Hangover

POJ - 1003 Hangover   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 1.00 3.71 0.04 5.19 0.00 Sample Output 3 card(s) 61 card(s) 1 card(s) 273 card(s)

 

方法-:



#include<stdio.h>

int main(){

    float x;

    while(scanf("%f",&x)==1){

        int i;

        float t=0.0;

        if(x==0.00) break;

        for(i=1;t<x;i++)  t+=1.0/(float)(i+1);                                     

        printf("%d card(s)\n",i-1);

    }

}







方法二:





#include<stdio.h>

int main(){

    int n;

    float b;

    for(;n=scanf("%f",&b),b;printf("%d card(s)\n",n-1))

        for(;b>0;b-=1./++n);

}