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http://codeforces.com/contest/466/problem/C

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_ ≤ 5·10 5 ) , showing how many
numbers are in the array. The second line contains _n_ integers _a_ [1] ,
_a_ [2] , …, _a_ [ _n_ ] (| _a_ [ _i_ ]| ≤  10 9 ) — 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







#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string.h>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<sstream>
#include<time.h>
#include<utility>
#include<malloc.h>

using namespace std;

long long sum[500005];

int main()
{
    sum[0]=0;
    int n;
    while(cin>>n)
    {
        memset(sum, 0, sizeof(sum));
        for(int i=1;i<=n;++i)
        {
            cin>>sum[i];
            sum[i]+=sum[i-1];
        }
        long long aver;
        if(sum[n]==0)
        {
            long long k = 0;
            for (int i = 1; i <= n; ++i)
            {
                if (sum [i] == 0)
                {
                    k++;
                }
            }
            cout <<(k-1)*(k-2)/2 <<endl;
            continue;
        }
        if(sum[n]%3==0 )
        {
            aver=sum[n]/3;
        }
        else
        {
            cout<<0<<endl;
            continue;
        }

        long long pc=0, ss=0;

        for(int i=1;i<=n;++i)
        {
            if (sum[i] == aver)
                pc++;
            if (sum[i] == aver *2)
                ss += pc;
        }
            cout<<ss*pc<<endl;
    }
    return 0;
}

赛后懂了。。。