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 = 5000 + 10;
int n;
int a[maxn];
int b[maxn];
int sum[maxn];

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    memset(sum, 0, sizeof sum);
    memset(b, 0, sizeof b);
    sort(a+1, a + n+1);
    for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++)
            b[a[j] - a[i]]++;

    double ans = 0;

    for (int i = 1; i <= 5000; i++) sum[i] = sum[i - 1] + b[i];

    for (int i = 0; i <= 5000; i++)
        for (int j = 0; j <= 5000; j++)
        {
            if (i + j <= 5000)
                ans = ans + 1.0*b[i] * b[j] * (sum[5000] - sum[i + j]);
        }
    
//    printf("%d\n", ans);
    double N = (double)n;
    double f = N*(N - 1)*N*(N - 1)*N*(N - 1) / 8.0;
    printf("%.6lf\n", 1.0*ans / f);
    return 0;
}