ZOJ 1494 Climbing Worm

Climbing Worm

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

题意：一只1英尺长的虫子爬一口n英尺长的井，每分钟爬u英尺，然后休息1分钟，在休息期间掉下d英尺，爬到井口要用多久。

注意：注意最后是一步d到达的，小心不要把退回来到n算进去

代码：

#include <stdio.h>


int main()
{
    int n,d,u;
    while(scanf("%d%d%d",&n,&d,&u))
    {
        if(n==0&&d==0&&u==0)
            break;
        int i;
        for(i=0;;i++)
        {
            if(i*(d-u)+d>=n)
                break;
        }
        i=2*i+1;
        printf("%d\n",i);
    }
    return 0;
}