hdoj1049

[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：

#include<stdio.h>
int main()
{
int n,u,d,T,i;
while(scanf("%d%d%d",&n,&u,&d)!=EOF)
{
T=0;
if(n==0&&u==0&&d==0)
break;
for(i=0;;i++)
{
if(n<=0)
{
printf("%d\n",T);
break;
}
else
{
n-=u;
T++;
if(n<=0)
{
printf("%d\n",T);
break;
}
n+=d;
T++;
}
}
}
return 0;
}

思路：简单的贪心题，蜗牛滑坡…………