A - Climbing Worm HDU - 1049

最新推荐文章于 2021-09-05 13:04:37 发布

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最新推荐文章于 2021-09-05 13:04:37 发布

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source by : weijiew.com

本文链接：https://blog.youkuaiyun.com/weixin_43924623/article/details/99708513

探讨一种算法挑战，涉及一只能量有限的虫子如何从一口深井中爬出。虫子每分钟能向上爬升一定距离，然后必须休息并会下滑。任务是计算虫子完全爬出井口所需的总时间。

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Climbing Worm

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24984 Accepted Submission(s): 17063

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

总结：直接模拟就好，和之前写的蜗牛爬金字塔类似，不过那个用找规律取余做的。

#include<iostream>
using namespace std;
int main(){
	int n,u,d;
	while(cin>>n>>u>>d&&n!=0&&u!=0&&d!=0){
		int sum=0;
		int k=0;
		while(1){
			sum+=u;
			k++;
			if(sum>=n){
				break;
			}else{
				sum-=d;
				k++;
			}
		}
		cout<<k<<endl;
	}
	return 0;
}