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「HD_ACM」Climbing Worm

2015-05-03 19:17 393 查看

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.

->一只虫子一不小心掉进一个n英寸深洞中。它有足够的能量来爬你每分钟英寸,但随后必须休息一分钟之前再次攀升。在休息期间,它滑落d英寸。攀爬的过程和休息然后重复。多久前的虫子爬出好吗?我们会永远数每分钟作为一个整体的一部分,如果虫子只是达到顶端的最后的攀爬,我们将假设蠕虫让出来。

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.

->将会有多个问题实例。每一行将包含三个正整数n,u和d。在上面的段落中提到这些给值。此外,你可以假设d < u和n < 100。n = 0表示值的输出。

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.

->每个输入实例应该生成一个整数,指示分钟的数量需要的虫子爬出来的。

题目分析

虫子一分钟向上爬，一分钟休息并向下滑落，虫子在爬上来后不下滑

代码分析

#include <stdio.h>
#include <stdlib.h>
int main()
{
int n,d,u;            //n:洞深 、d:向上爬的长度  、u:下滑的长度
int t;                   //所需要的天数
int s=0;             //实际爬行长度
while(scanf("%d %d %d" , &n ,&d ,&u) !=EOF && n !=0 && d != 0 &&  u != 0)
{
t=0;
s=0;
while(1)
{
s+=d;
t++;
if(s>=n) break;
s-=u;
t++;
}
printf("%d\n" , t);
}

return 0;
}

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