HangOver

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10608    Accepted Submission(s): 4521

Problem Description

How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.

 

The input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c will contain exactly three digits.

For each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples.

 

Sample Input

1.003.710.045.190.00

 

Sample Output

3 card(s)61 card(s)1 card(s)273 card(s)

 

#include<cstdio>#include<iostream>using namespace std;int main(){    double n;    double sum;    while(scanf("%lf",&n)&&n!=0.00)    {       // printf("%.2lf\n",n);        sum=0.00;        int i=2,j;        while(sum<n)        {            sum+=1.0/(i++);        }//注意这个测试数据0.04 1题目是最少！        j=i-2;        printf("%d card(s)\n",j);    }}