【CODEFORCES】 C. Number of Ways

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

51 2 3 0 3

output

2

input

40 1 -1 0

output

1

input

24 1

output

0

题解：此题并不算难，我们设lsum[i]是从1到i所有数组元素的和（前缀和），rsum[i]是从i到n所有数组元素的和，那么现在就是要找到一组i,j，i<j，j-i>=2，lsum[i]=lsum[j]=sum/3。看到题目数据较大，n方算法不划算，所以用一个c数组记录，c[i]表示i之后有多少个rsum[i]与sum/3相等，这样就把时间复杂度降为O（n）了。

感觉主要是前缀和...

#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;}