POJ 1003

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

1.003.710.045.190.00

Sample Output

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

Source

Mid-Central USA 2001

#include <iostream>      //刚接触算法，想在博客上记录下自己的POJ历程（虽然刚开始刷）using namespace std;//224K 16MSconst int size = 300;//快速排序void Qsort(int a[], int low, int high){if (low >= high){return;}int first = low;int last = high;int key = a[first];/*用字表的第一个记录作为枢轴*/while (first < last){while (first < last && a[last] >= key){--last;}a[first] = a[last];/*将比第一个小的移到低端*/while (first < last && a[first] <= key){++first;}a[last] = a[first];/*将比第一个大的移到高端*/}a[first] = key;/*枢轴记录到位*/Qsort(a, low, first - 1);Qsort(a, first + 1, high);}int main(){float c[100];for (int i = 0; i < 100; i++){  std::cin >> c[i];  if (c[i] == 0.00)  break;}for (int i = 0; i < 100; i++){float n;float sum; int t[size];for (int j = 0; j < size; j++){t[j] = 300;}int s;for (n = 2.0,s=0, sum = 0.00; n<size; n++,s++){sum = sum + 1/n;float x = sum - c[i];if(x>0){t[s] = int(n-1);}}Qsort(t, 0, sizeof(t) / sizeof(t[0]) - 1);if (c[i] == 0.00){break;}if (c[i] != 0.00){cout << t[0] << " card(s)" << endl;}}return 0;}