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)
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 134994   Accepted: 65614

题目大意:若将一叠卡片放在一张桌子的边缘,你能放多远?如果你有一张卡片,你最远能达到卡片长度的一半。(我们假定卡片都正放在桌 子上。)如果你有两张卡片,你能使最上的一张卡片覆盖下面那张的1/2,底下的那张可以伸出桌面1/3的长度,即最远能达到 1/2 + 1/3 = 5/6 的卡片长度。一般地,如果你有n张卡片,你可以伸出 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) 的卡片长度,也就是最上的一张卡片覆盖第二张1/2,第二张超出第三张1/3,第三张超出第四张1/4,依此类推,最底的一张卡片超出桌面1/(n + 1)。

现在给定伸出长度C(0.01至5.20之间),输出至少需要多少张卡片。

二分+离线

#include<bits/stdc++.h>
using namespace std;
const double delta= 1e-8;//设置精度
double len[1001];
int zero(double x)
{
	if(x>delta)
		return 1;
	else if(x<-delta)
		return -1;
	else
		return 0;
}
int tol=1;
void work()//离线打表
{   
	len[0]=0.0;
	for(;zero(len[tol-1]-5.20)<0;tol++)
	{
		len[tol]=len[tol-1]+1.0/(double)(tol+1);
	}
}
int main(int argc, char const *argv[])
{    
   //freopen("data.txt","r",stdin);
    double c;
    work();
    while(scanf("%lf",&c)!=EOF)
    {
    	if(0.00==c)break;
    	int l=0;
    	int r=tol-1;
    	int mid;
    	while(l<r)
    	{   
    		mid=(l+r)/2;
    		if(zero(len[mid]-c)<0)
    			l=mid+1;//因为是至少所以小于的时候左边界要加一
    		else 
    			r=mid;//大于等于的时候直接r等于mid
    	}
    	printf("%d card(s)
",r);
    	
    }	
 	return 0;
}


原文地址:https://www.cnblogs.com/kuroko-ghh/p/9363345.html