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

题意：

给一个长度为n的数组，将其分成连续的三段使三段的和相等，求有几种这样的组合

分析：

从头扫到尾，将所有的前缀和为（sum/3）的点统计起来，然后再从尾开始统计，找到统计所有后缀和为（sum/3）的节点 然后这种方案的数为

这个点之前所有前缀和为sum/3的个数

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=500010;
long long e[maxn];
int cnt[maxn];
int main(){
	int n;
	while(~scanf("%d",&n)){
		memset(cnt,0,sizeof(cnt));
		long long s=0,p=0;
		for(int i=1;i<=n;i++){
			scanf("%lld",&e[i]);
		    s+=e[i];
		}
		if(s%3){
			printf("0\n");
			continue;
		}
		s/=3;
		int num=0;
		for(int i=1;i<=n;i++){
			p+=e[i];
			if(p==s){
				cnt[num++]=i;
			}
		}
		p=0;
		long long ans=0;
		for(int i=n;i>=1;i--){
			p+=e[i];
			if(p==s){
				//返回第一个小于等于i-1的值 
				int pos=lower_bound(cnt,cnt+num,i-1)-cnt;
			//	printf("%d***%d\n",i-1,pos);
				ans+=pos;
			}
		}
		cout << ans << endl;
	}
	return 0;
}