现在的位置: 首页 > 算法 > 正文

poj3252-Round Number 组合数学

2017年05月14日 算法 ⁄ 共 3785字 ⁄ 字号 评论关闭

题目:

Round Numbers
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 8492   Accepted: 2963

Description

The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors', 'Ro, Sham, Bo', and a host of other names) in order to make arbitrary decisions such as who gets to be milked first.
They can't even flip a coin because it's so hard to toss using hooves.

They have thus resorted to "round number" matching. The first cow picks an integer less than two billion. The second cow does the same. If the numbers are both "round numbers", the first cow wins,
otherwise the second cow wins.

A positive integer N is said to be a "round number" if the binary representation of N has as many or more zeroes than it has ones. For example, the integer 9, when written in binary form, is 1001. 1001 has two zeroes and two ones; thus,
9 is a round number. The integer 26 is 11010 in binary; since it has two zeroes and three ones, it is not a round number.

Obviously, it takes cows a while to convert numbers to binary, so the winner takes a while to determine. Bessie wants to cheat and thinks she can do that if she knows how many "round numbers" are in a given range.

Help her by writing a program that tells how many round numbers appear in the inclusive range given by the input (1 ≤ Start < Finish ≤ 2,000,000,000).

Input

Line 1: Two space-separated integers, respectively Start and Finish.

Output

Line 1: A single integer that is the count of round numbers in the inclusive range Start..Finish

Sample Input

2 12

Sample Output

6

思路基本上和网上http://zhyu.me/acm/poj-3252.html做的很类似:

举例说明,

[2,12]区间的RoundNumbers(简称RN)个数:Rn[2,12]=Rn[0,12]-Rn[0,1]
即:Rn[start,finish]=Rn[0,finish]-Rn[0,start-1]
所以关键是给定一个X,求出Rn[0,X]
现在假设X=10100100 
这个X的二进制总共是8位,任何一个小于8位的二进制都小于X
第一部分,求出长度为[0,7]区间内的二进制是RoundNumber的个数
对于一个长度为Len的二进制(最高位为1),如何求出他的RoundNumbers呢(假设为用R(len)来表达),分为奇数和偶数两种情况
1、奇数情况:在Len=2k+1的情况下,最高位为1,剩下2k位,至少需要k+1为0
用C(m,n)表示排列组合数:从m个位置选出n个位置的方法
R(len)=C(2k,k+1)+C(2k,k+2)+...+C(2k,2k).
由于 A:C(2k,0)+C(2k,1)+...+C(2k,2k)=2^(2k)
B:C(2k,0)=C(2k,2k), C(2k,1)=C(2k,2k-1) ,,C(2k,i)=C(2k,2k-i)
于是  C(2k,0)+C(2k,1)+...+C(2k,2k)
= C(2k,0)+C(2k,1)+...+C(2k,k)+C(2k,k+1)+C(2k,K+2)+...+C(2k,2k)
= 2*R(len)+C(2k,k)
=2^(2k)
所以R(len)=1/2*{2^(2k)-C(2k,k)};
2. 偶数情况 len=2*k,类似可以推到 R(len)=1/2*(2^(2k-1));
第二部分,对于上面这个长度为8的例子:即X=10100100,首先如果本身是RoundNumbers,第二部分的结果总数+1
第一部分已经将长度小于8的部分求出。现在要求长度=8的RoundNumber数目
长度为8,所以第一个1不可改变
现在到第二个1,如果Y是前缀如100*****的二进制,这个前缀下,后面取0和1必然小于X,已经有2个0,一个1,剩下的5个数字中至少需要2个0,
所以把第二个1改为0:可以有C(5,2)+C(5,3)+C(5,4)+C(5,5)
现在第三个1,也就是前最为101000**,同样求出,至少需要0个0就可,所以有C(2,0)+C(2,1)+C(2,2)个RoundNumbers
。。。
将所有除了第一个1以外的1全部变为0,如上算出有多少个RoundNumbers,结果相加(由于前缀不一样,所以后面不管怎么组合都是唯一的)

将第一部分和第二部分的结果相加,就是最后的结果了。

唯一特别需要注意的是在计算组合数的时候很容易越界,虽然上面分析了计算结果在int范围内是没有问题的,但是计算组合数中间过程还是很可能越界,所以这里要特别注意。

解决方法是利用C(n,m)=C(n-1,m-1)+C(n-1,m)进行递归计算,而不是使用传统的乘法计算方式。为了更有效率一点,可以事先计算好n=1~32,m=1~32的组合数的结果然后存起来。

import java.util.*;

public class Combinatorics_RoundNumbers3252 {

	/**
	 * @param args
	 */
	public static void main(String[] args) {

		Scanner in=new Scanner(System.in);
		Init();
		while(in.hasNext())
		{
			int a=in.nextInt();
			int b=in.nextInt();
			//System.out.println(roundNumber(a-1)+" " +roundNumber(b));
			System.out.println(roundNumber(b)-roundNumber(a-1));
		}
		
	}
	
   static int c[][]=new int[35][35];
   public static void Init(){
	    for(int i=0;i<33;i++){
	        c[i][0]=c[i][i]=1;
	        for(int j=1;j<i;j++)
	            c[i][j]=c[i-1][j]+c[i-1][j-1];
	    }
	   
	}
	
	public static int roundNumber(int value)
	{
		char b[]=toBinary(value);
		int sum=0;
		for(int len=1;len<b.length;len++)
		{
			for(int j=(len+1)/2;j<len;j++)
			sum+=c[len-1][j];
		}
		int zeros=0;
		for(int i=1;i<b.length;i++)
		{
			if(b[i]=='1')
			{
				int k=(b.length+1)/2;
				int m=Math.max(0, k-(zeros+1));
				int n=b.length-i-1;
				for(int j=n;j>=m;j--)
					sum+=c[n][j];
			}
			else
			{
				zeros++;
			}
		}
		if(2*zeros>=b.length)
			sum++;
		return sum;
	}
	
	private static char[] toBinary(int value) {
		return Integer.toBinaryString(value).toCharArray();
	}

	/*public static int roundNumberOfLength(int len)
	{
		int k=len/2;
		if(len%2==0)
		{
			return (1<<(len-2));
		}
		else
		{
			return ((1<<(len-1))-choose(len-1,k))/2;
		}
	}*/

	
	public static int choose(int n, int m) {

		if(n==0)
			return 0;
		if(m==0||m==n)
			return 1;
		if(m>n)
			return 0;
		return choose(n-1,m-1)+choose(n-1,m);
	}
	

}

抱歉!评论已关闭.