Climbing Worm
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.
#include <iostream>
using namespace std;
int main()
{
int n;
cin>>n;
while (n)
{
int u,v,current=0,min=0;
cin>>u>>v;
while (current<n)
{
min++;
if (current+u>=n)
{
current+=u;
cout<<min<<endl;;
}
else
{
min++;
current+=u-v;
}
}
cin>>n;
}
return 0;
}
本文探讨了一个经典的井底爬虫问题,通过C++代码实现了解决方案。该问题涉及一只爬虫如何克服每分钟上升和休息时下滑的情况,最终爬出深度为n英寸的井。输入包含井深n、每分钟爬升高度u和休息时下滑距离d,输出为爬虫完全爬出井所需的总分钟数。
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