poj 1003:Hangover（水题，数学模拟）

Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 99450		Accepted: 48213

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.

Input

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.

Output

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

1.00

3.71

0.04

5.19

0.00

Sample Output

3 card(s)

61 card(s)

1 card(s)

273 card(s)

Source

Mid-Central USA 2001

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　　水题，模拟。

　　题意：求最小的n让1+1/2+1/3+...+1/n大于给的一个实数。

　　思路：找一个数sum记录结果，写一个循环不断累加 sum+=1.0/n，直到sum大于c，输出n-2.

　　代码：

 1 #include <iostream>

 2 

 3 using namespace std;  4 

 5 int main()  6 {  7     double c,sum;  8     int n,card;  9     while(cin>>c){ 10         if(int(c*100)==0) break; 11         sum = 0; 12         n=2; 13         while(sum<=c){  //循环模拟。直到超过

14             sum+=1.0/n; 15             n++; 16  } 17         cout<<n-2<<" card(s)"<<endl; 18  } 19     return 0; 20 }

 

Freecode : www.cnblogs.com/yym2013