ASCII值可以由N的数字组成的字母计数

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📜 ASCII值可以由N的数字组成的字母计数

📅 最后修改于: 2021-04-26 07:17:08 🧑 作者: Mango

给定整数N。您可以从该数字中选择任意两个数字(数字可以相同，但位置应不同)，并以两种可能的方式中的任何一种对其进行排序。对于上述每种方式，您都可以从中创建一个两位数的数字(可能包含前导零)。然后，将选择对应于ASCII值的字符等于该数字，即数字65个对应于“A”，66〜“B”等等。任务是计算以这种方式可以选择的英文字母(小写或大写)的数量。

例子：

Input: N = 656
Output: 2
Only the characters ‘A’ (65) and ‘B’ (66) are possible.

Input: N = 1623455078
Output: 27

为什么编程需要懂一点英语

方法：想法是观察到可能的字符总数为(26个小写字母+ 26个大写字母= 52)。因此，请检查这52个字符的出现，而不是从N生成两位数字的所有可能组合。

因此，计算N中每个数字的出现，然后针对每个字符(小写或大写)，找到其ASCII值，并检查是否可以从给定的数字中选取它。最后打印此类字母的计数。

下面是上述方法的实现：

C++

// C++ implementation of the approach 
#include
  
using namespace std;
  
// Function that returns true if num 
// can be formed with the digits in
// digits[] array 
bool canBePicked(int digits[], int num) 
{ 
  
    int copyDigits[10];
  
    // Copy of the digits array 
    for(int i =0 ; i < 10;i++)
        copyDigits[i]=digits[i]; 
          
    while (num > 0)
    {
        // Get last digit 
        int digit = num % 10; 
  
        // If digit array doesn't contain
        // current digit 
        if (copyDigits[digit] == 0)
            return false; 
  
        // One occurrence is used 
        else
            copyDigits[digit] -= 1; 
  
        // Remove the last digit 
        num = floor(num / 10); 
    }
  
    return true; 
}
  
// Function to return the count of
// required alphabets 
int countAlphabets(long n)
{
  
    int count = 0; 
  
    // To store the occurrences of 
    // digits (0 - 9) 
    int digits[10]= {0}; 
    while (n > 0)
    {
  
        // Get last digit 
        int digit = n % 10; 
  
        // Update the occurrence of the digit 
        digits[digit] += 1; 
  
        // Remove the last digit 
        n = floor(n / 10); 
    }
  
    // If any lowercase character can be 
    // picked from the current digits 
    for(int i = 97; i <= 122 ;i ++)
        if (canBePicked(digits, i)) 
            count += 1; 
  
    // If any uppercase character can be 
    // picked from the current digits 
    for(int i = 65; i < 91;i++) 
        if (canBePicked(digits, i)) 
            count += 1; 
  
    // Return the required count 
    // of alphabets 
    return count; 
}
  
// Driver code 
int main()
{
  
    long n = 1623455078; 
    cout<<(countAlphabets(n)); 
}
  
// This code is contributed by chitranayal

Java

// Java implementation of the approach
class GFG {
  
    // Function that returns true if num can be formed
    // with the digits in digits[] array
    static boolean canBePicked(int digits[], int num)
    {
        // Copy of the digits array
        int copyDigits[] = digits.clone();
        while (num > 0) {
  
            // Get last digit
            int digit = num % 10;
  
            // If digit array doesn't contain current digit
            if (copyDigits[digit] == 0)
                return false;
  
            // One occurrence is used
            else
                copyDigits[digit]--;
  
            // Remove the last digit
            num /= 10;
        }
  
        return true;
    }
  
    // Function to return the count of required alphabets
    static int countAlphabets(int n)
    {
        int i, count = 0;
  
        // To store the occurrences of digits (0 - 9)
        int digits[] = new int[10];
        while (n > 0) {
  
            // Get last digit
            int digit = n % 10;
  
            // Update the occurrence of the digit
            digits[digit]++;
  
            // Remove the last digit
            n /= 10;
        }
  
        // If any lowercase character can be picked 
        // from the current digits
        for (i = 'a'; i <= 'z'; i++)
            if (canBePicked(digits, i))
                count++;
  
        // If any uppercase character can be picked 
        // from the current digits
        for (i = 'A'; i <= 'Z'; i++)
            if (canBePicked(digits, i))
                count++;
  
        // Return the required count of alphabets
        return count;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int n = 1623455078;
        System.out.println(countAlphabets(n));
    }
}

Python3

# Python3 implementation of the approach 
import math
  
# Function that returns true if num 
# can be formed with the digits in
# digits[] array 
def canBePicked(digits, num): 
  
    copyDigits = [];
      
    # Copy of the digits array 
    for i in range(len(digits)):
        copyDigits.append(digits[i]); 
          
    while (num > 0): 
  
        # Get last digit 
        digit = num % 10; 
  
        # If digit array doesn't contain
        # current digit 
        if (copyDigits[digit] == 0): 
            return False; 
  
        # One occurrence is used 
        else:
            copyDigits[digit] -= 1; 
  
        # Remove the last digit 
        num = math.floor(num / 10); 
  
    return True; 
  
# Function to return the count of
# required alphabets 
def countAlphabets(n): 
  
    count = 0; 
  
    # To store the occurrences of 
    # digits (0 - 9) 
    digits = [0] * 10; 
    while (n > 0): 
  
        # Get last digit 
        digit = n % 10; 
  
        # Update the occurrence of the digit 
        digits[digit] += 1; 
  
        # Remove the last digit 
        n = math.floor(n / 10); 
  
    # If any lowercase character can be 
    # picked from the current digits 
    for i in range(ord('a'), ord('z') + 1): 
        if (canBePicked(digits, i)): 
            count += 1; 
  
    # If any uppercase character can be 
    # picked from the current digits 
    for i in range(ord('A'), ord('Z') + 1): 
        if (canBePicked(digits, i)): 
            count += 1; 
  
    # Return the required count 
    # of alphabets 
    return count; 
  
# Driver code 
n = 1623455078; 
print(countAlphabets(n)); 
  
# This code is contributed by mits

C#

// C# implementation of the approach
using System;
  
class GFG 
{
  
    // Function that returns true
    // if num can be formed with
    // the digits in digits[] array
    static bool canBePicked(int []digits, int num)
    {
        // Copy of the digits array
        int []copyDigits = (int[]) digits.Clone();
        while (num > 0)
        {
  
            // Get last digit
            int digit = num % 10;
  
            // If digit array doesn't 
            // contain current digit
            if (copyDigits[digit] == 0)
                return false;
  
            // One occurrence is used
            else
                copyDigits[digit]--;
  
            // Remove the last digit
            num /= 10;
        }
  
        return true;
    }
  
    // Function to return the count 
    // of required alphabets
    static int countAlphabets(int n)
    {
        int i, count = 0;
  
        // To store the occurrences 
        // of digits (0 - 9)
        int[] digits = new int[10];
        while (n > 0)
        {
  
            // Get last digit
            int digit = n % 10;
  
            // Update the occurrence of the digit
            digits[digit]++;
  
            // Remove the last digit
            n /= 10;
        }
  
        // If any lowercase character can be  
        // picked from the current digits
        for (i = 'a'; i <= 'z'; i++)
            if (canBePicked(digits, i))
                count++;
  
        // If any uppercase character can be  
        // picked from the current digits
        for (i = 'A'; i <= 'Z'; i++)
            if (canBePicked(digits, i))
                count++;
  
        // Return the required count of alphabets
        return count;
    }
  
    // Driver code
    public static void Main()
    {
        int n = 1623455078;
        Console.WriteLine(countAlphabets(n));
    }
}
  
// This code is contributed by 
// tufan_gupta2000

PHP

 0) 
    { 
  
        // Get last digit 
        $digit = $num % 10; 
  
        // If digit array doesn't contain
        // current digit 
        if ($copyDigits[$digit] == 0) 
            return false; 
  
        // One occurrence is used 
        else
            $copyDigits[$digit]--; 
  
        // Remove the last digit 
        $num = floor($num / 10); 
    } 
  
    return true; 
} 
  
// Function to return the count of
// required alphabets 
function countAlphabets($n) 
{ 
    $count = 0; 
  
    // To store the occurrences of 
    // digits (0 - 9) 
    $digits = array_fill(0, 10, 0); 
    while ($n > 0) 
    { 
  
        // Get last digit 
        $digit = $n % 10; 
  
        // Update the occurrence of the digit 
        $digits[$digit]++; 
  
        // Remove the last digit 
        $n = floor($n / 10); 
    } 
  
    // If any lowercase character can be 
    // picked from the current digits 
    for ($i = ord('a'); $i <= ord('z'); $i++) 
        if (canBePicked($digits, $i)) 
            $count++; 
  
    // If any uppercase character can be  
    // picked from the current digits 
    for ($i = ord('A'); 
         $i <= ord('Z'); $i++) 
        if (canBePicked($digits, $i)) 
            $count++; 
  
    // Return the required count 
    // of alphabets 
    return $count; 
} 
  
// Driver code 
$n = 1623455078; 
echo countAlphabets($n); 
  
// This code is contributed by Ryuga
?>

输出：