poj算法-使用C语言在VC++6.0实现poj1003

本文主要是介绍poj算法-使用C语言在VC++6.0实现poj1003，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

算法题：
poj1003:

Hangover

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.
例如，输入：
1.00
3.71
0.04
5.19
0.00
输出：
3 card(s)
61 card(s)
1 card(s)
273 card(s)

在桌子上悬放卡片问题，可以转化为求数学问题：输入浮点数c，其中（c大于等于0.01，小于等于5.20）,求1/2+1/3+1/4+…..+1/(n+1)>=c的最小值的问题。首先，输入浮点数c，判断其值范围是否合题意；其次，通过比较含n的数列累加和以及c的大小来判断循环是否结束。输出相应的结果。

#include <stdio.h>int result(double c){double i=2.0;double sum=0.0;while(1){sum=sum+1/i;if(sum>=c) break;i=i+1.0;}return (int)(i-1);
}int main(){double c;scanf("%lf",&c);while(c>0.01&&c<=5.20){printf("%d card(s)\n",result(c));scanf("%lf",&c);}
}