Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 103896 | Accepted: 50542 |
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.
![](http://poj.org/images/1003/hangover.jpg)
Input
Output
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)
题意:求sum(1/n)=给定数的n
思路:因为n只有可能300种,预处理比较即可,精度0.01,不需要设置eps
#include <iostream> double tc; double sum[300]; using namespace std; int main(){ ios::sync_with_stdio(false); double s=0; for(int i=0;i<300;i++){ s+=1.00/(i+2); sum[i]=s; } while(cin>>tc&&tc){ int i=0; for(int i=0;i<300;i++){ if(sum[i]>=tc){ cout<<i+1<<" card(s)"<<endl; break; } } if(i==300)cout<<"ERROR"<<endl; } return 0; }