HDU1056 HangOver

HangOver

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9412    Accepted Submission(s): 3971

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

1.003.710.045.190.00

 

Sample Output

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

 

Source

Mid-Central USA 2001 

import java.util.Scanner;public class Main{static final int maxn = 300;static double[] len = new double[maxn];public static void main(String[] args){Scanner cin = new Scanner(System.in);double a = 0.0;for(int i = 1; i < maxn; ++i){a += 1.0 / (i + 1.0);len[i] = a;}while(true){a = cin.nextDouble();if(a == 0.00) break;System.out.println(bSearch(a) + " card(s)");}}public static int bSearch(double a){int l = 1, r = maxn - 1, mid;while(l <= r){mid = (l + r) >> 1;if(len[mid] < a) l = mid + 1;else if(len[mid] > a) r = mid - 1;else return mid;}return l;}}

#include <stdio.h>#include <string.h>#define maxn 300double len[maxn];int bSearch(double a){    int l = 1, r = maxn - 1, mid;    while(l <= r){        mid = (l + r) >> 1;        if(len[mid] < a) l = mid + 1;        else if(len[mid] > a) r = mid - 1;        else return mid;    }    return l;}int main(){    double a = 0.0;    int i;    for(i = 1; i < maxn; ++i){        a += 1.0 / (i + 1.0);        len[i] = a;    }    while(scanf("%lf", &a), a != 0.00){        printf("%d card(s)\n", bSearch(a));    }    return 0;}