HDU-1049 Climbing Worm 当前位置: Home » algorithm » 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. <!--more--> ### 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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