Climbing Worm

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 19841 Accepted Submission(s): 13532

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

Sample Output

17

19

Source

East Central North America 2002

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Problem : 1049 ( Climbing Worm ) Judge Status : Accepted

RunId : 21270620 Language : G++ Author : hnustwanghe

Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;

int main(){
int u,n,d,time;
while(scanf("%d %d %d",&n,&u,&d)==3 && (n||u||d)){
time = 1;
while(n-u>0){
time+=2;
n=n-u+d;
}
printf("%d\n",time);
}
}

#include<iostream>

#include<cstring>

#include<cstdio>



using namespace std;

int main(){

    int u,n,d,time;

    while(scanf("%d %d %d",&n,&u,&d)==3 && (n||u||d)){

        time = 1;

        while(n-u>0){

            time+=2;

            n=n-u+d;

        }

        printf("%d\n",time);

    }

}

巴特西

思维体操： HDU1049Climbing Worm

Climbing Worm

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