BZOJ 1800 [Ahoi2009]fly 飞行棋

暴力枚举

矩形对角线必定是直径，可以枚举直径，然后算方案数，O(n^2)

也可以直接枚举四个顶点,O(n^4)

#include<cstdio>using namespace std;int sum[23];int main(){    int n;    scanf("%d",&n);    for(int i = 1; i <= n; i++)    {        scanf("%d",&sum[i]);        sum[i]+=sum[i-1];    }    int ans=0;    for(int a = 1; a <= n; a++)        for(int b = a+1; b <= n; b++)            for(int c = b+1; c <= n; c++)                for(int d = c+1; d <= n; d++)                    if(sum[b-1]-sum[a-1]==sum[d-1]-sum[c-1] && sum[c-1]-sum[b-1]==sum[a-1]+sum[n]-sum[d-1])                        ans++;    printf("%d\n",ans);}