codeforces 808D

来源:互联网 发布:淘宝pos机 编辑:程序博客网 时间:2024/05/22 05:25

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;    }}


原创粉丝点击