检查给定的解码字符串是否可被6整除

给定一个由小写字符组成的字符串str ，任务是根据给定的规则更改该字符串后，检查该字符串是否可被6整除：

• ‘a’更改为1 。
• ‘b’更改为2 。
• …
• ‘z’更改为26 。

例如，字符串“ abz”将变为1226 。

例子：

Input: str = “ab”
Output: Yes
“ab” is equivalent to 12 which is divisible by 6.

Input: str = “abc”
Output: No
123 is not divisible by 6.

方法：可以通过一个简单的数学技巧来解决：仅当数字的所有位数之和可被3整除且该数字的最后一位被2整除时，该数字才能被6整除。查找所形成数字的数字总和，并将其存储在可变总和中。另外，找到数字的最后一位并将其存储在lastDigit中。
现在，如果总和可被3整除，而lastDigit可被2整除，则打印“是”，否则打印“否” 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the sum
// of the digits of n
int sumDigits(int n)
{
    int sum = 0;
    while (n > 0) {
        int digit = n % 10;
        sum += digit;
        n /= 10;
    }
    return sum;
}
  
// Function that return true if the
// decoded string is divisible by 6
bool isDivBySix(string str, int n)
{
    // To store the sum of the digits
    int sum = 0;
  
    // For each character, get the
    // sum of the digits
    for (int i = 0; i < n; i++) {
        sum += (int)(str[i] - 'a' + 1);
    }
  
    // If the sum of digits is
    // not divisible by 3
    if (sum % 3 != 0)
        return false;
  
    // Get the last digit of
    // the number formed
    int lastDigit = ((int)(str[n - 1]
                           - 'a' + 1))
                    % 10;
  
    // If the last digit is
    // not divisible by 2
    if (lastDigit % 2 != 0)
        return false;
    return true;
}
  
// Driver code
int main()
{
    string str = "ab";
    int n = str.length();
  
    if (isDivBySix(str, n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
  
// Function to return the sum
// of the digits of n
static int sumDigits(int n)
{
    int sum = 0;
    while (n > 0) 
    {
        int digit = n % 10;
        sum += digit;
        n /= 10;
    }
    return sum;
}
  
// Function that return true if the
// decoded string is divisible by 6
static boolean isDivBySix(String str, int n)
{
    // To store the sum of the digits
    int sum = 0;
  
    // For each character, get the
    // sum of the digits
    for (int i = 0; i < n; i++) 
    {
        sum += (int)(str.charAt(i) - 'a' + 1);
    }
  
    // If the sum of digits is
    // not divisible by 3
    if (sum % 3 != 0)
        return false;
  
    // Get the last digit of
    // the number formed
    int lastDigit = ((int)(str.charAt(n - 1) - 
                                    'a' + 1)) % 10;
  
    // If the last digit is
    // not divisible by 2
    if (lastDigit % 2 != 0)
        return false;
    return true;
}
  
// Driver code
public static void main(String []args)
{
    String str = "ab";
    int n = str.length();
  
    if (isDivBySix(str, n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach 
  
# Function to return the sum 
# of the digits of n 
def sumDigits(n) : 
  
    sum = 0; 
    while (n > 0) :
        digit = n % 10; 
        sum += digit; 
        n //= 10; 
  
    return sum; 
  
# Function that return true if the 
# decoded string is divisible by 6 
def isDivBySix(string , n) : 
  
    # To store the sum of the digits 
    sum = 0; 
  
    # For each character, get the 
    # sum of the digits 
    for i in range(n) :
        sum += (ord(string[i]) - 
                ord('a') + 1); 
      
    # If the sum of digits is 
    # not divisible by 3 
    if (sum % 3 != 0) :
        return False; 
  
    # Get the last digit of 
    # the number formed 
    lastDigit = (ord(string[n - 1]) - 
                 ord('a') + 1) % 10; 
  
    # If the last digit is 
    # not divisible by 2 
    if (lastDigit % 2 != 0) :
        return False; 
    return True; 
  
# Driver code 
if __name__ == "__main__" : 
  
    string = "ab"; 
    n = len(string); 
  
    if (isDivBySix(string, n)) :
        print("Yes"); 
    else :
        print("No"); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to return the sum
// of the digits of n
static int sumDigits(int n)
{
    int sum = 0;
    while (n > 0) 
    {
        int digit = n % 10;
        sum += digit;
        n /= 10;
    }
    return sum;
}
  
// Function that return true if the
// decoded string is divisible by 6
static bool isDivBySix(String str, int n)
{
    // To store the sum of the digits
    int sum = 0;
  
    // For each character, get the
    // sum of the digits
    for (int i = 0; i < n; i++) 
    {
        sum += (int)(str[i] - 'a' + 1);
    }
  
    // If the sum of digits is
    // not divisible by 3
    if (sum % 3 != 0)
        return false;
  
    // Get the last digit of
    // the number formed
    int lastDigit = ((int)(str[n - 1] - 
                             'a' + 1)) % 10;
  
    // If the last digit is
    // not divisible by 2
    if (lastDigit % 2 != 0)
        return false;
    return true;
}
  
// Driver code
public static void Main(String []args)
{
    String str = "ab";
    int n = str.Length;
  
    if (isDivBySix(str, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
  
// This code is contributed by 29AjayKumar

输出：

Yes