poj2509---抽k根烟就换一支，求能抽烟的总数

#include <stdio.h>
#include <stdlib.h>int main()
{int now,k;while(scanf("%d %d",&now,&k) != EOF){int hav=0;while(now >= k){hav +=   now/k*k;now = now/k + now%k;}hav += now;printf("%d\n",hav);}return 0;
}

算法思路：目前有几支烟就放进now中，把now看成一个整体，now对k取余和取商，商即为k支烟为大小的整体有多少个，一次就have掉now/k*k

now=now/k+now%k