标签：hdu1056

HangOver

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.

1.00
3.71
0.04
5.19
0.00

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

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 300

double 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;
}

HDU1056 HangOver

标签：hdu1056

原文地址：http://blog.csdn.net/chang_mu/article/details/38845539