POJ-1003 Hangover

Hangover

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 150249 Accepted: 72569

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

http://poj.org/problem?id=1003

解释

这题就是个递推式子
就是说3张卡片 所能总的最大伸出量为1/2+1/3+1/4
现在给出 伸出量 问最少需要多少卡片
递推式为
res[0]=0.0;
res[i] =res[i-1]+1.0/(i+1); //i从1开始到276
因为276的时候就大于5.20了，后面的就不需要了
还有浮点数的比较 是两数相减小于x，x为精确度
1e-10，就是10的-10次方

代码

这里先进行了预处理，后面直接用

#include<iostream>
using namespace std;
int main()
{
    double a;double res[277]={
    0.00};for(int i=1;i<277;i++){
    res[i] =res[i-1]+1.0/(i+1);}while(cin>>a){
    if(a < 1e-10)// 判断a是否为0 break;for(int i=1;i<277;i++){
    if(res[i]-a>1e-10)// 判断res[i]是大于0 {
    cout<< i<<" card(s)" <<endl;break;}}}return 0;
}

人一我百！人十我万！永不放弃~~~怀着自信的心，去追逐梦想。