题目：
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
解析：
1. 题目，有一只虫子从井里底像上爬,井深n,每分钟爬d,每次休息一分钟，休息时往下下降u.输入 n d u 输出虫子爬出井的时间
#include<stdio.h> int main() { int n,u,d,time1,time2,count; while(scanf("%d %d %d",&n,&u,&d),n+u+d) { count=0; time1=time2=0; while(count<=n) { count=count+u; time1+=1;//累加向上爬的时间 if(count>=n)break; count=count-d; time2+=1;//累加休息的时间 } printf("%d\n",time1+time2); } return 0; }
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