[解题报告]HDU 1049 Climbing Worm
发布日期:2021-08-15 20:52:03 浏览次数:2 分类:技术文章

本文共 1518 字,大约阅读时间需要 5 分钟。

Climbing Worm

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

Total Submission(s): 9062    Accepted Submission(s): 5908

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
 

 

Source
 
 
 
 
水题,不要图方便把升降结合在一起,就照题意做
#include
int main(){ int d,u,n,x,time; while(scanf("%d %d %d",&n,&u,&d)!=EOF&&d
<100&&n>0) { x=0;time=0; while(1) { x+=u;time++; if(x>=n) break; x-=d;time++; } printf("%d\n",time); } return 0;}

 

转载于:https://www.cnblogs.com/TheLaughingMan/archive/2013/03/18/2966882.html

转载地址:https://blog.csdn.net/weixin_30779691/article/details/98105369 如侵犯您的版权,请留言回复原文章的地址,我们会给您删除此文章,给您带来不便请您谅解!

上一篇:Dell 1420N使用Kubuntu默认无线驱动后网络不稳定的解决方法
下一篇:Uva 10003,切木棍

发表评论

最新留言

很好
[***.229.124.182]2024年04月02日 04时35分53秒