2019牛客多校 Round5

Solved:4

Rank:122

补题:8/10 

 

A digits 2 

签到 把这个数写n遍

#include 
using namespace std;
int T;
int n;
int main(){
 
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++)printf("%d",n);
        puts("");
    }
    return 0;
}

digits 2

 

B generator 1

题意:求矩阵快速幂 幂超级大

题解:10进制模拟矩阵快速幂

#include 
using namespace std;
typedef long long ll;
ll mod;
 
struct node {
    ll c[2][2];
};
 
node mul(node x, node y) {
    node res;
    memset(res.c, 0, sizeof(res.c));
 
    for(int i = 0; i < 2; i++)
    for(int j = 0; j < 2; j++)
    for(int k = 0; k < 2; k++)
        res.c[i][j] = (res.c[i][j] + x.c[i][k] * y.c[k][j] % mod) % mod;
    return res;
}
 
node pow_mod(node x, ll y) {
    node res;
    res.c[0][0] = res.c[1][1] = 1LL;
    res.c[0][1] = res.c[1][0] = 0LL;
 
    while(y) {
        if(y & 1) res = mul(res, x);
        x = mul(x, x);
        y >>= 1;
    }
    return res;
}
 
char s[1000005];
int ss[1000005];
int main() {
    ll x0, x1, a, b;
    cin>>x0>>x1>>a>>b;
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    for(int i = 1; i <= len; i++) ss[i] = s[i] - '0';
    cin>>mod;
 
    node A;
    A.c[0][0] = a;
    A.c[0][1] = b;
    A.c[1][0] = 1LL;
    A.c[1][1] = 0LL;
    if(len == 1) {
        if(s[1] == '1') {
            printf("%lld\n", x1);
            return 0;
        }
    }
    ss[len]--;
    for(int i = len; i >= 1; i--) {
        if(ss[i] < 0) {
            ss[i] += 10;
            ss[i - 1]--;
        }
    }
 
 
    node now;
    now.c[0][0] = now.c[1][1] = 1LL; now.c[0][1] = now.c[1][0] = 0LL;
    for(int i = 1; i <= len; i++) {
        now = pow_mod(now, 10);
        int t = ss[i];
 
        node pp = pow_mod(A, 1LL * t);
        now = mul(now, pp);
    }
    ll ans = now.c[0][0] * x1 % mod + now.c[0][1] * x0 % mod;
    ans %= mod;
    printf("%lld\n", ans);
    return 0;
}

generator 1

 

C generator 2 (BSGS)

只是大概了解了下BSGS的大概思想.. 题是学弟补的 orz (我摸鱼了

#include 
using namespace std;
typedef long long ll;
const int maxn = 1000000;
int T;
ll x0,n,a,b,p;
ll ksm(ll a,ll b){
    ll res = 1;
    for(;b;b>>=1){
        if(b & 1)res = res*a%p;
        a = a * a % p;
    }
    return res;
}
pair<int,int> node[maxn];
int pos[maxn],val[maxn];
int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%lld%lld%lld%lld%lld",&n,&x0,&a,&b,&p);
        int Q;scanf("%d",&Q);
        if(a == 0){
            while(Q--){
                int v;scanf("%d",&v);
                if(v == x0)
                    puts("0");
                else if(v == b) puts("1");
                else puts("-1");
            }
            continue;
        }
        ll now = x0;
        int up = min(n,(ll)maxn);
        for(int i=0;i){
            node[i] = {now,i};
            now = (now * a + b) % p;
        }
        sort(node,node+up);
        int  m = 0;
        for(int i=0;i){
            val[m] = node[i].first;
            pos[m++] = node[i].second;
            while(i < up - 1 && node[i].first == node[i+1].first)i++;
        }
        int inv_a = ksm(a,p-2);
        int inv_b = (p-b)%p*inv_a%p;
        ll bb = 0, aa = 1;
        for(int i=0;i){
            aa = aa * inv_a % p;
            bb = (bb*a%p + b) % p;
        }
        int r = p / maxn + 1;
        while(Q--){
            int v;scanf("%d",&v);
            int id = lower_bound(val,val+m,v) - val;
            if(id < m && val[id] == v){
                printf("%d\n",pos[id]);continue;
            }
            if(n < maxn){
                puts("-1");continue;
            }
            bool flag = false;
            for(int i=1;i<=r;i++){
                v = (v-bb+p)%p*aa%p;
                int id = lower_bound(val,val+m,v) - val;
                if(id < m && val[id] == v){
                    int res = pos[id] + i * maxn;
                    if(res >= n)puts("-1");
                    else printf("%d\n",res);
                    flag = true;break;
                }
            }
            if(!flag)puts("-1");
        }
    }
    return 0;
}

generator 2

 

E independent set 1 (状压DP)

题意:最多26个点 一共有1<<26个子集 求所有子集最大独立集的和

题解:看题解补的.. 看到26个点以为int根本开不下状压的数组 结果可以用char开.. 因为每个状态的值不会超过26.. (原来还可以这样

　　 于是对于每一个状态 选取他最小的一个点 可能有两种状态转移过来 选了这个点的贡献 和不选这个点

　　 为什么是最小的就可以.. 因为当前状态是1个1个点慢慢选出来的 去掉这个点是子问题 

　　 还有为什么在算这个点的时候 要把和他相邻的点都扣掉.. 我最开始觉得有的点虽然和他相邻 但是不一定在那个状态下这个点产生了贡献 所以应该不扣

　　 扣掉的点肯定不会对最大独立集产生贡献 因为是要放入的点相邻的 然后扣掉是不扣的子问题 所以没区别了

#include 
using namespace std;
 
char dp[1 << 26];
int adj[30];
 
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        adj[u] |= (1 << v);
        adj[v] |= (1 << u);
    }
    for(int i = 0; i < n; i++) adj[i] = (~adj[i]), adj[i] ^= (1 << i);
    //puts("??");
    for(int i = 1; i < (1 << n); i++) {
        int t = __builtin_ctz(i);
        dp[i] = max(dp[i - (1 << t)], (char)(dp[i & adj[t]] + 1));
    }
    int ans = 0;
    for(int i = 1; i < (1 << n); i++) ans += dp[i];
    printf("%d\n", ans);
    return 0;
}

independent set 1

 

F maximum clique 1 (最大独立集)

题意:n个数 选一个最大的子集 使得子集的元素两两不矛盾 矛盾是指两个数的二进制至少有两位不一样

题解:不矛盾的话 显然两个数xor起来不是2的幂和0 而且我们发现如果两个数的二进制1的个数奇偶性一样的话 是一定不矛盾的

　　 于是根据二进制1的个数 建二分图 我跑的最大独立集 = 最小割

　　 输出匹配的话 如果在最后一次bfs中 dis为INF 且是二分图中连s点的集合的点 那么就是被割掉了

　　 dis不为INF的点 如果是连t点的集合中的点 也是被割掉了 因为这个点是肯定跑不到t点的

　　 map常数真的大.... 学了下出题人std判2的幂

#include 
using namespace std;
const int INF = 0x3f3f3f3f;

map<int, int> mp;
int n, s, t, maxflow, cnt;

struct node {
    int to, nex, val;
}E[1300000];
int head[5005];
int cur[5005];

void addedge(int x, int y, int va) {
    E[++cnt].to = y; E[cnt].nex = head[x]; head[x] = cnt; E[cnt].val = va;
    E[++cnt].to = x; E[cnt].nex = head[y]; head[y] = cnt; E[cnt].val = 0;
}

int dis[5005];
bool inque[5005];
int st[1000005];
bool spfa() {
    for(int i = 1; i <= t; i++) dis[i] = INF, inque[i] = 0, cur[i] = head[i];
    int top = 0;
    dis[s] = 0, inque[s] = 1;
    st[++top] = s;

    while(top) {
        int u = st[top];
        top--;
        inque[u] = 0;

        for(int i = head[u]; i; i = E[i].nex) {
            int v = E[i].to;
            if(E[i].val && dis[v] > dis[u] + 1) {
                dis[v] = dis[u] + 1;
                if(!inque[v]) {
                    inque[v] = 1;
                    st[++top] = v;
                }
            }
        }
    }
    return dis[t] != INF;
}

int vis;
int dfs(int x, int flow) {
    if(x == t) {
        vis = 1;
        maxflow += flow;
        return flow;
    }

    int used = 0;
    int rflow = 0;
    for(int i = cur[x]; i; i = E[i].nex) {
        cur[x] = i;
        int v = E[i].to;
        if(E[i].val && dis[v] == dis[x] + 1) {
            if(rflow = dfs(v, min(flow - used, E[i].val))) {
                used += rflow;
                E[i].val -= rflow;
                E[i ^ 1].val += rflow;
                if(used == flow) break;
            }
        }
    }
    return used;
}

void dinic() {
    maxflow = 0;
    while(spfa()) {
        vis = 1;
        while(vis) {
            vis = 0;
            dfs(s, INF);
        }
    }
}

bool is_power2(int x) {
    return x && (x & (x - 1)) == 0;
}

int a[5005];
int er[5005];
int main() {
    scanf("%d", &n);
    cnt = 1;
    s = n + 1;
    t = s + 1;
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        int tot = 0;
        int tmp = a[i];
        while(tmp) {
            if(tmp & 1) tot++;
            tmp >>= 1;
        }
        if(tot & 1) addedge(s, i, 1), er[i] = 1;
        else addedge(i, t, 1), er[i] = 0;
    }

    for(int i = 1; i <= n; i++) {
        for(int j = i + 1; j <= n; j++) {
            if(is_power2(a[i] ^ a[j])) {
                if(er[i]) addedge(i, j, 1);
                else addedge(j, i, 1);
            }
        }
    }

    dinic();
    int ans = n - maxflow;
    printf("%d\n", ans);

    int cn = 0;
    for(int i = 1; i <= n; i++) {
        if(dis[i] == INF) {
            if(!er[i]) {
                cn++;
                if(cn != ans) printf("%d ", a[i]);
                else {
                    printf("%d\n", a[i]);
                    break;
                }
            }
        } else {
            if(er[i]) {
                cn++;
                if(cn != ans) printf("%d ", a[i]);
                else {
                    printf("%d\n", a[i]);
                    break;
                }
            }
        }
    }

    return 0;
}

maximum clique 1

 

G subsequence 1 (DP)

题意:给s，t的数字串 求s中有多少子序列10进制下大于t

题解:dp i,j,k表示s的第i位匹配到t的第j位 k = 0表示这个子序列小于t的前j位 1表示等于 2表示大于 模拟即可

#include 
using namespace std;
typedef long long ll;
const ll mod = 998244353;
 
char s[3005];
char t[3005];
ll dp[3005][3005][3];
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int n, m;
        scanf("%d%d", &n, &m);
        scanf("%s", s + 1);
        scanf("%s", t + 1);
        for(int i = 1; i <= n; i++) {
            int t1 = s[i] - '0';
            for(int j = 0; j <= i; j++) dp[i][j][0] = dp[i][j][1] = dp[i][j][2] = 0;
            //dp[i][0][1] = 1;
            //dp[i][0][1] = 1;
            //dp[i][0][2] = 1;
            for(int j = 1; j <= i; j++) {
                int t2 = t[j] - '0';
                dp[i][j][0] = dp[i - 1][j][0];
                dp[i][j][1] = dp[i - 1][j][1];
                dp[i][j][2] = dp[i - 1][j][2];
 
 
                if(j == 1) {
                    if(t1 == 0) continue;
                    if(t1 == t2) dp[i][j][1]++;
                    else if(t1 > t2) dp[i][j][2]++;
                    else dp[i][j][0]++;
                } else if(j == m + 1) {
                    dp[i][j][2] += dp[i - 1][j - 1][0] + dp[i - 1][j - 1][1] + dp[i - 1][j - 1][2];
                    dp[i][j][1] = 0;
                    dp[i][j][0] = 0;
                } else if(j > m + 1) {
                    dp[i][j][2] += dp[i - 1][j - 1][2];
                }
                else {
                    if(t1 == t2) {
                        dp[i][j][1] += dp[i - 1][j - 1][1];
                        dp[i][j][2] += dp[i - 1][j - 1][2];
                        dp[i][j][0] += dp[i - 1][j - 1][0];
                    } else if(t1 > t2) {
                        dp[i][j][2] += dp[i - 1][j - 1][1] + dp[i - 1][j - 1][2];
                        dp[i][j][0] += dp[i - 1][j - 1][0];
                    } else {
                        dp[i][j][0] += dp[i - 1][j - 1][1] + dp[i - 1][j - 1][0];
                        dp[i][j][2] += dp[i - 1][j - 1][2];
                    }
                }
 
                dp[i][j][0] %= mod;
                dp[i][j][1] %= mod;
                dp[i][j][2] %= mod;
            }
        }
        ll ans = 0;
        /*
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= i; j++) {
                printf("%d %d %lld %lld %lld\n", i, j, dp[i][j][0], dp[i][j][1], dp[i][j][2]);
            }
        }*/
        for(int i = m; i <= n; i++) {
            ans = (ans + dp[n][i][2]) % mod;
        }
        printf("%lld\n", ans);
    }
    return 0;
}

subsequence 1

 

H subsequence 2 (模拟)

题意:给一个串的许多信息 每次给两种字母 然后给出在只剩下这两种字母的情况下原串的样子 输出原串

题解:每种信息更新这一种字母前有多少字母 最后组成原串

#include 
using namespace std;
 
char s[5];
char t[10005];
int num[15];
int pos[15][10005];
int ans[10005];
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= 10; i++) pos[i][0] = 0;
    for(int i = 1; i <= n; i++) ans[i] = 0;
    bool f = true;
    for(int i = 1; i <= m * (m - 1) / 2; i++) {
        scanf("%s", s + 1);
        int len2; scanf("%d", &len2);
        getchar();
        if(len2 == 0) continue;
        scanf("%s", t + 1);
        int c1 = s[1] - 'a' + 1;
        int c2 = s[2] - 'a' + 1;
        int cn1 = 0, cn2 = 0;
        if(c1 > c2) swap(c1, c2);
        for(int j = 1; j <= len2; j++) {
            if(t[j] - 'a' + 1 == c1) {
                cn1++;
                pos[c1][cn1] += cn2;
            } else if(t[j] - 'a' + 1 == c2) {
                cn2++;
                pos[c2][cn2] += cn1;
            } else {
                f = false;
                break;
            }
        }
        if(pos[c1][0] == 0) pos[c1][0] = cn1;
        else if(pos[c1][0] != cn1) f = false;
        if(pos[c2][0] == 0) pos[c2][0] = cn2;
        else if(pos[c2][0] != cn2) f = false;
    }
 
    for(int i = 1; i <= m; i++) {
        for(int j = 1; j <= pos[i][0]; j++) {
            //cout << pos[i][j] << " ";
 
            if(pos[i][j] + j > n) f = false;
            else {
                if(ans[pos[i][j] + j]) f = false;
                else ans[pos[i][j] + j] = i;
            }
        }
        //puts("");
    }
    //puts("f");
    //cout << f <
    for(int i = 1; i <= n; i++) if(!ans[i]) f = false;
    if(!f) {
        puts("-1");
    } else {
        for(int i = 1; i <= n; i++) {
            printf("%c", ans[i] - 1 + 'a');
        }
        puts("");
    }
 
    return 0;
}

subsequence 2

 

I three points 1 (计算几何)

队友补的 (啥东西写了400行???

#include 
#include 
#include <string.h>
#include 
#include 
#include 
 
using namespace std;
 
int gcd(int a, int b){
    return b == 0 ? a : gcd(b, a % b);
}
 
const double eps = 1e-8;
int cmp(double x) {
    if (fabs(x) < eps) return 0;
    if (x > 0) return 1;
    return -1;
}
 
const double pi = acos(-1.0);
 
inline double sqr(double x) {
    return x * x;
}
 
struct point {
    double x, y;
    point() {}
    point(double a, double b) : x(a), y(b) {}
    void input() {
        scanf("%lf%lf", &x, &y);
    }
    friend point operator + (const point &a, const point &b) {
        return point(a.x + b.x, a.y + b.y);
    }
    friend point operator - (const point &a, const point &b) {
        return point(a.x - b.x, a.y - b.y);
    }
    friend bool operator == (const point &a, const point &b) {
        return cmp(a.x - b.x) == 0 && cmp(a.y - b.y) == 0;
    }
    friend point operator * (const point &a, const double &b) {
        return point(a.x * b, a.y * b);
    }
    friend point operator * (const double &a, const point &b) {
        return point(a * b.x, a * b.y);
    }
    friend point operator / (const point &a, const double &b) {
        return point(a.x / b, a.y / b);
    }
    double norm(){
        return sqrt(sqr(x) + sqr(y));
    }
 
    friend bool operator < (const point a, const point b){
        if(a.x == b.x) return a.y < b.y;
        if(a.y == b.y) return a.x < b.x;
        return a.x < b.x;
    }
};
 
double det(const point &a, const point &b) {
    return a.x * b.y - a.y * b.x;
}
 
double dot(const point &a, const point &b) {
    return a.x * b.x + a.y * b.y;
}
 
double dist(const point &a, const point &b) {
    return  (a - b).norm();
}
 
point rotate_point(const point &p, double A){
    double tx = p.x, ty = p.y;
    return point(tx * cos(A) - ty * sin(A), tx * sin(A) + ty * cos(A));
}
 
struct line {
    point a, b;
    line() {}
    line(point x, point y) : a(x), b(y) {}
};
 
line point_make_line(const point a, const point b) {
    return line(a, b);
}
 
double dis_point_segment(const point p, const point s, const point t) {
    if(cmp(dot(p - s, t - s)) < 0) return (p - s).norm();
    if(cmp(dot(p - t, s - t)) < 0) return (p - t).norm();
    return fabs(det(s - p, t - p) / dist(s, t));
}
 
void PointProjLine(const point p, const point s, const point t, point &cp){
    double r = dot((t - s), (p - s)) / dot(t - s, t - s);
    cp = s + r * (t - s);
}
 
bool PointOnSegment(point p, point s, point t){
    return cmp(det(p - s, t - s)) == 0 && cmp(dot(p - s, p - t)) <= 0;
}
 
bool parallel(line a, line b){
    return !cmp(det(a.a - a.b, b.a - b.b));
}
 
bool line_make_point(line a, line b, point &res){
    if(parallel(a, b)) return false;
    double s1 = det(a.a - b.a, b.b - b.a);
    double s2 = det(a.b - b.a, b.b - b.a);
    res = (s1 * a.b - s2 * a.a) / (s1 - s2);
    return true;
}
 
line move_d(line a, const double &len){
    point d = a.b - a.a;
    d = d / d.norm();
    d = rotate_point(d, pi / 2);
    return line(a.a + d * len, a.b + d * len);
}
 
bool SegmentSamePoints(line i, line j){
    return max(i.a.x, i.b.x) >= min(j.a.x, j.b.x) &&
            max(j.a.x, j.b.x) >= min(i.a.x, i.b.x) &&
            max(i.a.y, i.b.y) >= min(j.a.y, j.b.y) &&
            max(j.a.y, j.b.y) >= min(i.a.y, i.b.y) &&
            cmp(det(j.a - i.a, i.b - i.a) * det(j.b - i.a, i.b - i.a)) <= 0 &&
            cmp(det(i.a - j.a, j.b - j.a) * det(i.b - j.a, j.b - j.a)) <= 0;
}
 
double Sabc(point a, point b, point c){
    return fabs(a.x * b.y + b.x * c.y + c.x * a.y - a.x * c.y - b.x * a.y - c.x * b.y) / 2.0;
}
 
struct section{
    double l, r;
    section() {}
    section(double a, double b) {
        l = min(a, b);
        r = max(a, b);
    }
    friend bool operator < (section a, section b){
        if(a.l == b.l) return a.r < b.r;
        return a.l < b.l;
    }
};
 
bool sjs(section a, section b){
    if(b < a) swap(a, b);
    if(a.l < b.l){
        return (a.r > b.l) || (b.r < a.r);
    }
    else{
        return (a.r < b.r);
    }
}
 
const int maxn = 1005;
struct polygon {
    int n;
    point a[maxn];
    polygon() {}
    
    double perimeter() {
        double sum = 0;
        a[n] = a[0];
        for(int i = 0; i < n; i++){
            sum += (a[i + 1] - a[i]).norm();
        }
        return sum;
    }
 
    double area(){
        double sum = 0;
        a[n] = a[0];
        for(int i = 0; i < n; i++){
            sum += det(a[i + 1], a[i]);
        }
        return sum / 2.0;
    }
 
    int Point_In(point t){
        int num = 0, i, d1, d2, k;
        a[n] = a[0];
        for(i = 0; i < n; i++){
            if(PointOnSegment(t, a[i], a[i + 1])) return 2;
            k = cmp(det(a[i + 1] - a[i], t - a[i]));
            d1 = cmp(a[i].y - t.y);
            d2 = cmp(a[i + 1].y - t.y);
            if(k > 0 && d1 <= 0 && d2 > 0) num++;
            if(k < 0 && d2 <= 0 && d1 > 0) num--;
        }
        return num != 0;
    }
 
    int Border_Int_Point_Num();
    int Inside_Int_Point_Num();
};
 
int polygon::Border_Int_Point_Num(){
    int num = 0;
    a[n] = a[0];
    for(int i = 0; i < n; i++){
        num += gcd(abs(int(a[i + 1].x - a[i].x)), abs(int(a[i + 1].y - a[i].y)));
    }
    return num;
}
 
int polygon::Inside_Int_Point_Num(){
    return int(area()) + 1 - Border_Int_Point_Num() / 2;
}
 
struct polygon_convex{
    vector P;
    polygon_convex(int Size = 0){
        P.resize(Size);
    }
};
 
bool comp_less(const point &a, const point &b){
    return cmp(a.x - b.x) < 0 || cmp(a.x - b.x) == 0 && cmp(a.y - b.y) < 0;
}
 
polygon_convex convex_hull(vector a){
    polygon_convex res(2 * a.size() + 5);
    sort(a.begin(), a.end(), comp_less);
    a.erase(unique(a.begin(), a.end()), a.end());
    int m = 0;
    for(int i = 0; i < a.size(); ++i){
        while(m > 1 && cmp(det(res.P[m - 1] - res.P[m - 2], a[i] - res.P[m - 2])) <= 0)
            --m;
        res.P[m++] = a[i];
    }
    int k = m;
    for(int i = int(a.size()) - 2; i >= 0; --i){
        while(m > k && cmp(det(res.P[m - 1] - res.P[m - 2], a[i] - res.P[m - 2])) <= 0)
            --m;
        res.P[m++] = a[i];
    }
    res.P.resize(m);
    if(a.size() > 1) res.P.resize(m - 1);
    return res;
}
 
int T;
double a, b, c;
double w, h;
 
bool w_h_swap = false;
 
bool wrong(point r){
    return !(cmp(r.x - 0) >= 0 && cmp(w - r.x) >= 0 && cmp(r.y - 0) >= 0 && cmp(h - r.y) >= 0);
}
 
// #define __DEBUG
 
int main(){
    scanf("%d", &T);
 
    #ifdef __DEBUG
    printf("%d\n", T);
    #endif
 
    while(T--){
        scanf("%lf%lf%lf%lf%lf", &w, &h, &a, &b, &c);
         
 
        #ifdef __DEBUG
            printf("%.0f %.0f %.0f %.0f %.0f ", w, h, a, b, c);
        #endif
 
        double l[3] = {a, b, c};
        sort(l, l + 3);
        w_h_swap = false;
 
        point A, B, C;
        double dgr;
 
        A = point(0, 0);
        C = point(-1, -1);
 
        if(wrong(C)){
            if(l[0] <= w)
                B = point(l[0], 0);
            else
                B = point(w, sqrt(sqr(l[0]) - sqr(w)));
            if(!wrong(B)){
                dgr = acos( (sqr(l[0]) + sqr(l[1]) - sqr(l[2])) / (2 * l[0] * l[1]) );
                C = rotate_point(B, dgr) / B.norm() * l[1];
                if(wrong(C)){
                    dgr = acos( (sqr(l[0]) + sqr(l[2]) - sqr(l[1])) / (2 * l[0] * l[2]) );
                    C = rotate_point(B, dgr) / B.norm() * l[2];
                }                
            }
        }
 
        if(wrong(C)){
            if(l[1] <= w)
                B = point(l[1], 0);
            else
               B = point(w, sqrt(sqr(l[1]) - sqr(w)));
            if(!wrong(B)){
                dgr = acos( (sqr(l[1]) + sqr(l[0]) - sqr(l[2])) / (2 * l[1] * l[0]) );
                C = rotate_point(B, dgr) / B.norm() * l[0];
                if(wrong(C)){
                    dgr = acos( (sqr(l[1]) + sqr(l[2]) - sqr(l[0])) / (2 * l[1] * l[2]) );
                    C = rotate_point(B, dgr) / B.norm() * l[2];
                }                
            }
        }
 
       if(wrong(C)){
            if(l[2] <= w)
                B = point(l[2], 0);
            else
                B = point(w, sqrt(sqr(l[2]) - sqr(w)));
            if(!wrong(B)){
                dgr = acos( (sqr(l[2]) + sqr(l[0]) - sqr(l[1])) / (2 * l[2] * l[0]) );
                C = rotate_point(B, dgr) / B.norm() * l[0];
                if(wrong(C)){
                    dgr = acos( (sqr(l[2]) + sqr(l[1]) - sqr(l[0])) / (2 * l[2] * l[1]) );
                    C = rotate_point(B, dgr) / B.norm() * l[1];
                }                
            }
        }
 
        if(wrong(C)){
            swap(w, h);
            w_h_swap = true;
        }
 
        if(wrong(C)){
            if(l[0] <= w)
                B = point(l[0], 0);
            else
                B = point(w, sqrt(sqr(l[0]) - sqr(w)));
            if(!wrong(B)){
                dgr = acos( (sqr(l[0]) + sqr(l[1]) - sqr(l[2])) / (2 * l[0] * l[1]) );
                C = rotate_point(B, dgr) / B.norm() * l[1];
                if(wrong(C)){
                    dgr = acos( (sqr(l[0]) + sqr(l[2]) - sqr(l[1])) / (2 * l[0] * l[2]) );
                    C = rotate_point(B, dgr) / B.norm() * l[2];
                }                
            }
        }
 
        if(wrong(C)){
            if(l[1] <= w)
                B = point(l[1], 0);
            else
               B = point(w, sqrt(sqr(l[1]) - sqr(w)));
            if(!wrong(B)){
                dgr = acos( (sqr(l[1]) + sqr(l[0]) - sqr(l[2])) / (2 * l[1] * l[0]) );
                C = rotate_point(B, dgr) / B.norm() * l[0];
                if(wrong(C)){
                    dgr = acos( (sqr(l[1]) + sqr(l[2]) - sqr(l[0])) / (2 * l[1] * l[2]) );
                    C = rotate_point(B, dgr) / B.norm() * l[2];
                }                
            }
        }
 
       if(wrong(C)){
            if(l[2] <= w)
                B = point(l[2], 0);
            else
                B = point(w, sqrt(sqr(l[2]) - sqr(w)));
            if(!wrong(B)){
                dgr = acos( (sqr(l[2]) + sqr(l[0]) - sqr(l[1])) / (2 * l[2] * l[0]) );
                C = rotate_point(B, dgr) / B.norm() * l[0];
                if(wrong(C)){
                    dgr = acos( (sqr(l[2]) + sqr(l[1]) - sqr(l[0])) / (2 * l[2] * l[1]) );
                    C = rotate_point(B, dgr) / B.norm() * l[1];
                }
            }
        }
 
        if(w_h_swap){
            swap(A.x, A.y);
            swap(B.x, B.y);
            swap(C.x, C.y);
        }
 
        point ans[3] = {A, B, C};
 
        point X, Y, Z;
 
        bool flag = false;
        for(int i = 0; i < 3; i++){
            for(int j = 0; j < 3; j++){
                if(j == i) continue;
                for(int k = 0; k < 3; k++){
                    if(k == i || k == j) continue;
                    X = ans[i];
                    Y = ans[j];
                    Z = ans[k];
                    if(cmp(dist(X, Y) - a) == 0 && 
                        cmp(dist(X, Z) - b) == 0 && 
                        cmp(dist(Y, Z) - c) == 0){
                        flag = true;
                        break;
                    }
                }
                if(flag) break;
            }
            if(flag) break;
        }
        printf("%.8f %.8f %.8f %.8f %.8f %.8f\n", X.x, X.y, Y.x, Y.y, Z.x, Z.y);
    }
    return 0;
}

three points 1

 

转载于:https://www.cnblogs.com/lwqq3/p/11286317.html