杭电1302 The Snail

本文探讨了一道经典的编程题目——蜗牛爬井问题。通过分析蜗牛每天爬升和夜间下滑的距离及其疲劳因素,使用C语言实现了一个通用解决方案来判断蜗牛能否成功爬出井口及所需天数。

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看到英语题是不是很头大,不要这样啦,其实这方面的英文题好简单的(我是说理解起来)首先要克服心理阴影,然后攻破它,算什么,对不对,你一定可以的大笑

The Snail

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 2 Accepted Submission(s) : 1
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Problem Description

A snail is at the bottom of a 6-foot well and wants to climb to the top. The snail can climb 3 feet while the sun is up, but slides down 1 foot at night while sleeping. The snail has a fatigue factor of 10%, which means that on each successive day the snail climbs 10% * 3 = 0.3 feet less than it did the previous day. (The distance lost to fatigue is always 10% of the first day's climbing distance.) On what day does the snail leave the well, i.e., what is the first day during which the snail's height exceeds 6 feet? (A day consists of a period of sunlight followed by a period of darkness.) As you can see from the following table, the snail leaves the well during the third day.

Day Initial Height Distance Climbed Height After Climbing Height After Sliding
1 0 3 3 2
2 2 2.7 4.7 3.7
3 3.7 2.4 6.1 -

Your job is to solve this problem in general. Depending on the parameters of the problem, the snail will eventually either leave the well or slide back to the bottom of the well. (In other words, the snail's height will exceed the height of the well or become negative.) You must find out which happens first and on what day.

Input

The input file contains one or more test cases, each on a line by itself. Each line contains four integers H, U, D, and F, separated by a single space. If H = 0 it signals the end of the input; otherwise, all four numbers will be between 1 and 100, inclusive. H is the height of the well in feet, U is the distance in feet that the snail can climb during the day, D is the distance in feet that the snail slides down during the night, and F is the fatigue factor expressed as a percentage. The snail never climbs a negative distance. If the fatigue factor drops the snail's climbing distance below zero, the snail does not climb at all that day. Regardless of how far the snail climbed, it always slides D feet at night.

Output

For each test case, output a line indicating whether the snail succeeded (left the well) or failed (slid back to the bottom) and on what day. Format the output exactly as shown in the example.

Sample Input

6 3 1 10
10 2 1 50
50 5 3 14
50 6 4 1
50 6 3 1
1 1 1 1
0 0 0 0

Sample Output

success on day 3
failure on day 4
failure on day 7
failure on day 68
success on day 20
failure on day 2


大概意思

一只蜗牛在井底,现在它要爬出井去,已知井的总高:H、初始速度:U、下滑距离:D、速度下滑的概率:F。当下滑高度小于0,就失败,大于井的高度就成功.

#include<stdio.h>
int main()
{
	float h,u,d,f,l,p;
	int day;
	scanf("%f%f%f%f",&h,&u,&d,&f);
	while(h)
	{
		day=1;
		l=u;
		p=u;
		while(l>=0&&l<=h)
		{
			l=l-d;
			if(l<0)
				break;
			//p=p-u*f/100.0;
			if(p>=0)
				p=p;
			else
				p=0;
			l=l+p;
			day++;
		}
		if(l<0)
			printf("failure ");
		else
			printf("success ");
		printf("on day %d\n",day);
		scanf("%f%f%f%f",&h,&u,&d,&f);
	}
	return 0;
}

希望大家留言评论,谢谢
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