Climbing Worm
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10022 Accepted Submission(s): 6603
Problem Description
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
题目的大概意思：就是井深n米，每分钟向上爬u米；每爬一分钟就休息一分钟，休息的时候向下滑落d米；输出：一共需要多少分钟爬出井外。
#include<stdio.h>
int main()
{
int a,b,c,i,t,k;
while(scanf("%d%d%d",&a,&b,&c)&&(a!=0&&b!=0&&c!=0))
{
i=1;k=b;t=0;
while(1)
{
if(k>=a)
{printf("%d ",i);break;}
k=k-c+b;
i=i+2;//要求的答案是总共所用的时间（向上爬的时间加上休息的时间 ） 所以是 i+2;而不是 i+1；
}
}
return 0;
}