二、Climbing Worm
来源:互联网 发布:淘宝模特多少钱一天 编辑:程序博客网 时间:2024/06/08 13:27
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
#include<iostream> using namespace std; int main() {int s,s1,s2,s3;while(cin>>s>>s1>>s2&&s){s3=0; for(int i=1;;i++) { if(s1+s3>=s) {cout<<i<<endl;break;} i++; s3=s1-s2+s3; if(s3>=s) {cout<<i<<endl;break;} }} return 0;}
- 二、Climbing Worm
- Climbing Worm
- Climbing Worm
- Climbing Worm
- Climbing Worm
- Climbing Worm
- Climbing Worm
- Climbing Worm
- Climbing Worm
- Climbing Worm
- Climbing Worm
- Climbing Worm
- Climbing Worm
- Climbing Worm
- Climbing Worm
- 1049 Climbing Worm
- HDU 1049 Climbing Worm
- hdoj1049 Climbing Worm
- [cocos2dx]Cocos2d-x在win7下android环境搭建
- 百度富文本编辑器 UEditor 1.4.3 getContent会报错:Uncaught TypeError
- 判断字符串是否回文
- 数位DP导学模板
- SpringMVC实现文件上传
- 二、Climbing Worm
- 网站迁站工作步骤
- Nginx访问图片403错误
- linux的sort命令用法
- Spark DAG之SubmitStage
- windows编程实验四
- 420
- Spring read-only="true" 只读事务的一些概念
- nyoj 155求高精度幂