An inch worm is at the bottom of a well n inches deep. It has enough energy to climb u inches every minute, but then has to rest a minute before climbing again. During the rest, it slips down d inches. The process of climbing and resting then repeats. How long before the worm climbs out of the well? We'll always count a portion of a minute as a whole minute and if the worm just reaches the top of the well at the end of its climbing, we'll assume the worm makes it out.

 
 
Input
There will be multiple problem instances. Each line will contain 3 positive integers n, u and d. These give the values mentioned in the paragraph above. Furthermore, you may assume d < u and n < 100. A value of n = 0 indicates end of output.

 
 
Output
Each input instance should generate a single integer on a line, indicating the number of minutes it takes for the worm to climb out of the well.

 
 
Sample Input
10 2 1
20 3 1
0 0 0
 
Sample Output
17
19
问下高手噶。。那个虫子是每分钟爬u英寸，每次都要休息1分钟，那时候向掉下d英寸，一共是n英寸，为什么能算出17 19分钟的啊，，我想了下不是20分钟么

这条虫子本身长1英寸，你考虑了没有？

#include <iostream>
using namespace std;
int main()
{
    int x,y,z;
    while(1)
    {
        int r=0,s=0,t=0;
        cin>>x>>y>>z;
        if(x+y+z==0) break;
        r=y-z;
        s=x-y;
        if(s%r==0)
        {
        t=2*(s/r);
        t+=1;
        }
        else
        {
            t=2*(s/r+1)+1;
        }
        cout<<t<<endl;

    }
    return 0;   
}

额..没有考虑诶。。不过，本身长1英寸的吧。那和总路程除去1英寸一样么？

出现输出越界

不一样的，你画图模拟一下过程，找出症结再说写程序。

哦，，妳这个是对的，，不过不大懂诶，能给的C的么

哦。。谢了。。那我看看

int a,b,c;
scanf("%d%d%d",&a,&b,&c);
int sum=(a-2)/(b-c)*2+1;
printf("%d",sum);

也来凑个热闹

#include<stdio.h>
int main()
{
  int n, u, d;
  while(scanf("%d%d%d", &n, &u, &d), n)
    printf("%d\n", (n - d - 1) / (u - d) + 1);
  return 0;
}