给你一个长为n的序列a

定义f(i,j)=(i-j)²+g(i,j)²

#include<bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

using ll = long long;

const int qq = ;

ll a[qq];

inline ll sqr(ll x){ return x*x;}

int main()

{

    int n;

    cin >> n;

    for(int i=;i <= n;i++)

    {

        scanf("%lld",a + i);

        a[i] += a[i-];

    }

    ll minans = INF;

    for(int i=;i <= n;i++)

    {

        int k=;

        for(int j=i-;j >= &&k<=;j--,k++)

        {

            ll ans = sqr(i-j) + sqr(a[i]-a[j]);

            if(ans < minans)

                minans = ans;

        }

    }

    cout << minans << endl;

    return ;

}