int main()
{
    int a,b,c,d,x,f;
    while(scanf("%d%d",&a,&b)!=-1)
    {
        f=0;
        x=a;
        if(a>b)
        {
            for(;;)
           {
               c=a/b;
               d=a%b;
               a=c+d;
               f+=c;
               if(a<b)
                {printf("%d\n",f+x);
                 break;}
           }
        }

        if(a<b)
            printf("%d\n",a);
        if(a==b)
            printf("%d\n",b+1);

    } return 0;
}

这个程序用来做什麽？

蜡烛问题   原题如下
ACMer the Programmer loves romance, so this year he decided to illuminate his room with candles.
ACMer has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. ACMer is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now ACMer wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
Input
There are more test cases. The single line contains two integers, a and b (1≤a≤1000; 2≤b≤1000).
Output
Print a single integer — the number of hours Vasily can light up the room for.
Sample Input
4 2
Sample Output
7

读了题了，还是没懂题目的意思？

大体意思是这样：
       输入第一个数代表现在有多少支蜡烛，每支蜡烛可燃烧一小时，输入的第二个数代表新蜡烛燃烧完后需要几支蜡烛的残余物可组成一个新的蜡烛，所组成的新蜡烛同样可燃烧一小时。需要输出的是这些蜡烛总共可燃烧多少个小时！

求大神帮忙  万分感谢

int candle(int n, int d){
if(n==d)  return (n+1);
else      return ((n/d)*d+candle(n-(n/d)*d+(n/d),d));
}
这是我写的一个递归函数，你可以参考。

谢谢  我现在想知道为什么出现这种错误情况

int _tmain(int argc, _TCHAR* argv[])
{
    int a,b,c,d,x,f;
    scanf("%d%d",&a,&b);
    if(a<1|| a>2000 || b<2 || b>2000)
        printf("请重新输入a(1-2000)，b(2-2000)：");
    f  = 0;
    x = a;
    if(a>b)
    {
        for(;;)
        {
            c=x/b;
            d=x%b;
            x=c+d;
            f+=c;
            if(x < b)break;
        }
        printf("%d\n",f+a);
    }

    else if(a<b)
        printf("%d\n",a);
    else
        printf("%d\n",b+1);
    return 0;
}
这样应该就可以了

楼上说的对 之前楼主那个版本的 三个if之间不存在非此即彼的关系 都要执行的