两个十六进制数的模

给定两个十六进制数N和K ，任务是找到N模K。

例子：

Input: N = 3E8, K = 13 
Output: C 
Explanation: 
Decimal representation of N( = 3E8) is 1000 
Decimal representation of K( = 13) is 19 
Decimal representation of (N % K) = 1000 % 19 = 12 ( = C). 
Therefore, the required output is C.

Input: N = 2A3, K = 1A 
Output: 19

方法：请按照以下步骤解决问题：

• 将十六进制数N和K分别转换为等效的十进制数，例如X和Y。
• 将十进制数(X％Y)转换为其等效的十六进制数，例如res
• 最后，打印res的值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to calculate modulus of
// two Hexadecimal numbers
void hexaModK(string s, string k)
{
     
    // Store all possible
    // hexadecimal digits
    map mp;
     
    // Iterate over the range ['0', '9']
    for(char i = 1; i <= 9; i++)
    {
        mp[i + '0'] = i;
    }
     
    mp['A'] = 10;
    mp['B'] = 11;
    mp['C'] = 12;
    mp['D'] = 13;
    mp['E'] = 14;
    mp['F'] = 15;
     
    // Convert given string to long
    long m = stoi(k, 0, 16);
     
    // Base to get 16 power
    long base = 1;
     
    // Store N % K
    long ans = 0;
     
    // Iterate over the digits of N
    for(int i = s.length() - 1;
            i >= 0; i--)
    {
         
        // Stores i-th digit of N
        long n = mp[s[i]] % m;
         
        // Update ans
        ans = (ans + (base % m * n
                      % m) % m) % m;
         
        // Update base
        base = (base % m * 16 % m) % m;
    }
     
    // Print the answer converting
    // into hexadecimal
    stringstream ss;
    ss << hex << ans;
    string su = ss.str();
    transform(su.begin(), su.end(),
              su.begin(), ::toupper);
    cout << (su);
}
 
// Driver Code
int main()
{
     
    // Given string N and K
    string n = "3E8";
    string k = "13";
     
    // Function Call
    hexaModK(n, k);
    return 0;
}
 
// This code is contributed by sallagondaavinashreddy7

Java

// Java program to implement
// the above approach
 
import java.util.*;
public class Main {
 
    // Function to calculate modulus of
    // two Hexadecimal numbers
    static void hexaModK(String N, String k)
    {
        // Store all possible
        // hexadecimal digits
        HashMap map
            = new HashMap<>();
 
        // Iterate over the range ['0', '9']
        for (char i = '0'; i <= '9'; i++) {
            map.put(i, i - '0');
        }
        map.put('A', 10);
        map.put('B', 11);
        map.put('C', 12);
        map.put('D', 13);
        map.put('E', 14);
        map.put('F', 15);
 
        // Convert given string to long
        long m = Long.parseLong(k, 16);
 
        // Base to get 16 power
        long base = 1;
 
        // Store N % K
        long ans = 0;
 
        // Iterate over the digits of N
        for (int i = N.length() - 1;
            i >= 0; i--) {
 
            // Stores i-th digit of N
            long n
                = map.get(N.charAt(i)) % m;
 
            // Update ans
            ans = (ans + (base % m * n % m) % m) % m;
 
            // Update base
            base = (base % m * 16 % m) % m;
        }
 
        // Print the answer converting
        // into hexadecimal
        System.out.println(
            Long.toHexString(ans).toUpperCase());
    }
 
    // Driver Code
    public static void main(String args[])
    {
        // Given string N and K
        String n = "3E8";
        String k = "13";
 
        // Function Call
        hexaModK(n, k);
    }
}

Python3

# Python3 program to implement
# the above approach
 
# Function to calculate modulus of
# two Hexadecimal numbers
def hexaModK(s, k) :
     
    # Store all possible
    # hexadecimal digits
    mp = {};
     
    # Iterate over the range ['0', '9']
    for i in range(1, 10) :
     
        mp[chr(i + ord('0'))] = i;
         
    mp['A'] = 10;
    mp['B'] = 11;
    mp['C'] = 12;
    mp['D'] = 13;
    mp['E'] = 14;
    mp['F'] = 15;
     
    # Convert given string to long
    m = int(k);
     
    # Base to get 16 power
    base = 1;
     
    # Store N % K
    ans = 0;
     
    # Iterate over the digits of N
    for i in range(len(s) - 1, -1, -1) :
         
        # Stores i-th digit of N
        n = mp[s[i]] % m;
         
        # Update ans
        ans = (ans + (base % m * n
                      % m) % m) % m;
         
        # Update base
        base = (base % m * 16 % m) % m;
     
    # Print the answer converting
    # into hexadecimal
    ans = hex(int(ans))[-1].upper()
     
    print(ans)
 
# Driver Code
if __name__ == "__main__" :
     
    # Given string N and K
    n = "3E8";
    k = "13";
     
    # Function Call
    hexaModK(n, k);
 
    # This code is contributed by AnkThon

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
  
class GFG{
 
// Function to calculate modulus of
// two Hexadecimal numbers
static void hexaModK(String N, String k)
{
     
    // Store all possible
    // hexadecimal digits
    Dictionary map = new Dictionary();
 
    // Iterate over the range ['0', '9']
    for(char i = '0'; i <= '9'; i++)
    {
        map.Add(i, i - '0');
    }
    map.Add('A', 10);
    map.Add('B', 11);
    map.Add('C', 12);
    map.Add('D', 13);
    map.Add('E', 14);
    map.Add('F', 15);
 
    // Convert given string to long
    long m = long.Parse(k);
 
    // Base to get 16 power
    long Base = 1;
 
    // Store N % K
    long ans = 0;
 
    // Iterate over the digits of N
    for(int i = N.Length - 1; i >= 0; i--)
    {
         
        // Stores i-th digit of N
        long n = map[N[i]] % m;
 
        // Update ans
        ans = (ans + (Base % m * n % m) % m) % m;
 
        // Update base
        Base = (Base % m * 16 % m) % m;
    }
 
    // Print the answer converting
    // into hexadecimal
    Console.WriteLine(ans.ToString("X"));
}
 
// Driver Code
public static void Main(String []args)
{
     
    // Given string N and K
    String n = "3E8";
    String k = "13";
 
    // Function Call
    hexaModK(n, k);
}
}
 
// This code is contributed by Princi Singh

输出：

C

时间复杂度： O(N)
辅助空间： O(1)