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  • NUC1003 Hangover

    Hangover

    时间限制: 1000ms 内存限制: 65536KB

    问题描述

        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 are 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.

    hangover.jpg

    输入描述

    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.

    输出描述

    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.

    样例输入
    1.00
    3.71
    0.04
    5.19
    0.00
    样例输出
    3 card(s)
    61 card(s)
    1 card(s)
    273 card(s)
    来源
    {Mid-Central USA 2001}


    问题分析:

    这个题与《POJ1003 UVALive2294 HDU1056 ZOJ1045 Hangover》完全相同,代码直接拿过来用就AC了。

    程序说明:

    参见参考链接。

    参考链接:POJ1003 UVALive2294 HDU1056 ZOJ1045 Hangover

    题记:

    程序写多了,似曾相识的也就多了。


    AC的C++程序如下:

    /* POJ1003 UVALive2294 HDU1056 ZOJ1045 Hangover */  
      
    #include <iostream>  
    #include <cstdio>  
      
    using namespace std;  
      
    const double one = 1.0;  
      
    int main()  
    {  
        double len, sum, d;  
        int i;  
      
        while((cin >> len) && len != 0.00) {  
            i = 1;  
      
            d = 2.0;  
            sum = one / d;  
            while(sum < len) {  
                d += 1.0;  
                sum += (one / d);  
                i++;  
            }  
      
            cout << i << " card(s)" << endl;  
        }  
      
        return 0;  
    } 














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  • 原文地址:https://www.cnblogs.com/tigerisland/p/7563785.html
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