Codeforces Round #266 (Div. 2) C

标签：sum Codeforces long 266 output input Div include array

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 ≤ 5·105), 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

#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。。。

标签：sum,Codeforces,long,266,output,input,Div,include,array
From： https://blog.51cto.com/u_15990681/6098468