hdu 1056 HangOver

HangOver

Time Limit: ²⁰⁰⁰⁄₁₀₀₀ MS (Java/Others) Memory Limit: ⁶⁵⁵³⁶⁄₃₂₇₆₈ K (Java/Others) Total Submission(s): 7186 Accepted Submission(s): 2884

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)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
using namespace std;
int main(void){
  double c; int n;
#ifndef ONLINE_JUDGE
  freopen("1056.in", "r", stdin);
#endif
  while (~scanf("%lf", &c)){
    if (fabs(c) <= 1e-9) break;
    double sum = 0; n = 0;
    for (double i = 2;; ++i){
      sum += 1.0/(i); n++;
      if (sum >= c) break;
    }
    printf("%d card(s)\n", n);
  }
  return 0;
}

我晕……我说怎么总是ＷＡ，原来我输出的时候，写的是cards(s)……次嗷……