Code Forces 21C Stripe 2

C. Stripe 2

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

Once Bob took a paper stripe of n squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into three pieces so that the sum of numbers from each piece is equal to the sum of numbers from any other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?

Input

The first input line contains integer n (1 ≤ n ≤ 105) — amount of squares in the stripe. The second line contains n space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.

Output

Output the amount of ways to cut the stripe into three non-empty pieces so that the sum of numbers from each piece is equal to the sum of numbers from any other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.

Examples

input

4
1 2 3 3

output

1

input

5
1 2 3 4 5

output

0

切割两个点，那么两个点到顶点的区间和肯定是总和的3分之1.对于每个点找到他右边有多少个满足条件的点，

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
int n;
long long int a[100005];
long long int s[100005];
int l[100005];
int r[100005];
long long int sum;
int main()
{
    scanf("%d",&n);
    s[0]=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        s[i]=s[i-1]+a[i];
        sum+=a[i];
    }
    if(sum%3!=0||n<=2)
    {printf("0\n");return 0;}
    int cnt=0;int cot=0;
    memset(l,0,sizeof(l));
    memset(r,0,sizeof(r));
    for(int i=n;i>=1;i--)
    {
        if(s[n]-s[i-1]==sum/3)
        {
            r[i]=r[i+1]+1;
        }
        else
            r[i]=r[i+1];
    }
    long long int ans=0;
    for(int i=1;i<=n;i++)
    {
        if(s[i]==sum/3)
        {
            int x=r[i+2];

            ans+=x;
        }
    }
    printf("%lld\n",ans);
    return 0;

}