hdu 2015 偶數求和

Total Submission(s): 36010    Accepted Submission(s): 15665

//之前自己想太複雜。。真的很簡單。。。#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;}//這有個借鑒的代碼

//  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;}