NYOJ 156 Hangover

Hangover

时间限制：1000 ms  |  内存限制：65535 KB

难度：1

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.

样例输入

1.00

3.71

0.04

5.19

0.00

样例输出

3 card(s)

61 card(s)

1 card(s)

273 card(s)

来源

POJ

上传者

iphxer

/***类型：入门级题目**题目来源：NYOJ  156**时间：2017/7/22**解决方案：*/#include<stdio.h>int main(){double c;while(scanf("%lf",&c)!=EOF&&c!=0.00)   //%lf {double sum = 0.0;int count = 0;if(c <= 0.5){printf("1 card(s)\n");}else{for(int i = 2; ; i++){sum = sum + 1.0/i;count ++;if(sum >= c){printf("%d card(s)\n",count);break;}}}} return 0;} //再次看一下%f  %lf  以及思考最大值5.20，可以表示最大数为？