Added mod_inverse_by_fermat_theorem.cpp in math directory.

math/mod_inverse_by_fermat_theorem.cpp

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+// C++ Program to find the modular inverse using Fermat's Little Theorem.
+// Fermat's Little Theorem state that => ϕ(m) = m-1, where m is prime number.
+/*
+ * (a * x) ≡ 1 mod m.
+ * x ≡ (a^(-1)) mod m.
+ *
+ * Using Euler's theorem we can modify the equation.
+ *
+ * (a^ϕ(m)) ≡ 1 mod m (Where '^' denotes the exponent operator)
+ * Here 'ϕ' is Euler's Totient Function. For modular inverse existence 'a' and 'm' must be relatively primes numbers.
+ * To apply Fermat's Little Theorem is necessary that 'm' must be a prime number.
+ * Generally in many competitive programming competitions 'm' is either 1000000007 (1e9+7) or 998244353.
+ *
+ * We considered m as large prime (1e9+7).
+ * (a^ϕ(m)) ≡ 1 mod m (Using Euler's Theorem)
+ * ϕ(m) = m-1 using Fermat's Little Theorem.
+ * (a^(m-1)) ≡ 1 mod m
+ * Now multiplying both side by (a^(-1)).
+ * (a^(m-1)) * (a^(-1)) ≡  (a^(-1)) mod m
+ * (a^(m-2)) ≡  (a^(-1)) mod m
+ *
+ * We will find the exponent using binary exponentiation. Such that the algorithm works in O(log(m)) time.
+ *
+ */
+
+ #include<iostream>
+ #include<vector>
+
+ // m is large prime number.
+ const long long int m = 1000000007;
+
+// Recursive function to calculate exponent in O(log(n)) using binary exponent.
+long long int binExpo(long long int a, long long int b) {
+    a %= m;
+    long long int res = 1;
+    while (b > 0) {
+        if (b%2) {
+            res = res * a % m;
+        }
+        a = a * a % m;
+        // Dividing b by 2 is similar to right shift.
+        b >>= 1;
+    }
+    return res;
+}
+
+int main() {
+    // Take input of  a. (A number for which we want to find modular inverse with m)
+    long long int a;
+    std::cout << "Give input a for computing ((a^(-1))%(m)) : ";
+    std::cin >> a;
+
+    std::cout << "The modular inverse of a with mod m is (a^(m-2)) equivalent (a^(-1)) mod m : ";
+    std::cout << binExpo(a, m-2) << std::endl;
+}
+
+
+
+