Round Numbers （排列组合）

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 ofN 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 ： 1 ans = num（last） - num（first - 1）；

2 打表组合数

3 num（x）计算 <= x 的 round number 的个数

a ： 计算小于x二进制位数的个数

b ： 计算等于x二进制位数的个数

此时从高位向后找，除首位外，如果遇到1，先把他当做0，他后面位从全部是0遍历到

不满足 J>=0 || zero >= one

4 注意 因为是 first - 1； 当 first == 1 时是不对的。

#include 
#include 
#include 
using namespace std;

typedef long long ll;

ll C[40][40];

void cmn()
{
    for(int i = 0; i <= 35; i++)
    for(int j = 0; j <= i; j++)
    {
       if(j == 0 || i == j)  C[i][j] = 1;
       else
       C[i][j] = C[i-1][j] + C[i-1][j-1];
    }
}

ll num(ll x)
{
   if(x == 0) return 0;
   ll bin[32],ans = 0,k = 0;
   ll zero = 0,one = 0;
   while(x)
   {
       bin[k++] = x % 2;
       if(x % 2 == 1)  one ++;
       else  zero ++;
       x /= 2;
   }

   for(int i = k-1;i > 0; i--)
   {
       if(i % 2 == 0) ans += (1 << (i-1)) / 2; // 如果i是偶数除去首位必须是1，其余满足的占1/2
       else  
       ans += ((1 << (i-1)) - C[i-1][(i-1) / 2]) / 2;
   }

   if(zero >= one) ans ++; // 4 的问题就出在这

   zero = 0,one = 1;
   for(int i = k-2; i >= 0; i--)
   {
       if(bin[i] == 0)
           zero++;
       else
       {
           for(int j = i;j >= 0 && zero + j + 1>= one + i - j; j--)//
               ans += C[i][j];
           one ++;
       }
   }
   return ans;
}
int main()
{
  cmn();
  ll m,n;
  while(cin >> m >> n)
  {
     ll tmp = num(n) - num(m-1);
     cout << tmp << endl;
  }
  return 0;
}