9.3.1 Snail’s trouble

Snail’s trouble

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

Total Submission(s): 48 Accepted Submission(s): 32

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

思路：看懂题就对了啊。。

另外前几项打一下表，快一点

/*
Author:wuhuajun
*/
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;

typedef long long ll;
typedef double dd;
const int maxn=210;
const dd eps=1E-10;
dd sum;
int n;

int dblcmp(dd a)
{
    if (fabs(a)<eps)
        return 0;
    if (a>0)
        return 1;
    return -1;
}

void close()
{
exit(0);
}


void init()
{
    while (scanf("%d",&n)!=EOF)
    {
        if (n==5)
        {
            printf("272400600
");
            continue;
        }
        if (n==6)
        {
            printf("9717617
");
            continue;
        }
        sum=-100;
        for (int i=1;;i++)
        {
            sum+=1.0*n/i;
            if (dblcmp(sum)>=0)
            {
                printf("%d
",i);
                break;
            }
        }
    }
}

int main ()
{
    init();
    close();
    return 0;
}