codeforces 808D - Array Division

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探讨如何通过移动数组中的一个元素实现将数组分割成两个部分，使两部分的元素之和相等。利用前缀和技巧，结合二分查找提高效率。

Vasya has an array a consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position).

Inserting an element in the same position he was erased from is also considered moving.

Can Vasya divide the array after choosing the right element to move and its new position?

Input

The first line contains single integer n (1 ≤ n ≤ 100000) — the size of the array.

The second line contains n integers a1, a2... an (1 ≤ ai ≤ 109) — the elements of the array.

Output

Print YES if Vasya can divide the array after moving one element. Otherwise print NO.

Examples

input
3
1 3 2

output
YES

input
5
1 2 3 4 5

output
NO

input
5
2 2 3 4 5

output
YES

Note

In the first example Vasya can move the second element to the end of the array.

In the second example no move can make the division possible.

In the third example Vasya can move the fourth element by one position to the left.

题目：问能否移动一位或者不移动，使得数组能分成前后两块（两个数组的和相等）

先前缀和，然后枚举每一个元素解决

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1000100;
const ll inf = 99999999999999;

int n;
ll sum[maxn],a[maxn];

bool ok(ll x,int l,int r)
{
    while(l<=r)
    {
         int  mid=(l+r)/2;
         if(sum[mid]==x)return true;
         else if(sum[mid]<x)l=mid+1;
         else r=mid-1;
    }
    return false;
}
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        sum[i]=sum[i-1]+a[i];
    }
    if(sum[n]&1)
        cout<<"NO"<<endl;
    else
    {
        ll ans=sum[n]/2;
        int  flag=0;
        for(int i=1;i<=n;i++)
        {
            if(ok(ans-a[i],1,i-1)||ok(ans+a[i],i,n))
            {
                flag=1; break;
            }
        }
        if(flag)
            cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
}