Climbing Worm（語言訓練題）
ime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23349 Accepted Submission(s): 15984
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
問題簡述：
有一扇高n英尺的牆，一條長1英尺的小蟲子在牆下以u英尺每秒的速度向上爬，每爬一秒就休息一秒，休息時小蟲子會下滑d秒。多次輸入n,u,d,計算要多少秒小蟲子才能爬上牆，結果一行輸出，當遇到n=0時結束輸出。
程式說明：
用while語句實現迴圈，中間插入if語句防止計算過度.
程式實現：
#include <iostream>
using namespace std;
int main()
{
int n, u, d, s =0, k =0;
while (cin >>