Snail’s trouble

Snail’s trouble

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 74 Accepted Submission(s): 51

Problem Description

Once upon a time, there was a poor snail. Every day, it tried very hard to crawl forward, while there was a keeper who’d like to maltreat this poor little snail. The snail was crawling on a one-meter rubber band at first, and it can move on k cm every minute. But after that, the keeper stretches the rubber band quickly, and it would be elongated one meter, during that the distances ratio between the snail and the two endpoints remain unchanged. In the next minute, little snail tried to keep moving forward again.
“Can I finally get to the endpoint?”
The snail often asked himself such a question because he was afraid he would never succeed. Now, we hope you can tell this poor snail when he would reach the endpoint.

 

Input

Every line of the input contains an integer number k, indicating that the snail moved forward k cm every minute. (5 <= k <= 100)

 

Output

Output an integer t, indicating that the snail doesn’t get to the endpoint until t-1 minutes later, while t minutes later, it finally succeed.

 

Sample Input

10
100

 

Sample Output

12367
1

 

Source

2009 Multi-University Training Contest 2 - Host by TJU

 

Recommend

gaojie

 

/*
题意：一个蜗牛初始的时候在一块长1米的橡胶上，速度为一分钟k厘米，在每分钟的末尾橡胶会很快的
    拉长一米，但是蜗牛在橡胶上的比例位置不变。蜗牛第一次到达末尾的时间是多少

初步思路：简单的模拟嘛，但是很好时间，所以就打表
*/
#include<bits/stdc++.h>
using namespace std;
int n;
int a[]={,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,};
int main(){
    // freopen("in.txt","r",stdin);
    // for(int n=5;n<=100;n++){
        // int t=0;
        // double sum=100;
        // double w=0;
        // while(true){
            // t++;
            // w+=n;//蜗牛走的
            // if(w>=sum){
                // cout<<t<<",";
                // break;
            // }
            // double tmp=sum+100;
            // w=tmp*w*1.0/sum;
            // sum=tmp;
        // }
    // }
    while(scanf("%d",&n)!=EOF){
        printf("%d\n",a[n-]);
    }
    return ;
}