codeforces Round 21 808DArray Division

codeforces Round 21 808D Array Division

D. Array Division

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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... a_n (1 ≤ a_i ≤ 10⁹) — the elements of the array.

Output

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

Examples

input

31 3 2

output

YES

题意： 在数组中移动一个数 使得分组可以分割成两个数组 使得两个数组之和相等

题解 首先分两种情况

1. 往后面移动一个数到前面  (维护前缀和 看看后面有没有符合条件的数)

2. 移掉一个数            (移掉第i个数 看看连续的数能不能符合条件)

#include <stdio.h>#include <bits/stdc++.h>#define mod 1000000007typedef long long ll;using namespace std;ll s[100005];ll n,avg;map<ll,ll>mp;int main(){ cin>>n;for(int i=1;i<=n;i++){cin>>s[i];avg+=s[i];mp[s[i]]=i;   //记录最后一次出现的位置 }mp[0]=n+1;if(avg%2==1){cout<<"NO";return 0;}avg/=2;ll sum=0;for(int i=1;i<=n;i++){   //把数往前面移动 sum+=s[i];if(sum>avg) break;if(mp[avg-sum]>i){    //如果后面的数存在 且下标大于i     cout<<"YES"<<endl;return 0;}}int r=1;sum=0;     for(int i=1;i<=n;i++){   //把第i个数往后面掉    sum-=s[i];    while(sum<avg) sum+=s[r++];    while(sum>avg) sum-=s[--r];    if(r<=i) break;    if(sum==avg){    cout<<"YES"<<endl;return 0;    }    sum+=s[i];     }    cout<<"NO";return 0;}