Hangover(poj 1003)
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1 问题描述
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
0.00Sample Output
3 card(s)
61 card(s)
1 card(s)
273 card(s)大致意思
若将一叠卡片放在一张桌子的边缘,你能放多远?如果你有一张卡片,你最远能达到卡片长度的一半。(我们假定卡片都正放在桌 子上。)如果你有两张卡片,你能使最上的一张卡片覆盖下面那张的1/2,底下的那张可以伸出桌面1/3的长度,即最远能达到 1/2 + 1/3 = 5/6 的卡片长度。一般地,如果你有n张卡片,你可以伸出 1/2 + 1/3 + 1/4 + … + 1/(n + 1) 的卡片长度,也就是最上的一张卡片覆盖第二张1/2,第二张超出第三张1/3,第三张超出第四张1/4,依此类推,最底的一张卡片超出桌面1/(n + 1)。现在给定伸出长度C(0.00至5.20之间),输出至少需要多少张卡片。
2 算法思路
非常简单的题目,第i次的长度为
3 算法实现
#include <iostream>#include <cstdio>using namespace std;int main(){ double val; while(cin >> val) { // double无法直接判断相等,可以乘100转成int判断是否是0 if((int)(val*100) == 0) return 0; int i; double len = 0.00; for(i = 1; true; ++ i) { len += 1.0/(i+1); // 大于输入的值就结束 if(len > val) { printf("%d card(s)\n", i); break; } } } return 0;}