【HDU】 1049

Climbing Worm

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20468    Accepted Submission(s): 14008

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

 

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题解：深度n，每秒上升u，之后休息一秒下降d，问多少秒>=n;

AC代码：

#include<cstdio>#include<iostream>using namespace std;int main(){int n,u,d;while(scanf("%d %d %d",&n,&u,&d)!=EOF&&n){int t=(n-u)/(u-d);if(t*(u-d)<(n-u)){t++;}t*=2;t++;cout<<t<<endl;}return 0;}