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  • [POJ1003]Hangover

    [POJ1003]Hangover

    试题描述

    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'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)

    数据规模及约定

    见“输入

    题解

    先预处理一下 S[k] = 1/1 + 1/2 + 1/3 + ... + 1/k 的前缀和,因为输入最多是 5.20,看到样例非常良心,所以应该不会比 273 大多少,我就预处理了前 100 个。

    每个询问二分答案就好了。(或者直接扫一遍也能过)

    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cmath>
    #include <stack>
    #include <vector>
    #include <queue>
    #include <cstring>
    #include <string>
    #include <map>
    #include <set>
    using namespace std;
    
    const int BufferSize = 1 << 16;
    char buffer[BufferSize], *Head, *Tail;
    inline char Getchar() {
        if(Head == Tail) {
            int l = fread(buffer, 1, BufferSize, stdin);
            Tail = (Head = buffer) + l;
        }
        return *Head++;
    }
    int read() {
        int x = 0, f = 1; char c = getchar();
        while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
        while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
        return x * f;
    }
    
    #define maxn 1010
    double S[maxn];
    
    int main() {
    	for(int i = 1; i <= maxn - 10; i++) S[i] = S[i-1] + 1.0 / (double)(i + 1);
    	
    	while(1) {
    		double x;
    		scanf("%lf", &x);
    		if(x == 0.0) break;
    		int l = 1, r = 1000;
    		while(l < r) {
    			int mid = l + r >> 1;
    			if(S[mid] < x) l = mid + 1; else r = mid;
    		}
    		printf("%d card(s)
    ", l);
    	}
    	
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/xiao-ju-ruo-xjr/p/5808973.html
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