Expressing a fraction as a natural number under modulo ‘m’

Given two integers A and B where A is not divisible by B, the task is to express A / B as a natural number modulo m where m = 1000000007.
Note: This representation is useful where we need to express Probability of an event, Area of Curves and polygons etc.

Examples:

Input: A = 2, B = 6
Output: 333333336

Input: A = 4, B = 5
Output: 600000005



Approach: We know that, A / B can be written as A * (1 / B) i.e. A * (B ^ -1).

It is known that the modulo(%) operator satisfies the relation:

(a * b) % m = ( (a % m) * (b % m) ) % m

So, we can write:

(b ^ -1) % m = (b ^ m-2) % m (Fermat's little theorem)

Therefore the result will be:

( (A mod m) * ( power(B, m-2) % m) ) % m

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define m 1000000007
  
// Function to return the GCD of given numbers
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Recursive function to return (x ^ n) % m
ll modexp(ll x, ll n)
{
    if (n == 0) {
        return 1;
    }
    else if (n % 2 == 0) {
        return modexp((x * x) % m, n / 2);
    }
    else {
        return (x * modexp((x * x) % m, (n - 1) / 2) % m);
    }
}
  
// Function to return the fraction modulo mod
ll getFractionModulo(ll a, ll b)
{
    ll c = gcd(a, b);
  
    a = a / c;
    b = b / c;
  
    // (b ^ m-2) % m
    ll d = modexp(b, m - 2);
  
    // Final answer
    ll ans = ((a % m) * (d % m)) % m;
  
    return ans;
}
  
// Driver code
int main()
{
    ll a = 2, b = 6;
  
    cout << getFractionModulo(a, b) << endl;
  
    return 0;
}

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Java

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// Java implementation of the approach
  
import java.io.*;
  
class GFG {
      
  
  
static long m  = 1000000007;
  
// Function to return the GCD of given numbers
 static long gcd(long a, long b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Recursive function to return (x ^ n) % m
static long modexp(long x, long n)
{
    if (n == 0) {
        return 1;
    }
    else if (n % 2 == 0) {
        return modexp((x * x) % m, n / 2);
    }
    else {
        return (x * modexp((x * x) % m, (n - 1) / 2) % m);
    }
}
  
// Function to return the fraction modulo mod
 static long getFractionModulo(long a, long b)
{
    long c = gcd(a, b);
  
    a = a / c;
    b = b / c;
  
    // (b ^ m-2) % m
    long  d = modexp(b, m - 2);
  
    // Final answer
    long ans = ((a % m) * (d % m)) % m;
  
    return ans;
}
  
// Driver code
  
    public static void main (String[] args) {
        long a = 2, b = 6;
  
    System.out.println(getFractionModulo(a, b));
    }
}
// This code is contributed by inder_verma

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C#

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//C#  implementation of the approach
  
using System;
  
public class GFG{
      
  
static long m = 1000000007;
  
// Function to return the GCD of given numbers
static long gcd(long a, long b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Recursive function to return (x ^ n) % m
static long modexp(long x, long n)
{
    if (n == 0) {
        return 1;
    }
    else if (n % 2 == 0) {
        return modexp((x * x) % m, n / 2);
    }
    else {
        return (x * modexp((x * x) % m, (n - 1) / 2) % m);
    }
}
  
// Function to return the fraction modulo mod
static long getFractionModulo(long a, long b)
{
    long c = gcd(a, b);
  
    a = a / c;
    b = b / c;
  
    // (b ^ m-2) % m
    long d = modexp(b, m - 2);
  
    // Final answer
    long ans = ((a % m) * (d % m)) % m;
  
    return ans;
}
  
// Driver code
      
    static public void Main (){
          
        long a = 2, b = 6;
        Console.WriteLine(getFractionModulo(a, b));
    }
}

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Output:

333333336


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Improved By : inderDuMCA, Sach_Code