HDU-1049 Climbing Worm

简介: HDU-1049 Climbing Worm

题目:

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

解析:

1. 题目,有一只虫子从井里底像上爬,井深n,每分钟爬d,每次休息一分钟,休息时往下下降u.输入 n d u 输出虫子爬出井的时间

#include<stdio.h>
int main()
{
    int n,u,d,time1,time2,count;
    while(scanf("%d %d %d",&n,&u,&d),n+u+d)
    {
        count=0;
        time1=time2=0;
        while(count<=n)
        {
            count=count+u;
            time1+=1;//累加向上爬的时间
            if(count>=n)break;
            count=count-d;
            time2+=1;//累加休息的时间
        }
        printf("%d\n",time1+time2);
    }
    return 0;
}

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