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=;
while(now >= k)
{
hav += now/k*k;
now = now/k + now%k;
}
hav += now;
printf("%d\n",hav);
}
return ;
}

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

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