POJ1003解题报告

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题目概览

标题

Hangover

要求

• Time Limit: 1000MS

• Memory Limit: 10000K

题目描述

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.

输入

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.

输出

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.

输入示例

1.00

3.71

0.04

5.19

0.00

输出示例

3 card(s)

61 card(s)

1 card(s)

273 card(s)

解题报告

分析

本题较简单，可以采用map+vector的方式提高效率，vector用于保存原始的数据顺序，map可以根据数据大小自动排序。

在计算的时候可以将每一步的计算结果与map的迭代器里的数比较，如果符合，则将取出map中的下一个数，这样能仅进行一遍计算就能得到全部结果。

最后输出的时候键的顺序采用vector里保存的顺序（即原始顺序），值则采用map中的值。

源码

#include<map>
#include<vector>
#include<string>
#include<iostream>
using namespace std;
int main() {
    string item;
    vector<double> input;
    map<double, int> sorted;
    double num;
    while (true) {
        cin >> item;
        if (item == "0.00") {
            break;
        }
        num = item[0] - '0' + (item[2] - '0') * 0.1 + (item[3] - '0') * 0.01;
        input.push_back(num);
        sorted.insert(make_pair(num, 0));
    }
    double result = 0;
    map<double, int>::iterator it = sorted.begin();
    for (int i = 2; it != sorted.end(); ++i) {
        result += 1.0 / i;
        while (result > it->first) {
            it->second = i - 1;
            ++it;
            if (it == sorted.end()) {
                break;
            }
        }
    }
    for (vector<double>::iterator itv = input.begin(); itv != input.end(); ++itv) {
        cout << sorted[*itv] << " card(s)" << endl;
    }
}