HDU 1049 Climbing Worm

Climbing Worm

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19860    Accepted Submission(s): 13549

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
/***类型：贪心算法**题目来源：HDU  1049**时间：2017/7/29**解决方案：*/#include<stdio.h>int main(){int n,u,d;  //深度  上升  下降while(scanf("%d%d%d",&n,&u,&d)!=EOF){if(n==0&&u==0&&d==0){break;}int t = ((n-d-1)/(u-d))*2+1;// t = 上取整[（n-u）/（u-d）]*2+1          //而 X/Y的上取整=(X+Y-1)/Y,要求X与Y为正整数          printf("%d\n",t);  } return 0;}