CF 808D D. Array Division（二分）

题目：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.

题意：给你一组数，问知否可以最多移动一个数，使得这一串数可以分成两个部分，每一部分所有数的和相等。

思路：显然由数据范围暴力肯定会超。所以想到二分。首先显然，对于每一个数把它移到第一个或者最后一个是最优的。这样二分就行了。

代码：

#include<bits/stdc++.h>
using namespace std;
int n;
typedef long long ll;
ll a[100005],sum[100005];//sum数组存前缀和
int check(int x)
{

int ans;
ans=lower_bound(sum+1,sum+1+x-1,sum
/2-a[x])-sum;//移到开头的情况，只需要在1到x-1的前缀和中找是否有 sum
/2-a[x]就可以了
if(ans<x&&sum[ans]==sum
/2-a[x]) return 1;
ans=lower_bound(sum+1+x+1,sum+1+n,sum
/2+a[x])-sum;//同理，移到末尾的情况
if(ans<=n&&sum[ans]==sum
/2+a[x]) return 1;
return 0;
}
int main(){
cin>>n;
sum[0]=0;
int flag=0;
for(int i=1;i<=n;i++)
{
scanf("%I64d",&a[i]);
sum[i]=sum[i-1]+a[i];
}
if(sum
%2)//总和必为偶数
{
printf("NO\n");
return 0;
}
int ans=lower_bound(sum+1,sum+1+n,sum
/2)-sum;//二分查找不移动的情况下可不可以找到总和的一半
if(ans<n&&sum[ans]==sum
/2)
{
printf("YES\n");
return 0;
}
for(int i=1;i<=n;i++)
{
if(check(i))
{
flag=1;
break;
}
}
if(flag) printf("YES\n");
else printf("NO\n");
return 0;
}