3513: [MUTC2013]idiots

3513: [MUTC2013]idiots

链接

分析：

　　考虑求不合法的方案数，可以枚举一条最大的边，另外两条边要求长度和小于等于这条边。

　　设dp[i]表示任选两条，长度为和为i的方案数，这是一个卷积，FFT优化。然后维护一个前缀和。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cctype>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
using namespace std;
typedef long long LL;
 
inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}
const int N = 270005;
const double Pi = acos(-1.0);
struct Com{ 
    double x, y; 
    Com() {} Com(double _x,double _y) { x = _x, y = _y; }
}A[N];
Com operator + (const Com &A,const Com &B) { return Com(A.x + B.x, A.y + B.y); }
Com operator - (const Com &A,const Com &B) { return Com(A.x - B.x, A.y - B.y); }
Com operator * (const Com &A,const Com &B) { return Com(A.x * B.x - A.y * B.y, A.x * B.y + A.y * B.x); }
 
int rev[N], cnt[N], dp[N];
 
void FFT(Com *a,int n,int ty) {
    for (int i = 0; i < n; ++i) if (i > rev[i]) swap(a[i], a[rev[i]]);
    Com w, w1, u, t;
    for (int m = 2; m <= n; m <<= 1) {
        w1 = Com(cos(2 * Pi / m), ty * sin(2 * Pi / m));
        for (int i = 0; i < n; i += m) {
            w = Com(1, 0);
            for (int k = 0; k < (m >> 1); ++k) {
                u = a[i + k], t = w * a[i + k + (m >> 1)];
                a[i + k] = u + t, a[i + k + (m >> 1)] = u - t;
                w = w * w1;
            }
        }
    }
}
void solve() {
    int n = read(), Mx = 0, len = 1, lg = 0;
    for (int x, i = 1; i <= n; ++i) x = read(), cnt[x] ++, Mx = max(Mx, x);
    while (len <= Mx + Mx) len <<= 1, lg ++;
    for (int i = 0; i < len; ++i) {
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (lg - 1));
        A[i] = Com(cnt[i], 0);
    }
    FFT(A, len, 1);
    for (int i = 0; i < len; ++i) A[i] = A[i] * A[i];
    FFT(A, len, -1);
    for (int i = 0; i < len; ++i) dp[i] = (A[i].x / (double)len + 0.5);
    LL ans = 0, now = 0, tot = 1ll * n * (n - 1) * (n - 2) / 6;
    for (int i = 0; i < len; ++i) {
        now += dp[i];
        if (i % 2 == 0) now -= cnt[i / 2];
        ans += 1ll * now * cnt[i];
    }
    ans /= 2;
    printf("%.7lf\n", (1.0 - 1.0 * ans / (double)tot));
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i < len; ++i) A[i].x = A[i].y = dp[i] = rev[i] = 0;
}
int main() {
    for (int T = read(); T--; solve());
    return 0;
}