HangOver

Problem 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.

 

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.

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

 

Sample Output

 

Source

Mid-Central USA 2001

 

 

很简单的一道题吧。直接从1/2+...+1/(n+1)是否大于输入。

 

#include <stdio.h>int main(){    double sum,length ;    int n ;    while(scanf("%lf", &length) != EOF && length != 0)    {        n = 1 ;        sum = 0 ;        while(sum < length)        {            sum += 1.0/(n+1) ;                n++ ;        }        printf("%d card(s)\n", n-1) ;    }    return 0 ;}