top();
if (x == 0)break;
q.pop();
q.push(x / 2);
}
LL sum = 0;
while (!q.empty()) {
sum += q.top();
q.pop();
}
cout << sum;
} 7-4 双重子串
simple 里有
7-5 对决
思路：
将某两个人的比赛看成一个节点，因为每个人都要按已有的顺序比赛，所以给每个人先比的比赛到后比的比赛建一条边，然后做一遍关键路径 ，即完成所有任务所需要的最长时间。
代码：
#include
using namespace std;
typedef pair PII;
typedef long long LL;
const int N = 1e6+10, M = 1e6+10, mod = 1e9 + 7;
int n, m, x;
map mp;
int h[N], e[M], ne[M], idx;
int d[M],f[M];
int cnt;
void add(int a, int b) {
e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
int get(int a, int b) {
if (a > b)swap(a, b);
if (mp.count(make_pair(a,b)) == 0)mp[make_pair(a, b)] = ++cnt;
return mp[{a, b}];
}
int main()
{
cin >> n;
memset(h, -1, sizeof h);
for (int i = 1; i <= n; i++) {
cin >> x;
int lst = get(i, x);
for (int j = 1; j <= n - 2; j++) {
cin >> x;
x = get(i, x);
add(lst, x);
d[x]++;
lst = x;
}
}
queue q;
for (int i = 1; i <= n * (n - 1) / 2; i++)if (!d[i])q.push(i),f[i] = 1;
int res = 0;
while (q.size()) {
int t = q.front();
q.pop();
res++;
for (int i = h[t]; ~i; i = ne[i]) {
int j = e[i];
if (--d[j] == 0)q.push(j);
f[j] = max(f[j], f[t] + 1);
}
}
if (res != (n - 1) * n / 2)puts("-1");
else {
int num = 0;
for (int i = 1; i <= (n - 1) * n / 2; i++)num = max(num, f[i]);
cout << num;
}
return 0;
} 7-6 最大公约数
思路：
这题有个结论：最大公约数一定是数组求和的一个约数
做法：
数组求和sum，计算每个约数能不能满足要求（看看能不能在m次操作内把这个数组变为全0）
代码：
#include
using namespace std;
typedef pair PII;
typedef long long LL;
const int N = 200010, M = N * 2, mod = 1e9 + 7;
int n, m;
int a[N],b[N];
int sum;
bool check(int x) {
int s = 0,t = 0;
for (int i = 0; i < n; i++) {
b[i] = a[i] % x;
s += b[i];
}
if (s % x != 0)return false;
t = s / x;
sort(b, b + n);
for (int i = n - 1; t ; i--,t--) {
s -= b[i];
}
if (s <= m)return true;
else return false;
}
int main() {
cin >> n >> m;
for (int i = 0; i < n; i++) {
cin >> a[i];
sum += a[i];
}
int res = 0;
for (int i = 1; i <= sum / i; i++) {
if (sum % i == 0) {
if (check(i))res = max(res, i);
if (check(sum / i))res = max(res, sum / i);
}
}
cout << res;
} 7-7 填数字
思路：
DP f[i][j] 表示前i位数，对13取模余j的方案数
初始化第一位，如果是问号，初始化0~9，否则是几就初始化几
转移方程： f[i][(j * 10 + s[i] - '0') % 13] += f[i - 1][j] ;
思考一个点：19对13取余得6 ， 192对13取余 等价于19对13取余的结果6 乘10 再加 2 再对13取余
代码：
#include
#include
#include
using namespace std;
const int N = 1e5+10,mod = 1e9 + 7;
int n;
int f[N][13];
int main()
{
string s;
cin>>s;
s = " " + s;
if(s[1] == '?'){
for(int i=0;i<10;i++)f[1][i] = 1;
}else f[1][s[1] - '0'] = 1;
for(int i = 2;i < s.size();i++){
bool fg = (s[i] == '?');
for(int j = 0;j < 13;j++){
if(f[i - 1][j]){
if(fg){
for(int k = 0;k < 10;k ++){
f[i][(j*10 + k) % 13] = (f[i][(j*10 + k) % 13] + f[i - 1][j]) % mod;
}
}else f[i][(j * 10 + s[i] - '0') % 13] = (f[i][(j * 10 + s[i] - '0') % 13] + f[i - 1][j]) %mod;
}
}
}
cout< 7-8 放小球
这道题，我们一定能通过一种方式将最终解构造出来
从后往前遍历
如果当前箱子的倍数箱子放小球数对2取余不满足这个属性，那就在这个箱子放，否则不放
代码：
#include
using namespace std;
typedef pair PII;
typedef long long LL;
const int N = 200010, M = N * 2, mod = 1e9 + 7;
int n, m;
int a[N],b[N];
vector res;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
for (int i = n; i >= 1; i--) {
int sum = 0;
for (int k = 2; i * k <= n; k++)sum += b[i * k];
if (sum % 2 == 0) {
if (a[i]) {
res.push_back(i);
b[i] = 1;
}
}
else {
if(!a[i]) {
res.push_back(i);
b[i] = 1;
}
}
}
cout << res.size() << endl;
for (int i = 0; i < res.size(); i++) {
cout << res[i] << " ";
}
} 7-9 给树染色
#include
using namespace std;
typedef pair PII;
typedef long long LL;
const int N = 100010, M = N * 2, mod = 1e9 + 7;
int n, m;
int tol[N];
int h[N], e[M], ne[M], idx;
LL sum = 1;
void add(int a, int b) {
e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
void dfs1(int cur, int fa) {
int x;
if (cur == 1)x = m - 1;
else x = m - 2;
for (int i = h[cur]; i != -1; i = ne[i]) {
int j = e[i];
if (j == fa)continue;
tol[j] = x--;
dfs1(j, cur);
}
}
void dfs2(int u, int fa) {
sum = (sum * tol[u]) % mod;
for (int i = h[u]; i != -1; i = ne[i]) {
int j = e[i];
if (j == fa)continue;
dfs2(j, u);
}
}
int main() {
cin >> n >> m;
memset(h, -1, sizeof h);
for (int i = 1; i < n; i++) {
int a, b;
cin >> a >> b;
add(a, b), add(b, a);
}
tol[1] = m;
dfs1(1, 0);
dfs2(1, 0);
cout << sum;
} 7-10 摆放小球
#include
using namespace std;
typedef long long LL;
const int N = 2010, mod = 1e9 + 7;
int n, a, b, x;
int fact[N], infact[N];
int qmi(int a, int k, int p) {
int res = 1;
while (k) {
if (k & 1)res = (LL)res * a % p;
k >>= 1;
a = (LL)a * a % p;
}
return res;
}
int C(int a, int b) {
return (LL)fact[a] * infact[b] % mod * infact[a - b] % mod;
}
int main() {
fact[0] = infact[0] = 1;
for (int i = 1; i < N; i++) {
fact[i] = (LL)fact[i - 1] * i % mod;
infact[i] = (LL)infact[i - 1] * qmi(i, mod - 2, mod) % mod;
}
scanf("%d%d", &n, &x);
for (int i = 1; i <= x; i++) {
if (i > n - x + 1)cout << 0 << endl;
else cout << (LL)C(n - x + 1, i) * C(x - 1,i - 1) % mod < 7-11 区间的个数
#include
using namespace std;
typedef pair PII;
typedef long long LL;
const int N = 1e5+10, M = 1e6+10, mod = 1e9 + 7;
int n;
LL m;
LL a[N];
int main()
{
cin >> n >> m;
for (int i = 0; i < n; i++)cin >> a[i];
int l = 0, r = 0;
LL res = 0;
LL sum = 0;
while (r < n) {
while(sum < m && r < n) {
sum += a[r];
r++;
}
if (sum >= m)res += n - r + 1;
sum -= a[l++];
while (sum >= m) {
sum -= a[l++];
res += n - r + 1;
}
}
cout << res;
return 0;
} 7-12 选零食
#include
using namespace std;
typedef long long LL;
typedef pair PII;
const int N = 55;
int n, m;
int a[N];
int res;
int main() {
cin >> n >> m;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
int buy = min(n, m);
for (int i = 1; i <= buy; i++) {
int put = m - i;
for (int jl = 0; jl <= i; jl++) {
int sum = 0;
priority_queue, greater> q;
int jr = n - i + jl + 1;
for (int j = 1; j <= jl; j++) {
q.push(a[j]);
sum += a[j];
}
for (int j = jr; j <= n; j++) {
q.push(a[j]);
sum += a[j];
}
int x = put;
while (!q.empty() && q.top() < 0 && x > 0) {
sum -= q.top();
q.pop();
x--;
}
res = max(res, sum);
}
}
cout << res;
return 0;
} 7-13 小球之间的距离
#include
using namespace std;
typedef long long LL;
const int N = 2e5 + 10, mod = 1e9 + 7;
int n, m, k;
int qmi(int a, int k, int p)
{
int res = 1;
while (k)
{
if (k & 1) res = (LL)res * a % p;
a = (LL)a * a % p;
k >>= 1;
}
return res;
}
int C(int a, int b, int p)
{
if (b > a) return 0;
int res = 1;
for (int i = 1, j = a; i <= b; i++, j--)
{
res = (LL)res * j % p;
res = (LL)res * qmi(i, p - 2, p) % p;
}
return res;
}
int main() {
cin >> n >> m >> k;
int tmp = (LL)m * m % mod;
LL sum = 0,sum2 = 0;
for (int i = 1; i < n; i++) {
sum = (sum + (LL)i * (n - i) % mod) % mod;
}
sum = sum * tmp % mod;
tmp = (LL)n * n % mod;
for (int i = 1; i < m; i++) {
sum2 = (sum2 + (LL)i * (m - i) % mod) % mod;
}
sum2 = sum2 * tmp % mod;
sum = ((sum + sum2) % mod * C(n * m - 2, k - 2, mod)) % mod;
cout << sum;
return 0;
} 7-14 1还是2
#include
using namespace std;
typedef pair PII;
typedef long long LL;
const int N = 1e5+10, M = 1e6+10, mod = 1e9 + 7;
int n, m;
int p[N];
int find(int x) {
if (x != p[x])p[x] = find(p[x]);
return p[x];
}
int main()
{
cin >> n >> m;
for (int i = 1; i <= n; i++)p[i] = i;
for (int i = 0; i < m; i++) {
int a, b, c;
cin >> a >> b >> c;
a = find(a), b = find(b);
if (a != b)p[a] = b;
}
set st;
for (int i = 1; i <= n; i++) {
p[i] = find(p[i]);
st.insert(p[i]);
}
cout << st.size();
return 0;
} 7-15 青春猪头之我没学过C语言
#include
using namespace std;
typedef pair PII;
typedef long long LL;
const int N = 1e5+10, M = 1e6+10, mod = 1e9 + 7;
int n;
LL x, y;
LL fun(LL x) {
if (x % 4 == 1)return 1;
else if (x % 4 == 2)return x + 1;
else if (x % 4 == 3)return 0;
else return x;
}
int main()
{
cin >> x >> y;
LL res = fun(max((LL)0,x - 1)) ^ fun(y);
cout << res;
return 0;
} 7-16 打怪兽
#include
#include
#include
#include
#include
#include
using namespace std;
#define clr(a,b) memset(a,b,sizeof(a))
const int maxn=700+2;
const int inf=0x3f3f3f3f;
typedef long long ll;
struct Node
{
int maxw,cnt;
Node(){}
Node(int maxw,int cnt):maxw(maxw),cnt(cnt){}
bool operator > (Node t)
{
return maxw>t.maxw;
}
}tree[maxn<<2][maxn<<2];
int m,n,q,r,mat[maxn][maxn];
Node ans;
void push_upy(int deep,int k)
{
tree[deep][k].maxw=max(tree[deep][k<<1].maxw,tree[deep][k<<1|1].maxw);
if(tree[deep][k].maxw==tree[deep][k<<1].maxw)
tree[deep][k].cnt+=tree[deep][k<<1].cnt;
if(tree[deep][k].maxw==tree[deep][k<<1|1].maxw)
tree[deep][k].cnt+=tree[deep][k<<1|1].cnt;
return ;
}
void push_upx(int deep,int k)
{
tree[deep][k].maxw=max(tree[deep<<1][k].maxw,tree[deep<<1|1][k].maxw);
if(tree[deep][k].maxw==tree[deep<<1][k].maxw)
tree[deep][k].cnt+=tree[deep<<1][k].cnt;
if(tree[deep][k].maxw==tree[deep<<1|1][k].maxw)
tree[deep][k].cnt+=tree[deep<<1|1][k].cnt;
return ;
}
void bulid_y(int ly,int ry,int deep,int k,int flag)
{
tree[deep][k].maxw=-inf;
if(ly==ry)
{
if(flag)
{
scanf("%d",&tree[deep][k].maxw);
mat[r][ly]=tree[deep][k].maxw;
tree[deep][k].cnt=1;
}
else {push_upx(deep,k);}//更新x方向
return ;
}
int mind=(ly+ry)>>1;
//bulid_y(ly,mind,deep,k<<1,flag);
bulid_y(ly,mind,deep,k<<1,flag);
bulid_y(mind+1,ry,deep,k<<1|1,flag);
push_upy(deep,k);
return ;
}
void bulid_x(int lx,int rx,int deep)
{
if(lx==rx)
{
r=lx;
bulid_y(1,n,deep,1,1);
return ;
}
int mind=(lx+rx)>>1;
bulid_x(lx,mind,deep<<1);
bulid_x(mind+1,rx,deep<<1|1);
bulid_y(1,n,deep,1,0);//同区域y轴方向
return ;
}
void query_y(int ly,int ry,int Ly,int Ry,int deep,int k)
{
if(Ly<=ly&&ry<=Ry)
{
if(ans.maxw==tree[deep][k].maxw)
{
ans.cnt+=tree[deep][k].cnt;
}
else if(tree[deep][k]>ans)
{
ans=tree[deep][k];
}
return;
}
int mind=(ly+ry)>>1;
if(Ly<=mind)
query_y(ly,mind,Ly,Ry,deep,k<<1);
if(Ry>mind)
query_y(mind+1,ry,Ly,Ry,deep,k<<1|1);
return ;
}
void query_x(int lx,int rx,int Lx,int Rx,int Ly,int Ry,int deep)
{
if(Lx<=lx&&rx<=Rx)
{
query_y(1,n,Ly,Ry,deep,1);
return;
}
int mind=(lx+rx)>>1;
if(Lx<=mind)
query_x(lx,mind,Lx,Rx,Ly,Ry,deep<<1);
if(Rx>mind)
query_x(mind+1,rx,Lx,Rx,Ly,Ry,deep<<1|1);
return ;
}
int main()
{
int t;
cin >> t;
while(t--)
{
scanf("%d%d",&m,&n);
clr(tree,0);
clr(mat,0);
bulid_x(1,m,1);
scanf("%d",&q);
int x1,y1,x2,y2;
while(q--)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
ans.maxw=0;
ans.cnt=0;
query_x(1,m,x1,x2,y1,y2,1);
printf("%d\n",ans.maxw);
}
}
return 0;
} 7-17 背上书包去旅行
#include
using namespace std;
typedef long long LL;
const int N = 20,M = 1 << N, mod = 1e9 + 7;
int n, m;
int w[N][N];
int f[M][N];
struct NODE {
int a, b, c;
}node[N];
int main()
{
cin >> n;
int i = 0;
for (; i < n; i++)
cin >> node[i].a >> node[i].b >> node[i].c;
node[i].a = node[0].a, node[i].b = node[0].b, node[i].c = node[0].c;
n++;
for (i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
w[i][j] = abs(node[j].a - node[i].a) + abs(node[j].b - node[i].b) + max(0, node[j].c - node[i].c);
}
}
memset(f, 0x3f, sizeof(f));
f[1][0] = 0;
for (i = 1; i < (1 << n); i++) {
for (int j = 0; j < n; j++) {
if (i >> j & 1) {
for (int k = 0; k < n; k++) {
if (i >> k & 1) {
f[i][j] = min(f[i][j], f[i - (1 << j)][k] + w[k][j]);
}
}
}
}
}
cout << f[(1 << n) - 1][n - 1];
return 0;
} 7-18 向前走
#include
using namespace std;
typedef long long LL;
const int N = 200010;
int n,w;
LL s[N],t[N];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%lld",&s[i]);
s[i]+=s[i-1];
t[i] = max(s[i],t[i-1]);
//cout< 7-19 热水器
#include
using namespace std;
typedef long long LL;
const int N = 200010;
LL n,w;
LL s[N];
int main()
{
scanf("%lld%lld",&n,&w);
int l,r,c,maxn=0;
for(int i=0;iw){
fg=false;
break;
}
}
if(fg)printf("Yes");
else printf("No");
return 0;
}
7-20 走方格
#include
using namespace std;
typedef long long LL;
const int N = 2010, mod = 1e9 + 7;
int n, m;
char ch[N][N];
LL f[N][N];
LL s1[N][N],s2[N][N],s3[N][N];
int main()
{
scanf("%d%d", &n, &m);
getchar();
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++)scanf("%c", &ch[i][j]);
getchar();
}
f[1][1] = 1,s1[1][1]= s2[1][1] = s3[1][1] =1;
for(int i=1;i<=n;i++)
for (int j = 1; j <= m; j++) {
if (ch[i][j] == '#') { continue; }
f[i][j] += ((s1[i - 1][j] + s2[i][j - 1])%mod + s3[i - 1][j - 1])%mod;
s1[i][j] = (f[i][j] + s1[i - 1][j])%mod;
s2[i][j] = (f[i][j] + s2[i][j - 1]) % mod;
s3[i][j] = (f[i][j] + s3[i - 1][j - 1]) % mod;
}
/*for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
cout << f[i][j]<<" ";
}
cout << endl;
}*/
cout << f[n][m];
return 0;
}
7-21 朋友圈
#include
using namespace std;
typedef long long LL;
const int N = 200010, mod = 1e9 + 7;
int n, q, x;
int a, b, c;
int p[N];
map res[N];
int siz[N];
int find(int x) {
if (x != p[x])p[x] = find(p[x]);
return p[x];
}
int main()
{
cin >> n >> q;
for (int i = 1; i <= n; i++) {
siz[i] = 1;
scanf("%d", &x);
res[i][x]++;
p[i] = i;
}
while (q--) {
scanf("%d%d%d" ,&c, &a, &b);
if (c == 1) {
a = find(a), b = find(b);
if (a != b) {
if (siz[a] > siz[b]) {
p[b] = a;
siz[a] += siz[b];
for (auto it : res[b]) {
res[a][it.first] += it.second;
}
}
else {
p[a] = b;
siz[b] += siz[a];
for (auto it : res[a]) {
res[b][it.first] += it.second;
}
}
}
}
else {
a = find(a);
printf("%d\n", res[a][b]);
}
}
return 0;
}
7-22 走方格
#include
using namespace std;
typedef long long LL;
const int N = 200010, mod = 1e9 + 7;
int n, m, a, b;
int fact[N], infact[N];
int qmi(int a, int k, int p)
{
int res = 1;
while (k)
{
if (k & 1) res = (LL)res * a % p;
a = (LL)a * a % p;
k >>= 1;
}
return res;
}
LL com(int n, int m) {
return ((LL)fact[n] * infact[m] % mod * infact[n - m]) % mod;
}
int main()
{
scanf("%d%d%d%d", &n, &m, &a, &b);
fact[0] = infact[0] = 1;
for (int i = 1; i < N; i++)
{
fact[i] = (LL)fact[i - 1] * i % mod;
infact[i] = (LL)infact[i - 1] * qmi(i, mod - 2, mod) % mod;
}
LL res = 0;
for (int i = 0; i < n - a; i++) {
LL t = ((LL)com(b + i - 1, b - 1) * com(n + m - b - i - 2, n - i - 1))%mod;
res = (res + t)%mod;
}
cout << res;
return 0;
}
7-23 和与积
#include
using namespace std;
typedef long long LL;
LL S, P;
int main()
{
cin >> S >> P;
LL j = 0;
for (LL i = 1; i <= P/i; i++) {
j = S - i;
if ((LL)i * j == P) {
cout << "Yes";
return 0;
}
}
cout << "No";
return 0;
}
7-24 增加硬币
#include
using namespace std;
typedef long long LL;
const int N = 110;
double f[N][N][N];
int d = 0;
double dp(int a, int b, int c) {
if (f[a][b][c] >= 0)return f[a][b][c];
f[a][b][c] = 0;
if (a >= 100 || b >= 100 || c >= 100)return f[a][b][c];
int sum = a + b + c;
f[a][b][c] += (dp(a + 1, b, c) + 1) * (1.0 * a) / sum;
f[a][b][c] += (dp(a, b + 1, c) + 1) * (1.0 * b) / sum;
f[a][b][c] += (dp(a, b, c + 1) + 1) * (1.0 * c) / sum;
return f[a][b][c];
}
int main()
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
memset(f, -1, sizeof f);
printf("%.9lf", dp(a, b, c));
return 0;
}
7-25 日落大道
#include
using namespace std;
typedef long long LL;
typedef pair PII;
const int N = 2010;
int n, m;
char ch;
char s[N][N];
int d[N][N];
vector mp[30];
int dx[4] = { 0,1,0,-1 }, dy[4] = { 1,0,-1,0 };
int main()
{
scanf("%d%d", &n, &m);
getchar();
int x, y, resx, resy;
memset(d, 0x3f, sizeof d);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
scanf("%c", &ch);
s[i][j] = ch;
if (ch >= 'a' && ch <= 'z')mp[ch - 'a'].push_back({i,j});
else if (ch == 'S')x = i, y = j;
else if (ch == 'G')resx = i, resy = j;
}
getchar();
}
queue q;
q.push({ x,y });
d[x][y] = 0;
while (q.size()) {
PII t = q.front();
q.pop();
for (int i = 0; i < 4; i++) {
int a = t.first + dx[i], b = t.second + dy[i];
if (a<1 || a>n || b<1 || b>m || s[a][b] == '#')continue;
if (d[a][b] > d[t.first][t.second] + 1) {
d[a][b] = d[t.first][t.second] + 1;
q.push({ a,b });
}
if (s[a][b] == 'G')break;
if (s[a][b] >= 'a' && s[a][b] <= 'z') {
for (auto it : mp[s[a][b] - 'a']) {
if (it.first == a && it.second == b)continue;
if (d[it.first][it.second] > d[a][b] + 1) {
d[it.first][it.second] = d[a][b] + 1;
q.push({ it.first,it.second });
}
mp[s[a][b] - 'a'].clear();
}
}
}
}
if (d[resx][resy] == 0x3f3f3f3f)printf("-1");
else printf("%d", d[resx][resy]);
return 0;
}
7-26 ICPC保定站
#include
using namespace std;
typedef long long LL;
const int N = 50;
typedef pair PII;
int n, t, k;
int w[N];
LL weights[1 << 24], cnt = 0;
LL res;
void dfs1(int u, LL s) {
if (u == k) {
weights[cnt++] = s;
return;
}
dfs1(u + 1, s);
if (s + w[u] <= t)dfs1(u + 1, s + w[u]);
}
void dfs2(int u, LL s) {
if (u == n) {
int l = 0, r = cnt - 1;
while (l < r) {
int mid = l + r + 1 >> 1;
if (weights[mid] <= (LL)t - s)l = mid;
else r = mid - 1;
}
if (weights[l] + s <= t)
res = max(res, weights[l] + s);
return;
}
dfs2(u + 1, s);
if (w[u] + s <= t)dfs2(u + 1, s + w[u]);
}
int main()
{
scanf("%d%d", &n, &t);
for (int i = 0; i < n; i++)scanf("%d", &w[i]);
sort(w, w + n);
reverse(w, w + n);
k = n / 2;
dfs1(0, 0);
sort(weights, weights + cnt);
cnt = unique(weights, weights + cnt) - weights;
dfs2(k, 0);
cout << res;
return 0;
}
7-27 简单的异或
#include
#include
#include
#include
using namespace std;
const int N = 300010;
int n, m;
int a[N], tr[N];
int lowbit(int x)
{
return x & -x;
}
void add(int x, int v)
{
for (int i = x; i <= n; i += lowbit(i)) tr[i] ^= v;
}
int query(int x)
{
int res = 0;
for (int i = x; i; i -= lowbit(i)) res ^= tr[i];
return res;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
for (int i = 1; i <= n; i++) add(i, a[i]);
while (m--)
{
int k, x, y;
scanf("%d%d%d", &k, &x, &y);
if (k == 2) printf("%d\n", query(y) ^ query(x - 1));
else add(x, y);
}
return 0;
}
7-28 选我啊！
#include
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
struct stu {
LL a, b, c;
}st[N];
LL resa;
int n;
bool cmp(stu s1, stu s2) {
return s1.c > s2.c;
}
int main()
{
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d%d", &st[i].a, &st[i].b);
resa += st[i].a;
st[i].c = st[i].a * 2 + st[i].b;
}
sort(st, st + n, cmp);
int res = 0;
for (int i = 0; i < n; i++) {
if (resa < 0)break;
resa -= st[i].c;
res++;
}
cout << res;
return 0;
}
7-29 豪华酒店
#include
using namespace std;
typedef long long LL;
const int N = 4e5+10;
int n,m;
int a,b,c;
struct node{
int x;
int money;
}s[N];
int cnt;
LL sum = 0;
LL res=0,pre = 0;
bool cmp(node n1,node n2){
if(n1.x==n2.x)return n1.money>n2.money;
return n1.x 7-30 摆放小球
#include
using namespace std;
typedef long long LL;
const int N = 105;
int n, m, a, b, k;
int st[N];
pair f[N], s[N];
int sum = 0;
void dfs(int u) {
if (u == k + 1) {
int res = 0;
for (int i = 1; i <= m; i++) {
if (st[f[i].first]!=0 && st[f[i].second] != 0)res++;
}
sum = max(sum, res);
return;
}
st[s[u].first]++;
dfs(u + 1);
st[s[u].first]--;
st[s[u].second]++;
dfs(u + 1);
st[s[u].second]--;
}
int main()
{
memset(st, 0, sizeof st);
scanf("%d%d", &n, &m);
for (int i = 1; i <= m; i++) {
scanf("%d%d", &a, &b);
f[i].first = a, f[i].second = b;
}
scanf("%d", &k);
for (int i = 1; i <= k; i++) {
scanf("%d%d", &a, &b);
s[i].first = a, s[i].second = b;
}
dfs(1);
cout << sum;
//if(a[n]==0)cout<<"Yes";
//else cout<<"No";
return 0;
}
7-31 最大公约数的和
#include
#define re register int
long long n, ans, f[4000010];
int main() {
scanf("%lld", &n);
for (re i = n; i; --i) {
f[i] = n / i * (n / i);
for (int j = i << 1; j <= n; j += i)f[i] -= f[j];
ans += (f[i] * i - i)/2;
}
printf("%lld", ans);
return 0;
} 7-32 一道简单题
#include
using namespace std;
int n;
int main()
{
cin >> n;
map mp;
for (int i = 0; i <= 30; i++) {
for (int j = 0; j <= 30; j++) {
if (i == j)continue;
int x = (1 << i) + (1 << j);
mp[x] = 1;
}
}
int minn = 1e9 + 10;
for (auto it : mp) {
minn = min(minn, abs(it.first - n));
}
cout << minn;
return 0;
}
7-33 一个无限大的集合
#include
using namespace std;
const int N = 1e6 + 10, mod = 998244353;
typedef long long LL;
int f[N];
int p;
int main() {
f[1] = 1;
LL res = 1;
cin >> p;
for (int i = 2; i <= p; i++) {
f[i] = (f[i - 1] + f[i - 2]) % mod;
res = (res + f[i]) % mod;
}
cout << res;
}