北大ACM 1003 Hangover

最新推荐文章于 2018-02-09 22:11:27 发布

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本文详细解析了如何使用一定数量的卡片在桌面上实现最大过挂长度，通过数学公式计算所需卡片数量，以达到指定长度的过挂效果。

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Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 91219		Accepted: 44197

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)


#include <iostream>
#define MAX 10
using namespace std;

float data[MAX];
int result[MAX];
int count = 0;

bool InputData();
bool GetCard();
bool Output();

int main()
{
	if(!InputData())
		cout << "数据录入错误" << endl;
	if(!GetCard())
		cout << "取得结果错误" << endl;
	if(!Output())
		cout << "输出结果错误" << endl;
	return 1;
}

bool Output()
{
	for(int i=0; i < count; i++)
	{
		cout <<result[i] << " card(s)" << endl;
	}
	return true;
}

bool GetCard()
{
	float sum = 0;
	bool flag = false;
	for(int i=0; i < count; i++)
	{
		for(float n=2; ; n++)
		{
			sum += 1/n;
			if(sum >= data[i])
			{
				result[i] = n-1;
				break;
			}
		}
		sum = 0;
		if(count-1 == i)
			flag = true;
	}
	return flag;
}

bool InputData()
{
	float temp = 0;
	bool flag = false;
	int i = 0;
	while(cin >> temp)
	{
		if(temp == 0.00)
		{
			flag =  true;
			break;
		}else{
			data[i] = temp;
			i++;
			count++;
		}
	}
	return flag;
}