hdoj 1049 Climbing Worm 【水题】

题目大意：一只蜗牛在深为n的井中， 每分钟往上爬u， 但是之后要歇一分钟，在这歇的一分钟中里下降d， 问什么时候蜗牛能爬出来，n为0的时候也算是出来了

Climbing Worm

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12272    Accepted Submission(s): 8275

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 2 120 3 10 0 0

 

Sample Output

1719

水题直接上代码：

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