用n个置位和m个未置位找到最小的数字

给定两个非负数n和m 。问题是找到在其二进制表示中具有n个置位位数和m个未置位位数的最小数字。

约束： 1 <= n，0 <= m，(m + n)<= 31

注意：在二进制表示形式中，前1(或最左1)位之前的0位被计数

例子：

Input : n = 2, m = 2
Output : 9
(9)10 = (1001)2
We can see that in the binary representation of 9 
there are 2 set and 2 unsets bits and it is the
smallest number. 

Input : n = 4, m = 1
Output : 23

方法：以下是步骤：

1. 计算num =(1 <<(n + m))–1。这将产生一个具有(n + m)个位数的数字num ，并且全部被设置。
2. 现在，在num的n到(n + m-1)范围内切换位，即从最右边的第n位切换到最右边的(n + m-1)位，然后返回切换后的数字。请参阅这篇文章。

C/C++

// C++ implementation to find the smallest number
// with n set and m unset bits
#include 
  
using namespace std;
  
// function to toggle bits in the given range
unsigned int toggleBitsFromLToR(unsigned int n,
                                unsigned int l,
                                unsigned int r)
{
    // for invalid range
    if (r < l)
        return n;
  
    // calculating a number 'num' having 'r'
    // number of bits and bits in the range l
    // to r are the only set bits
    int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
  
    // toggle bits in the range l to r in 'n'
    // and return the number
    return (n ^ num);
}
  
// function to find the smallest number
// with n set and m unset bits
unsigned int smallNumWithNSetAndMUnsetBits(unsigned int n,
                                           unsigned int m)
{
    // calculating a number 'num' having '(n+m)' bits
    // and all are set
    unsigned int num = (1 << (n + m)) - 1;
  
    // required smallest number
    return toggleBitsFromLToR(num, n, n + m - 1);
}
  
// Driver program to test above
int main()
{
    unsigned int n = 2, m = 2;
    cout << smallNumWithNSetAndMUnsetBits(n, m);
    return 0;
}

Java

// Java implementation to find the smallest number
// with n set and m unset bits
  
class GFG 
{
    // Function to toggle bits in the given range
    static int toggleBitsFromLToR(int n, int l, int r)
    {
        // for invalid range
        if (r < l)
            return n;
   
        // calculating a number 'num' having 'r'
        // number of bits and bits in the range l
        // to r are the only set bits
        int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
   
        // toggle bits in the range l to r in 'n'
        // and return the number
        return (n ^ num);
    }
      
    // Function to find the smallest number
    // with n set and m unset bits
    static int smallNumWithNSetAndMUnsetBits(int n, int m)
    {
        // calculating a number 'num' having '(n+m)' bits
        // and all are set
        int num = (1 << (n + m)) - 1;
   
        // required smallest number
        return toggleBitsFromLToR(num, n, n + m - 1);
    }
      
    // driver program
    public static void main (String[] args) 
    {
        int n = 2, m = 2;
        System.out.println(smallNumWithNSetAndMUnsetBits(n, m));
    }
}
  
// Contributed by Pramod Kumar

Python3

# Python3 implementation to find
# the smallest number with n set
# and m unset bits
  
# function to toggle bits in the
# given range
def toggleBitsFromLToR(n, l, r):
  
    # for invalid range
    if (r < l):
        return n
   
    # calculating a number 'num'
    # having 'r' number of bits
    # and bits in the range l
    # to r are the only set bits
    num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1)
   
    # toggle bits in the range
    # l to r in 'n' and return the number
    return (n ^ num)
  
# function to find the smallest number
# with n set and m unset bits
def smallNumWithNSetAndMUnsetBits(n, m):
  
    # calculating a number 'num' having
    # '(n+m)' bits and all are set
    num = (1 << (n + m)) - 1
   
    # required smallest number
    return toggleBitsFromLToR(num, n, n + m - 1);
  
   
# Driver program to test above
n = 2
m = 2
  
ans = smallNumWithNSetAndMUnsetBits(n, m)
print (ans)
  
# This code is contributed by Saloni Gupta

C#

// C# implementation to find the smallest number
// with n set and m unset bits
using System;
  
class GFG
{ 
    // Function to toggle bits in the given range
    static int toggleBitsFromLToR(int n, int l, int r)
    {
        // for invalid range
        if (r < l)
            return n;
  
        // calculating a number 'num' having 'r'
        // number of bits and bits in the range l
        // to r are the only set bits
        int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
  
        // toggle bits in the range l to r in 'n'
        // and return the number
        return (n ^ num);
    }
      
    // Function to find the smallest number
    // with n set and m unset bits
    static int smallNumWithNSetAndMUnsetBits(int n, int m)
    {
        // calculating a number 'num' having '(n+m)' bits
        // and all are set
        int num = (1 << (n + m)) - 1;
  
        // required smallest number
        return toggleBitsFromLToR(num, n, n + m - 1);
    }
      
    // Driver program
    public static void Main () 
    {
        int n = 2, m = 2;
        Console.Write(smallNumWithNSetAndMUnsetBits(n, m));
    }
}
  
// This code is contributed by Sam007

输出：

9

对于较大的n和m值，可以使用long int和long long int数据类型来生成所需的数字。