POJ 3252 Round Numbers 數學題解

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      

總體來說，十分困難的一道數學counting problem。

利用組合數學去做這些題目總是需要非常費力去總結規律的。

也許是數學思維還需要多鍛煉吧。

具體是找出規律，按照二進制數位去數這樣的題目，當然是不能模拟的。

#include <iostream>
#include <stdio.h>
#include <vector>
#include <algorithm>
#include <functional>
#include <bitset>
using namespace std;

int cTbl[33][33];

int calCombinate(int up, int down)
{
	down = min(down, up - down);
	int ans = 1;
	for (int i = 1; i <= down; i++)
	{
		ans *= (up - i + 1);
		ans /= i;
	}
	return ans;
}

void genTbl()
{
	cTbl[0][0] = 1;
	for (int i = 1; i < 33; i++)
	{
		cTbl[i][0] = 1;
		for (int j = 1; j <= i; j++)
		{
			cTbl[i][j] = calCombinate(i, j);
		}
	}
}

int calZeros(int n)
{
	bitset<33> bn = n;
	int len = 32;
	while (!bn[len]) len--;

	int ans = 0;
	for (int i = 1; i < len; i++)
	{
		for (int j = (i+2)>>1; j <= i; j++)
			ans += cTbl[i][j];
	}
	int zeros = 0, half = (len+2) >> 1;
	for (int i = len-1; i >= 0; i--)
	{
		if (bn[i])//前面選擇好01了，改為為1，變為0，然後選擇餘下的0有多少個
		{
			for (int j = half-zeros-1; j <= i; j++)
			{
				if (j < 0) continue;
				ans += cTbl[i][j];
			}
		}
		else zeros++;
	}
	return ans;
}

int main()
{
	genTbl();
	int a, b;
	while (scanf("%d %d", &a, &b) != EOF)
	{
		printf("%d\n", calZeros(b+1) - calZeros(a));
	}
	return 0;
}