poj算法-使用C语言在VC++6.0实现poj1003

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);
    }
}

代码可以在VC++6.0中正确运行，也可以在poj中正确运行。