HDU-1049-Climbing Worm(C++ && 編程初學者的題......)

標籤：acm   c++   hdu   最水的題   這樣水真的好嗎   

Climbing Worm Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14196    Accepted Submission(s): 9560


Problem DescriptionAn 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.
 
InputThere 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.
 
OutputEach 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

 
SourceEast Central North America 2002 
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沒什麼好解釋的,題目意思很明確，講的是一個小蟲子想要爬出水井的故事！（我也得儘快爬出水井......）

#include<iostream>#include<cstdio>using namespace std;int main(){    int n,u,d;    while(scanf("%d%d%d",&n,&u,&d)!=EOF)    {        if(n==0)            break;        int s=0,t=0;        while(s<n)        {            s+=u;            t++;            if(s>=n)            {                break;            }            s-=d;            t++;        }        printf("%d\n",t);    }    return 0;}

 

HDU-1049-Climbing Worm(C++ && 編程初學者的題......)