poj 1003 Hangover

Hangover

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 104363 Accepted: 50832

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

1.003.710.045.190.00

Sample Output

3 card(s)61 card(s)1 card(s)
273 card(s)
题目大意：一块板放在桌子上能悬空一半的长度，两块板能悬空2/1+1/3，依次类推，给出一个长度，求最少要几块板思路：按照题目模拟就行了2014,12,5
#include<stdio.h>double a[500]={0};double f(int n){int i,j=2;double sum=0.0;for(i=0;i<n;i++,j++)sum+=1.0/j;return sum;}int main(){int i;double m;for(i=2;i<300;i++)a[i]=f(i);while(scanf("%lf",&m),m){if(m<0.5)printf("1 card(s)\n");else{for(i=2;;i++){if(a[i]>m){printf("%d card(s)\n",i);break;}}}}return 0;}