codeforcodeforces 808D——Array Division（查找，set）

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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.

对于每一个数，如果那你想要把它换到另一个部分，则对两个部分差值的改变是2*a[i]

从前往后扫一遍，再从后往前扫一遍。

每次查找能不能通过移动当前数完成，然后把差值记录

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <string>#include <vector>#include <map>#include <set>#include <iomanip>using namespace std;#define _ ios::sync_with_stdio(false)const int MAXN = 100010;const int INF=0x7fffffff;long long a[MAXN];long long sum=0;long long s1[MAXN];set<long long > s;int main(){int n;scanf("%d",&n);s1[0]=0;for(int i=1;i<=n;i++){scanf("%lld",a+i);sum+=a[i];s1[i]=s1[i-1]+a[i];}int ok=0;s.clear();for(int i=1;i<=n;i++){if(s.count(2*a[i])){ok=1;break;}long long x=sum-2*s1[i];if(x==0){ok=1;break;}if(x>=0)s.insert(x);}s.clear();for(int i=n;i>=1;i--){if(s.count(2*a[i])){ok=1;break;}long long x=2*s1[i]-sum;if(x==0){ok=1;break;}if(x>=0)s.insert(x);}if(ok)puts("YES");else puts("NO");}

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