FZU Problem 1002 HangOver【】

Problem 1002 HangOver

Accept: 2670    Submit: 11015
Time Limit: 1000 mSec    Memory Limit : 32768 KB

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.

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) 

#include<iostream>#include<cstdio>#include<math.h>#include<cstring>#include<climits>#include<string>#include<queue>#include<stack>#include<set>#include<map>#include<list>#include<vector>#include<sstream>#include<algorithm>using namespace std;#define rep(i,j,k)for(i=j;i<k;i++)#define per(i,j,k)for(i=j;i>k;i--)#define ms(x,y)memset(x,y,sizeof(x))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ll long longconst int INF=0x7ffffff;const int M=1e5+10;double i,j,k,m,q;char s[M];int a[M];int b[M];double n,sum;int main(){      while(~scanf("%lf",&n)){        if(n==0)break;       int ans=0;       sum=0;       for(i=1;;i++){         sum+=1.0/(i+1.0);         ans++;         if(sum>=n)break;       }       printf("%d card(s)\n",ans);    }    return 0;}