HDOJ 1049-Climbing Worm【数学】

悠悠 2022-08-19 01:42 117阅读 0赞

# Climbing Worm #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 15450 Accepted Submission(s): 10488**

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

Source

[East Central North America 2002][]

解题思路

有一个小虫在一个深为N的井里,他每一分钟就会向上跳u英尺，跳完一下就会休息一分钟，这一分钟就会向下滑d英尺，问青蛙需要多长时间才能跳出井。

#include<stdio.h>
    #include<string.h>
    using namespace std;
    int main()
    {
    	int n,u,d;
    	while(scanf("%d%d%d",&n,&u,&d)&&(n||u||d))
    	{
    		int t;
    		t=0;
    		int wie=0;
    		while(1)
    		{
    			if(wie>=n)
    			break;
    			wie+=u;
    			t++;
    			if(wie>=n)
    			break;
    			wie-=d;
    			t++; 
    		}
    		printf("%d\n",t);
    	}
    	return 0;
    }

[East Central North America 2002]: http://acm.hdu.edu.cn/search.php?field=problem&key=East+Central+North+America+2002&source=1&searchmode=source