cf 808D Array Division(二分思维)@

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... 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

31 3 2

output

YES

input

51 2 3 4 5

output

NO

input

52 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：有n个物品的价值，问能否移动一个物品到任意位置使这个数组分成两部分，并且和相等；

解：计算这个数组的前缀后缀和，枚举物品，二分前缀和后缀和；

Vasya will erase some element and insert it into an arbitrary position

some有一些和某个的意思，因为这里用的是it所以是一些的意思

#include <iostream>#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N = 1e5+10;int n;LL a[N], b[N], s1[N], s2[N];int judge(LL x,int flag,int l,int r){    if(flag)    {        while(l<=r)        {            int mid=(l+r)/2;            if(s1[mid]==x) return 1;            else if(s1[mid]<x) l=mid+1;            else r=mid-1;        }    }    else    {        while(l<=r)        {            int mid=(l+r)/2;            if(s2[mid]==x) return 1;            else if(s2[mid]<x) l=mid+1;            else r=mid-1;        }    }    return 0;}int main(){    scanf("%d", &n);    LL sum=0;    memset(s1,0,sizeof(s1));    memset(s2,0,sizeof(s2));    for(int i=1; i<=n; i++)    {        scanf("%lld", &a[i]);        s1[i]=s1[i-1]+a[i];        sum+=a[i];    }    for(int i=1; i<=n; i++)    {        s2[i]=s2[i-1]+a[n-i+1];    }    if(sum%2!=0)    {        puts("NO");        return 0;    }    for(int i=1; i<=n; i++)    {        if(s1[i]==sum/2||s2[i]==sum/2)        {            puts("YES");            return 0;        }        else        {            LL x=sum/2-a[i];            if(judge(x,1,0,i-1))            {                puts("YES");                return 0;            }            if(judge(x,0,0,n-i-1))            {                puts("YES");                return 0;            }        }    }    puts("NO");    return 0;}