codeforces 466C

C. Number of Ways

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 integers a[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.

Examples

input

5
1 2 3 0 3

output

2

input

4
0 1 -1 0

output

1

input

2
4 1

output

0

题目的意思为输入一个数n，然后输入n个数，计算有几种方法让这n个数分为和相对的三份。
n的数字为十万级，需要定义全局变量才能运行，而且n需要定义为long long型，求解思路为：设定一个记录和的数组，然后将其中和的1/3和2/3的数据记录在另一个数组fenlie中，设置一个权值k=0，遇到1/3，k++，遇到2/3，sumn+=k。

#include<iostream>
#define N 500005
using namespace std;
long long sum[N];
long long number[N];
long long fenlie[N];
int main()
{
    long long i, n, k=0;
    long long sumn = 0;
    //int *sum = new int[N];
    //int *number = new int[N];
    //int *fenlie = new int[N];
    sum[0] = 0;
    cin >> n;
    for (i = 1;i <= n;i++)
    {
        cin >> number[i];
        sum[i] += sum[i - 1] + number[i];
    }
    if (sum[n] % 3 != 0)
        cout << '0' << endl;
    else
    {
        long long temp = sum[n] / 3;
        long long  j = 0;
        for (i = 1;i <= n;i++)
        {
            if (sum[i] == temp || sum[i] == 2 * temp)
            {
                fenlie[j] = sum[i];
                j++;
            }
        }
        if (temp == 0)
            sumn = (j - 1)*(j - 2) / 2;
        else
        {
            for (i = 0;i < j;i++)
            {
                if (fenlie[i] == temp)
                        k++;
                else if (fenlie[i] == 2 * temp)
                    sumn += k;
            }
            }
        cout << sumn << endl;
    }
    /*delete sum;
    delete fenlie;
    delete number;*/
    return 0;
}