3252 Round Numbers [a,b]中二进制中0的个数大1的数的个数

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本文介绍了一种名为“RoundNumbers”的游戏策略，该游戏涉及二进制数中零和一的数量比较。文章详细解释了如何判断一个正整数是否为“RoundNumber”，并提供了一个程序示例来计算指定范围内这类数字的数量。

Round Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3629		Accepted: 1186

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

Source

USACO 2006 November Silver

#include<iostream>
#include<cstdio>
using namespace std;
__int64 C[40][40];
void init()//求组合数
{
    for(int i=0;i<40;i++)
    {
        C[i][0]=1;
        for(int j=1;j<=i;j++)
        {
            C[i][j]=C[i-1][j]+C[i-1][j-1];
        }
    }
}
int cal(int n)//[0,n) 中的 round number
{
    int len=0;
    for(int x=n;x;len++,x>>=1);//求n的二进制长度
    int sum=1;//0
    //[0,2^(len-1)) 中的 round number
    for(int i=2;i<len;i++)
    {
        for(int j=0;j<i-1&&((j+1)<<1)<=i;j++)
        {
            sum+=C[i-1][j];
        }
    }
    //[2^(len-1),n) 中的 round number
    int y=1;
    for(int i=len-2;i>=0;i--)
    {
        if(n&(1<<i))
        {
            for(int j=0;j<=i;j++)
            {
                if(((j+y)<<1)<=len) sum+=C[i][j];
                else break;
            }
            y++;
        }
    }
    return sum;
}
int main()
{
    init();
    int s,t;
    while(scanf("%d%d",&s,&t)==2)
    {
        int cnt=cal(t+1)-cal(s);
        printf("%d/n",cnt);
    }
    return 0;
}
// 1 6
http://hi.baidu.com/solofancy/blog/item/63a41763568b35d5e6113a4b.html