按字典顺序将数字打印为字符串“ A”和“ B”

给定数字N，任务是打印与该数字相对应的字符串“ A”和“ B”。

如果我们将所有数字表示为“ A”和“ B”的字符串，如下所示，

1 = A
2 = B
3 = AA
4 = AB
5 = BA
6 = BB
7 = AAA
8 = AAB
9 = ABA
10 = ABB
.....
.....
1000000 = BBBABAAAABAABAAAAAB

例子：

Input: N = 30
Output: BBBB

Input: N = 55
Output: BBAAA

Input: N = 100
Output: BAABAB

方法：

1. 这个表示有点与二进制表示有关。
2. 首先，我们必须找到打印对应于给定数量的字符串所需的字符数，这是得到的字符串的长度。
3. 长度为1的2个数字，长度为2的4个数字，长度为3的8个数字，依此类推…
4. 因此，“ A”和“ B”的k个长度字符串可以表示从(pow(2，k)-2)+1到pow(2，k + 1)-2的pow(2，k)个数字，即AA …A(k次)至BB…B(k次)。
5. 因此，为打印相应的字符串，首先，计算字符串的长度，使其为k。现在计算

   N = M - (pow(2, k)-2);
   

6. 现在运行以下循环以打印相应的字符串。

   while (k>0)
   {
       num = pow(2, k-1);
       
       if (num >= N)
           cout <<"A";
       else{
           cout <<"B";
           N -= num;
       }
       k--;
   }
   

下面是上述方法的实现：

C++

// C++ program to implement the above approach
  
#include 
#include 
using namespace std;
  
// Function to calculate number of characters
// in corresponding string of 'A' and 'B'
int no_of_characters(int M)
{
  
    // Since the minimum number
    // of characters will be 1
    int k = 1;
  
    // Calculating number of characters
    while (true) {
  
        // Since k length string can
        // represent at most pow(2, k+1)-2
        // that is if k = 4, it can
        // represent at most pow(2, 4+1)-2 = 30
        // so we have to calculate the
        // length of the corresponding string
        if (pow(2, k + 1) - 2 < M)
            k++;
        else
            break;
    }
  
    // return the length of
    // the corresponding string
    return k;
}
  
// Function to print corresponding
// string of 'A' and 'B'
void print_string(int M)
{
    int k, num, N;
  
    // Find length of string
    k = no_of_characters(M);
  
    // Since the first number that can be represented
    // by k length string will be (pow(2, k)-2)+1
    // and it will be AAA...A, k times,
    // therefore, N will store that
    // how much we have to print
    N = M - (pow(2, k) - 2);
  
    // At a particular time,
    // we have to decide whether
    // we have to print 'A' or 'B',
    // this can be check by calculating
    // the value of pow(2, k-1)
    while (k > 0) {
        num = pow(2, k - 1);
  
        if (num >= N)
            cout << "A";
        else {
            cout << "B";
            N -= num;
        }
        k--;
    }
  
    // Print new line
    cout << endl;
}
  
// Driver code
int main()
{
  
    int M;
  
    M = 30;
    print_string(M);
  
    M = 55;
    print_string(M);
  
    M = 100;
    print_string(M);
  
    return 0;
}

Java

// Java implementation of the approach 
import java.util.*;
  
class GFG
{
      
// Function to calculate number of characters 
// in corresponding string of 'A' and 'B' 
static int no_of_characters(int M) 
{ 
  
    // Since the minimum number 
    // of characters will be 1 
    int k = 1; 
  
    // Calculating number of characters 
    while (true)
    { 
  
        // Since k length string can 
        // represent at most pow(2, k+1)-2 
        // that is if k = 4, it can 
        // represent at most pow(2, 4+1)-2 = 30 
        // so we have to calculate the 
        // length of the corresponding string 
        if ((int)Math.pow(2, k + 1) - 2 < M) 
            k++; 
        else
            break; 
    } 
  
    // return the length of 
    // the corresponding string 
    return k; 
} 
  
// Function to print corresponding 
// string of 'A' and 'B' 
static void print_string(int M) 
{ 
    int k, num, N; 
  
    // Find length of string 
    k = no_of_characters(M); 
  
    // Since the first number that can be represented 
    // by k length string will be (pow(2, k)-2)+1 
    // and it will be AAA...A, k times, 
    // therefore, N will store that 
    // how much we have to print 
    N = M - ((int)Math.pow(2, k) - 2); 
  
    // At a particular time, 
    // we have to decide whether 
    // we have to print 'A' or 'B', 
    // this can be check by calculating 
    // the value of pow(2, k-1) 
    while (k > 0) 
    { 
        num = (int)Math.pow(2, k - 1); 
  
        if (num >= N) 
            System.out.print("A"); 
        else 
        { 
            System.out.print( "B"); 
            N -= num; 
        } 
        k--; 
    } 
  
    // Print new line 
    System.out.println(); 
} 
  
// Driver code 
public static void main(String args[])
{ 
  
    int M; 
  
    M = 30; 
    print_string(M); 
  
    M = 55; 
    print_string(M); 
  
    M = 100; 
    print_string(M); 
  
} 
}
  
// This code is contributed by Arnab Kundu

Python3

# Python 3 program to implement
# the above approach
from math import pow
  
# Function to calculate number of characters
# in corresponding string of 'A' and 'B'
def no_of_characters(M):
      
    # Since the minimum number
    # of characters will be 1
    k = 1
  
    # Calculating number of characters
    while (True):
          
        # Since k length string can
        # represent at most pow(2, k+1)-2
        # that is if k = 4, it can
        # represent at most pow(2, 4+1)-2 = 30
        # so we have to calculate the
        # length of the corresponding string
        if (pow(2, k + 1) - 2 < M):
            k += 1
        else:
            break
  
    # return the length of
    # the corresponding string
    return k
  
# Function to print corresponding
# string of 'A' and 'B'
def print_string(M):
      
    # Find length of string
    k = no_of_characters(M)
  
    # Since the first number that can be 
    # represented by k length string will 
    # be (pow(2, k)-2)+1 and it will be 
    # AAA...A, k times, therefore, N will 
    # store that how much we have to print
    N = M - (pow(2, k) - 2)
  
    # At a particular time,
    # we have to decide whether
    # we have to print 'A' or 'B',
    # this can be check by calculating
    # the value of pow(2, k - 1)
    while (k > 0):
        num = pow(2, k - 1)
  
        if (num >= N):
            print("A", end = "")
        else:
            print("B", end = "")
            N -= num
        k -= 1
          
    print("\n", end = "")
  
# Driver code
if __name__ == '__main__':
    M = 30;
    print_string(M)
  
    M = 55
    print_string(M)
  
    M = 100
    print_string(M)
      
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach 
using System;
  
class GFG 
{ 
      
    // Function to calculate number of characters 
    // in corresponding string of 'A' and 'B' 
    static int no_of_characters(int M) 
    { 
      
        // Since the minimum number 
        // of characters will be 1 
        int k = 1; 
      
        // Calculating number of characters 
        while (true) 
        { 
      
            // Since k length string can 
            // represent at most pow(2, k+1)-2 
            // that is if k = 4, it can 
            // represent at most pow(2, 4+1)-2 = 30 
            // so we have to calculate the 
            // length of the corresponding string 
            if ((int)Math.Pow(2, k + 1) - 2 < M) 
                k++; 
            else
                break; 
        } 
      
        // return the length of 
        // the corresponding string 
        return k; 
    } 
      
    // Function to print corresponding 
    // string of 'A' and 'B' 
    static void print_string(int M) 
    { 
        int k, num, N; 
      
        // Find length of string 
        k = no_of_characters(M); 
      
        // Since the first number that can be represented 
        // by k length string will be (pow(2, k)-2)+1 
        // and it will be AAA...A, k times, 
        // therefore, N will store that 
        // how much we have to print 
        N = M - ((int)Math.Pow(2, k) - 2); 
      
        // At a particular time, 
        // we have to decide whether 
        // we have to print 'A' or 'B', 
        // this can be check by calculating 
        // the value of pow(2, k-1) 
        while (k > 0) 
        { 
            num = (int)Math.Pow(2, k - 1); 
      
            if (num >= N) 
                Console.Write("A"); 
            else
            { 
                Console.Write( "B"); 
                N -= num; 
            } 
            k--; 
        } 
      
        // Print new line 
        Console.WriteLine(); 
    } 
      
    // Driver code 
    public static void Main() 
    { 
      
        int M; 
      
        M = 30; 
        print_string(M); 
      
        M = 55; 
        print_string(M); 
      
        M = 100; 
        print_string(M); 
      
    } 
} 
  
// This code is contributed by Ryuga

PHP

 0) 
    {
        $num = pow(2, $k - 1);
  
        if ($num >= $N)
            echo "A";
        else {
            echo "B";
            $N -= $num;
        }
        $k--;
    }
  
    // Print new line
    echo "\n";
}
  
// Driver code
$M;
  
$M = 30;
print_string($M);
  
$M = 55;
print_string($M);
  
$M = 100;
print_string($M);
  
// This code is contributed 
// by Akanksha Rai
?>

输出：

BBBB
BBAAA
BAABAB