D. Array Divisiontime 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.
Examplesinput
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.
从一个数列中至多移动一个数字插至任意位置，问是否有可能使得数列可以被分为相等的两部分.
一开始用的map<int,pair<int,int> > 虽然A了可后来发现map的话不能判重会出错的啊，果然被hack了,
用一个map<LL,int> 表示某一个数出现的次数然后模拟即可
#include<bits/stdc++.h>
using namespace std;
#define LL long long
int a[100005];
map<LL,int> M;
int main()
{
int n,i,j,k;
LL sumn=0,ans;
cin>>n;
for(i=1;i<=n;++i){
scanf(“%d”,a+i);
sumn+=a[i];
++M[a[i]];
}
if(sumn & 1) {
puts(“NO”);
return 0;
}
LL tmp=sumn/2;
if(M[tmp]>0) {puts(“YES”);return 0;}
for(i=1;i<=n;++i){
tmp-=a[i];
–M[a[i]];
if(M[tmp]>0) {
puts(“YES”);
return 0;}
} tmp=sumn/2;
for(i=1;i<=n;++i) ++M[a[i]];
for(i=n;i>=1;i–){
tmp-=a[i];
–M[a[i]];
if(M[tmp]>0) {puts(“YES”);return 0;}
}
puts(“NO”);
return 0;
}
