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
問題鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=1049
問題簡述：輸入三個數(shù)池摧，分別是總路程狮杨、每分鐘前進路程乐严、每分鐘下滑距離答憔，
前進一分鐘后下滑一分鐘冠蒋，求要多少時間到達硼端。
問題分析：用循環(huán)判斷到達是為幾分鐘避消。
程序說明：先判斷是否能到達，再用for循環(huán)求幾分鐘時滿足條件缤言。
AC通過的C++程序如下：
include<iostream>
using namespace std;
int main()
{
int n, u, d,s=0;
while (cin >> n >> u >> d)
{
if (u == 0 || u < d)
{
break;
}
else {
int work = 0;
for(s = n / u;work <n;)
{
s++;
work = u (s/2) - (d((s/2) - 1));
}
}
s--;
cout << s << endl;
s = 0;
}
}
