haoj Climbing Worm 【水题】

Climbing Worm

[b]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 14959 Accepted Submission(s): 10137

[/b]

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

10 2 1
20 3 1
0 0 0

[align=left]Sample Output[/align]

17
19代码：[code]#include<cstdio>
#include<cstring>
int aa,bb;
int main()
{
int h,w,c;
while(scanf("%d%d%d",&h,&w,&c)!=EOF)
{
if(h==0&&w==0&&c==0)
break;
int aa=1;
h-=w;
while(h>0)
{
h-=w;
aa++;
h+=c;
aa++;
}
printf("%d\n",aa);
}
return 0;
}

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