Modulus of two Hexadecimal Numbers
Given two hexadecimal numbers N and K, the task is to find N modulo K.
Examples:
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
Approach: Follow the steps below to solve the problem:
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void hexaModK(string s, string k)
{
map< char , int > mp;
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;
long m = stoi(k, 0, 16);
long base = 1;
long ans = 0;
for ( int i = s.length() - 1;
i >= 0; i--)
{
long n = mp[s[i]] % m;
ans = (ans + (base % m * n
% m) % m) % m;
base = (base % m * 16 % m) % m;
}
stringstream ss;
ss << hex << ans;
string su = ss.str();
transform(su.begin(), su.end(),
su.begin(), :: toupper );
cout << (su);
}
int main()
{
string n = "3E8" ;
string k = "13" ;
hexaModK(n, k);
return 0;
}
|
Java
import java.util.*;
public class Main {
static void hexaModK(String N, String k)
{
HashMap<Character, Integer> map
= new HashMap<>();
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 );
long m = Long.parseLong(k, 16 );
long base = 1 ;
long ans = 0 ;
for ( int i = N.length() - 1 ;
i >= 0 ; i--) {
long n
= map.get(N.charAt(i)) % m;
ans = (ans + (base % m * n % m) % m) % m;
base = (base % m * 16 % m) % m;
}
System.out.println(
Long.toHexString(ans).toUpperCase());
}
public static void main(String args[])
{
String n = "3E8" ;
String k = "13" ;
hexaModK(n, k);
}
}
|
Python3
def hexaModK(s, k) :
mp = {};
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 ;
m = int (k);
base = 1 ;
ans = 0 ;
for i in range ( len (s) - 1 , - 1 , - 1 ) :
n = mp[s[i]] % m;
ans = (ans + (base % m * n
% m) % m) % m;
base = (base % m * 16 % m) % m;
ans = hex ( int (ans))[ - 1 ].upper()
print (ans)
if __name__ = = "__main__" :
n = "3E8" ;
k = "13" ;
hexaModK(n, k);
|
C#
using System;
using System.Collections.Generic;
class GFG{
static void hexaModK(String N, String k)
{
Dictionary< char ,
int > map = new Dictionary< char ,
int >();
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);
long m = long .Parse(k);
long Base = 1;
long ans = 0;
for ( int i = N.Length - 1; i >= 0; i--)
{
long n = map[N[i]] % m;
ans = (ans + (Base % m * n % m) % m) % m;
Base = (Base % m * 16 % m) % m;
}
Console.WriteLine(ans.ToString( "X" ));
}
public static void Main(String []args)
{
String n = "3E8" ;
String k = "13" ;
hexaModK(n, k);
}
}
|
Javascript
<script>
function hexaModK(s, k)
{
var mp = new Map();
for ( var i = 1; i <= 9; i++)
{
mp.set(String.fromCharCode(
i + '0' .charCodeAt(0)), i);
}
mp.set( 'A' , 10);
mp.set( 'B' , 11);
mp.set( 'C' , 12);
mp.set( 'D' , 13);
mp.set( 'E' , 14);
mp.set( 'F' , 15);
var m = parseInt(k, 16);
var base = 1;
var ans = 0;
for ( var i = s.length - 1;
i >= 0; i--)
{
var n = mp.get(s[i]) % m;
ans = (ans + (base % m * n % m) % m) % m;
base = (base % m * 16 % m) % m;
}
document.write(ans.toString(16).toUpperCase());
}
var n = "3E8" ;
var k = "13" ;
hexaModK(n, k);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
01 Jun, 2021
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