Hang this page electrically OJ 11 // 2015 outputs the average seek sequence.

Problem Description

Has a length of n (n <= 100) the number of columns, is defined as the number of columns 2 increments from the start of the even-ordered, asks you to obtain a mean value according to the number per m order, if the last number is less than m, places The actual number averaged. The average output programmed sequence.

Input

A plurality of input data sets, each representing one line, comprising two positive integers n and m, n and m are the meanings described above.

Output

For each set of input data, it outputs a mean value sequence, each output per line.

Sample Input

3 2
4 2

Sample Output

3 6
3 7

#include<stdio.h>
void main()
{
	int n,m,i,d,s,a;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		s=0;//偶数总和 
		a=0;//偶数 
		for(i=1;i<=n;i++)
		{
			a=i*2;
			s=s+a;
			if(i%m==0)//判读i是否到达m个，满足m个以后进行后面操作 
			{
				if(i==n)
				{
					printf("%d\n",s/m);
					break;
				}//如果i以及是最后一个数，那么就直接输出偶数总和除以m，并跳出来，注意换行符 
				printf("%d ",s/m);//否则就输出s/m，但不需要换行符 
				s=0;//将s清空，准备下一组m个偶数的求和 
			}
			else if(i==n)//如果m比n要大，这时候i就等于了n 
			{
				d=n%m;//求出余数 比如n=4，m=5；n%m=4，就相当于是m 
				printf("%d\n",s/d);
			}

			}
		}
	}