HDU 1056HangOver

HangOver

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14230 Accepted Submission(s): 6253

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
1.00
3.71
0.04
5.19
0.00

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

---

思路

水题—题意：就是求从1/2+1/3+1/4+…….+1/(1+n) 加多少次大于或等于你输入的数字！

---

代码

#include <iostream>#include <stdio.h>using namespace std;int main(){    double n;    while(cin>>n)    {        if(n==0)            break;        double m=2.0;        double sum=0.0;        int count=0;        for(int i=0;;i++)        {            sum+=1.0/m;            m++;            count++;            if(sum>=n)                break;        }        cout<<count<<" "<<"card(s)"<<endl;    }    return 0;}