HUD 1049 Climbing Worm

最新推荐文章于 2021-09-05 13:04:37 发布

原创最新推荐文章于 2021-09-05 13:04:37 发布 · 349 阅读

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本文探讨了一个经典的编程问题：一只虫子如何在爬升和滑落的过程中最终爬出一口特定深度的井。通过模拟虫子的爬升和休息周期，找到解决问题的方法并给出了解决方案的代码实现。

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

这道题我是找规律模拟出来的

#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{
    int n,u,d,i,j;
    while(~scanf("%d",&n),n)
    {
        scanf("%d%d",&u,&d);
        int x=u-d;
        int sum=0,s=0;
        while(1)
        {
            if(s<n-u)
                {
                    s+=x;
                    sum+=2;
                }
            if(s>=n-u&&s<n)
            {
                 s+=u;
                 sum++;
            }
           if(s>=n)
                break;
        }
        printf("%d\n",sum);
    }
    return 0;
}