H - Climbing Worm 문제 풀이 보고서(장우)

H - Climbing Worm
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
Submit 
Status 
Practice 
HDU 1049
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
        
        
        
        

 
제목: 고도 n, 분당 기어오르는 높이 u, 휴식 시 분당 내려가는 높이를 제시한다.먼저 1분만 올라가고 1분만 쉬세요. 이렇게 유추해서 높이 n까지 올라갈 때까지.제발 몇 분 걸렸어?n=0시에 끝납니다.또 초등학생 수학 문제야, 유목유!!!
[cpp]  view plain copy
#include  
using namespace std;  
int main()  
{  
    int n,u,d;//n은 올라갈 높이를 표시하고, u는 분당 올라갈 높이를 표시하며, d는 1분을 올라갈 때마다 쉬는 1분 안에 내려갈 높이를 나타낸다
    while(cin>>n>>u>>d)  
    {  
if(n==0)//n=0시에 끝납니다
            break;  
        int min;//기록에 걸린 시간
        int dis=0;//올라간 높이를 기록합니다
        for(min=1;;min++)  
        {  
//먼저 1분만 올라가고 1분만 더 쉬세요
            if(min%2==0)    
            {  
                dis-=d;  
            }  
            else  
            {  
                dis+=u;  
if (dis>=n)//높이 n에 도달하면 끝납니다
                    break;  
            }  
        }  
        cout<
    }  
    return 0;  
}