题目链接：点击打开链接

暴力出奇迹。

正解应该是近期点对。以i点为x轴，sum[i]（前缀和）为y轴，求随意两点间的距离。

先来个科学的暴力代码：

#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
#define N 100050
#define ll __int64
ll a[N], sum[N];
ll n;
int main(){
	ll u, v, i, j, que;
	sum[0] = 0;
	while(~scanf("%I64d",&n)){
		for(i=1;i<=n;i++)scanf("%I64d",&a[i]),sum[i] = sum[i-1]+a[i];
		ll ans = a[2]*a[2]+1;
		for(i = 1; i <= n; i++){
			if(i*i>=ans)break;
			ll tmp = ans;
			for(j = i+1; j<=n; j++){
				tmp = min(tmp, (sum[j]-sum[j-i])*(sum[j]-sum[j-i]));
			}
			ans = min(ans, tmp+i*i);
		}
		printf("%I64d\n",ans);
	}
	return 0;
}