Number of Ways

Description

You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.

More formally, you need to find the number of such pairs of indices i, j(2 ≤ i ≤ j ≤ n - 1), that .

Input

The first line contains integer n(1 ≤ n ≤ 5·10⁵), showing how many numbers are in the array. The second line contains n integersa[1], a[2], ..., a[n](|a[i]| ≤  10⁹) — the elements of array a.

Output

Print a single integer — the number of ways to split the array into three parts with the same sum.

Sample Input

Input

5
1 2 3 0 3

Output

2

Input

4
0 1 -1 0

Output

1

Input

2
4 1

Output

0
大意：给你一串数让你分成平均三分，问共有多少种分法，主要是复杂度问题，用一个for把复杂度降到O（n），先求出平均值，让后计算为平均值的结点和两倍平均值的结点，如果前面已经有结点为平均值，那么两倍结点的那个就加上前面共有多少一倍结点的，就把复杂度降低了orz自己太笨想不到

#include<cstdio>

#include<cstring>

#include<iostream>

using namespace std;

const int MAX = 520000;

long long a[MAX];

int main() 

{

   int n;

   int x;

   scanf("%d",&n);

   a[0] = 0;

      for(int i = 1; i <= n ; i++){

        scanf("%d",&x);

        a[i] = a[i-1] + x;

    }

long long  flag ,flag1;

flag = flag1 = 0;

if(a[n]%3 == 0){

long long k = a[n]/3;

long long m = a[n]/3 * 2;

for(int i = 1; i < n; i++){

    if(a[i] == m)

    flag1 += flag;

    if(a[i] == k)

    flag++;

   }

}

cout << flag1;

return 0;

}

View Code