CodeForces 626D Jerry's Protest

计算前两盘A赢，最后一盘B赢的情况下，B获得的球的值总和大于A获得的球总和值的概率。

存储每一对球的差值有几个，然后处理一下前缀和，暴力枚举就好了......

#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<queue>
#include<algorithm>
#include<iostream>
using namespace std;

const int maxn =  + ;
int n;
int a[maxn];
int b[maxn];
int sum[maxn];

int main()
{
    scanf("%d", &n);
    for (int i = ; i <= n; i++) scanf("%d", &a[i]);
    memset(sum, , sizeof sum);
    memset(b, , sizeof b);
    sort(a+, a + n+);
    for (int i = ; i <= n; i++)
        for (int j = i + ; j <= n; j++)
            b[a[j] - a[i]]++;

    double ans = ;

    for (int i = ; i <= ; i++) sum[i] = sum[i - ] + b[i];

    for (int i = ; i <= ; i++)
        for (int j = ; j <= ; j++)
        {
            if (i + j <= )
                ans = ans + 1.0*b[i] * b[j] * (sum[] - sum[i + j]);
        }

//    printf("%d\n", ans);
    double N = (double)n;
    double f = N*(N - )*N*(N - )*N*(N - ) / 8.0;
    printf("%.6lf\n", 1.0*ans / f);
    return ;
}