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.

Sample test(s) input

5
1 2 3 0 3

output

2

input

4
0 1 -1 0

output

1

input

2
4 1

output

0

題解：此題並不算難，我們設lsum[i]是從1到i所有陣列元素的和（字首和），rsum[i]是從i到n所有陣列元素的和，那麼現在就是要找到一組i,j，i<j，j-i>=2，lsum[i]=lsum[j]=sum/3。看到題目資料較大，n方演算法不划算，所以用一個c陣列記錄，c[i]表示i之後有多少個rsum[i]與sum/3相等，這樣就把時間複雜度降為O（n）了。

感覺主要是字首和...

#include <iostream>
#include <cstdio>

using namespace std;

int a[500003],c[500003],n;
long long lsum[500003],rsum[500003],sum,ans;
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        sum+=a[i];
        lsum[i]=a[i]+lsum[i-1];
    }
    if (sum%3!=0)
    {
        cout <<0<<endl;
        return 0;
    }
    for (int i=1;i<=n;i++) rsum[i]=sum-lsum[i]+a[i];
    for (int i=n;i>=1;i--) if (rsum[i]==sum/3) c[i]=c[i+1]+1;
                           else c[i]=c[i+1];
    for (int i=1;i<=n-1;i++) if (lsum[i]==sum/3) ans+=c[i+2];
    cout <<ans<<endl;
    return 0;
}