hdu 2015 偶数求和

偶数求和Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36010    Accepted Submission(s): 15665


Problem DescriptionInputOutputSample InputSample OutputAuthorSourceRecommend//之前自己想太复杂。。真的很简单。。。#include<stdio.h>int main(){ int a=2,n,m,i,j; while(scanf("%d%d",&n,&m)!=EOF) { a=2; j=0; for(i=0;i<n/m;i++) { if(j==0) { printf("%d",a+m-1); j++; a+=2*m; } else { printf(" %d",a+m-1); a+=2*m; j++; } } if(n%m==0) printf("\n"); else printf(" %d\n",a+n%m-1); } return 0;}//这有个借鉴的代码

//  http://www.cnblogs.com/jian1573/archive/2011/05/08/2040585.html#include <stdio.h>int main(void){    int i, n, m, b, c;    while (scanf("%d%d", &n, &m) != EOF)    {        b = 2;        c = 0;        for (i = 0 ; i < n / m ; i++)        {            printf(c++ ? " %d" : "%d", b + m - 1);            b += m * 2;        }        printf(n % m ? " %d\n" : "\n", b + n % m - 1);    }    return 0;}