Round Numbers(组合数学)

Round Numbers

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 10484 Accepted: 3831

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

Source

USACO 2006 November Silver

看了好久的题解才看懂了,求小于这个数的二进制中0大于1的个数

如100100010 首先可以求出二进制位数小于这个数的(其值一定比这个数小如10010000),当位数相同时可以分成几部分求如[100000000-100100000)[100100000-1001000100)求的值就是组合数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;

typedef long long LL;

const int MAX = 1e5+10;

int Bin[35];

int c[33][33];

void ComBinnations()//打表,组合数
{
for(int i=0;i<=32;i++)
{
for(int j=0;j<=i;j++)
{
if(!j||i==j)
{
c[i][j]=1;
}
else
{
c[i][j]=c[i-1][j]+c[i-1][j-1];
}

}
}
}

void GetBite(int n)//分解二进制
{
Bin[0]=0;
while(n)
{
Bin[++Bin[0]]=n%2;
n/=2;
}
}

int  cal(int n)
{
int sum=0;
GetBite(n);
for(int i=1;i<Bin[0];i++)//比这个数的位数小
{
for(int j=(i+1)/2;j<i;j++)
{
sum+=c[i-1][j];
}
}
int ans=0;
for(int i=Bin[0]-1;i>=1;i--)//相等的时候
{
if(Bin[i])
{
for(int j=(Bin[0]+1)/2-1-ans;j<i;j++)
{
sum+=c[i-1][j];
}
}
else
{
ans++;
}
}
return sum;
}

int main()
{
int n,m;
ComBinnations();
while(~scanf("%d %d",&n,&m))
{
printf("%d",cal(m+1)-cal(n));
}
return 0;
}