hdu-1049-Climbing Worm

http://acm.hdu.edu.cn/showproblem.php?pid=1049

Climbing Worm

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20187    Accepted Submission(s): 13805

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 
题意：一条虫子每分钟往上爬u英寸，爬一分钟休息一分钟，休息过程中下降d英寸，问多久可以爬出高度为n英寸的井。
如果最后一段距离不需要花一分钟，按一分钟算。
简单题啊，直接上代码= =
#include<iostream>#include<stdio.h>using namespace std;#define maxn 1000005int n;int main(){int n,u,d,ans;while(cin>>n>>u>>d){if(n==0){break;}int tmpu=u-d;int tmpn=n;tmpn=n-u;if(tmpn%tmpu){ans=2*(tmpn/tmpu+1)+1;}else {ans=2*(tmpn/tmpu)+1;}cout<<ans<<endl;}}