hdu1056 HangOver

HangOver

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13654    Accepted Submission(s): 5966

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

1.00 3.71 0.04 5.19 0.00

 

Sample Output

3 card(s) 61 card(s) 1 card(s) 273 card(s)

 

水题

//
// Created by Admin on 2017/4/4.
//

#include <cstdio>

int main(){
    double a[1010],sum=0.0,temp;
    for (int i = 1; i < 300; ++i) {
        a[i]=1.0/(i+1)+sum;
        sum+=1.0/(i+1);
    }
    //for (int i = 1; i < 300; ++i) printf("%.2f\n",a[i]);
    while (scanf("%lf",&temp),temp){
        for (int i = 1; i < 300; ++i) {
            if(a[i]>=temp){
                printf("%d card(s)\n",i);
                break;
            }
        }
    }
    return 0;
}