HDU ACM 1049 Climbing Worm

Climbing Worm

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8319    Accepted Submission(s): 5352

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

 

Source

East Central North America 2002

 

 

#include<stdio.h>
int main()
{
    int n, u, d;
    while(scanf("%d%d%d", &n, &u, &d) != EOF && n != 0)
          printf("%d\n", 2*((n-u)/(u-d))+1+2*((n-u)%(u-d) > 0? 1:0));
     return 0;
}

结题报告：

如果说这题提交前不敢确定它百分百正确，是否说是我对不住他呢？ 但遗憾的是我不知道EXe.Memory怎么样能够为0.