HDUACM2015

偶数求和

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 70404    Accepted Submission(s): 30026

Problem Description

有一个长度为n(n<=100)的数列，该数列定义为从2开始的递增有序偶数，现在要求你按照顺序每m个数求出一个平均值，如果最后不足m个，则以实际数量求平均值。编程输出该平均值序列。

 

Input

输入数据有多组，每组占一行，包含两个正整数n和m，n和m的含义如上所述。

 

Output

对于每组输入数据，输出一个平均值序列，每组输出占一行。

 

Sample Input

  

   3 2
4 2
  

 

Sample Output

  

   3 6
3 7
  

 

Author

lcy

 

Source

C语言程序设计练习（三）

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  2016 2017 2018 2019 2021 

#include<stdio.h>
int main(){
	int n,m;
	
	while(scanf("%d%d",&n,&m) != EOF&&n<=100){
		int re = n/m;
		int la = n - re*m;
		int num = 2;
		for(int j = 1;j <= re;j++){
			int sum,ave ;
			sum = (num + (m-1)*2+num)*m/2;
			ave = sum/m;	
			num = m*2 + num;
			if(j == 1){
				printf("%d",ave);
			}else{
				printf(" %d",ave);
			}
		}
		if(la!=0){
			int lave;
			lave = (num + (la-1)*2+num)*la/2;
			printf(" %d\n",lave);
		}else{
				printf("\n");
		}
		
	}
	
	return 0;
}