YTUOJ-偶数求和

Description

Input

Output

Sample Input

3 2
4 2

Sample Output

3 6
3 7

HINT

Source

代码如下：

#include <iostream>
using namespace std;
int main()
{
    int n,m;
    while (cin>>n>>m)
    {
        if (n==0&&m==0)
            break;
        int i,count = 0,t = 0,sum = 0;
        for (i=1; i<=n; i++)
        {
            count++;
            t += 2; 
            sum += t;
            if (i == n) 
            {
                cout<<sum/count;
            }
            else if (count == m)
            {
                cout<<sum/m<<" "; 
                count = 0;
                sum =0;
            }
        }
        cout<<endl;
    }
    return 0;
}

运行结果：

YTUOJ-偶数求和

原文：http://blog.csdn.net/liuchang54/article/details/45476515