Codefoces 429 D. Tricky Function

나체의 최근점은...
D. Tricky Function
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The selection will be made with the help of a single problem. Blatnatalag, a friend of Iahub, managed to get hold of the problem before the contest. Because he wants to make sure Iahub will be the one qualified, he tells Iahub the following task.
You're given an (1-based) array a with n elements. Let's define function f(i, j) (1 ≤ i, j ≤ n) as (i - j)2 + g(i, j)2. Function g is calculated by the following pseudo-code:

int g(int i, int j) {
    int sum = 0;
    for (int k = min(i, j) + 1; k <= max(i, j); k = k + 1)
        sum = sum + a[k];
    return sum;
}

Find a value mini ≠ j  f(i, j).
Probably by now Iahub already figured out the solution to this problem. Can you?
Input
The first line of input contains a single integer n (2 ≤ n ≤ 100000). Next line contains n integers a[1], a[2], ..., a[n] ( - 104 ≤ a[i] ≤ 104).
Output
Output a single integer — the value of mini ≠ j  f(i, j).
Sample test(s)
input

4
1 0 0 -1

output

1

input

2
1 -1

output

2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int LL;

const int maxn=100100;
const LL INF=1LL<<62;

LL a[maxn];
int n;

struct POINT
{
    LL x,y;
}p[maxn],temp[maxn];

bool cmpxy(POINT a,POINT b)
{
    if(a.x!=b.x) return a.x<b.x;
    return a.y<b.y;
}

bool cmpy(POINT a,POINT b)
{
    return a.y<b.y;
}

inline LL square(LL x)
{
    return x*x;
}

LL dist(POINT a,POINT b)
{
    return square(a.x-b.x)+square(a.y-b.y);
}

LL Close_pair(int left,int right)
{
    LL d=INF;
    if(left==right) return INF;
    if(left+1==right) return dist(p[left],p[right]);
    int mid=(left+right)/2;
    d=min(Close_pair(left,mid),Close_pair(mid+1,right));

    int k=0;
    for(int i=left;i<=right;i++)
    {
        if(square(p[i].x-p[mid].x)<=d)
        {
            temp[k++]=p[i];
        }
    }
    sort(temp,temp+k,cmpy);
    for(int i=0;i<k;i++)
    {
        for(int j=i+1;j<k&&square(temp[i].y-temp[j].y)<d;j++)
        {
            d=min(d,dist(temp[i],temp[j]));
        }
    }

    return d;
}

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%I64d",a+i);
        p[i].x=i;
        p[i].y=p[i-1].y+a[i];
    }
    sort(p+1,p+1+n,cmpxy);
    printf("%I64d
",Close_pair(1,n));
    return 0;
}