acm第二期训练题2

Climbing Worm

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23456 Accepted Submission(s): 16050

Problem De ion
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

问题链接

题目理解
普通的蜗牛爬井题，注意不要加上最后一次下滑的时间就ok。

#include<iostream>
using namespace std;
int main()
{
	int n, u, d, t = 0,s = 0 ;//time&S
	while (cin >> n >> u >> d)
	{
		
		if (n == 0 )break;
		while (1)
		{
			s += u;
			t++;
			if (s >= n)
			{
				cout << t;
				break;
			}
			s -= d;
			t++;
		}
		t = 0; 
		s = 0;
		cout << endl;
	}

}