B - Number of Ways

B - Number of Ways

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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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

小结：

        ....题目看清楚才发现这是一道彻彻底底的水题，原来以为要用DP做的...

以下是AC代码：

#include <iostream>
#include <cstdio>
typedef long long ll;
using namespace std;
const int maxn = 500005;
ll sum[maxn];
int main()
{
	int n;
	scanf("%d", &n);
	sum[0] = 0;
	for (int i = 0; i < n; i++)
	{
		int x;
		scanf("%d", &x);
		sum[i+1] = sum[i] + x;
	}

	ll ans = 0;
	if (sum[n] % 3 == 0)
	{
		ll one = sum[n] / 3, two = sum[n] / 3 * 2;
		ll cnt = 0;
		for (int i = 1; i < n; i++)
		{
			if (sum[i] == two)
				ans += cnt;
			if (sum[i] == one)
				cnt++;
		}
	}
	cout << ans << endl;
	return 0;
}