HDU 1049 Climbing Worm（水题）

Climbing Worm

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7047    Accepted Submission(s): 4455

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

 

 

 

水题。

总长为n，上升一秒走u,休息一秒下降d.

相当于每两秒走（u-d)；

先n-u,得到过了多少个u-d后超过n-u；

int t=(n-u)/(u-d);
        if(t*(u-d)<(n-u)) t++;
        t*=2;
        t++;

就是最后一秒可以一步到达~~~

代码：

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