Educational Codeforces Round 21 D （ 二分 ）

D. Array Division

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.

大致题意：给你n个数假设编号为1~n，你可以选择其中一个数移动到别的位置或者不选。然后选择一个断点k将这n个数分为前后两部分，1~k，k+1~n，问能否使得前后两部分的数的总和相等，如果可以输出YES，否则NO。

思路：如果直接暴力枚举所有情况肯定是会超时的。我们可以枚举移动的点i的位置，然后对该情况下的断点的位置进行二分，这样时间复杂度就降到了nlogn，具体看代码。

代码如下

#include<iostream>#include<algorithm>#include<string>#include<cstdio>#include<cstring>#include<queue>#include<vector>#include<cstring>#include <sstream> #include<cmath>#include<map>#define LL long long  #define ULL unsigned long long  using namespace std;int s[100005];LL pre[100005];//前缀和，注意开lnoglongbool find(int l,int r,LL x){    while(l<=r)    {        int mid=(l+r)/2;        if(pre[mid]==x) return 1;        if(pre[mid]<x)        l=mid+1;        else         r=mid-1;    }    return 0;}int main(){    memset(pre,0,sizeof(pre));    int n;    scanf("%d",&n);    LL sum=0;    for(int i=1;i<=n;i++)    {        scanf("%d",&s[i]);        sum+=s[i];        pre[i]=pre[i-1]+s[i];    }    if(sum%2)//剪枝    {       printf("NO\n");       return 0;    }    sum/=2;    for(int i=1;i<=n;i++)    {        if(find(i+1,n,sum+s[i]))//枚举i，假设需要将它移动到断点k后，那么需要找到pre[k]=sum+s[i]        {            printf("YES\n");            return 0;        }    }    for(int i=1;i<=n;i++)    {        if(find(1,i-1,sum-s[i]))//枚举i，假设需要将它移动到断点k前,那么需要找到pre[k]=sum-s[i]        {            printf("YES\n");            return 0;        }    }    printf("NO\n");    return 0; }