hdu1049 Climbing Worm（C语言）

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 2 120 3 10 0 0

 

Sample Output

1719

 

Source

East Central North America 2002

C语言AC代码

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

思路：长度为n，每分钟走u，休息一分钟下溜d，求走完所需的时间，求的是到达n的时间，到达n时的那次下溜就不用管，所一（n-d-1）就是除去下溜要走的路，u-d就是除去下溜能走的距离，两倍即为所用时间，最后减一是到达重点下溜的时间。