Modulo operations

#include <iostream>
#include <cstdio>
using namespace std;

/// Greatest Common Divisor
int gcd(int a, int b) {
    if (a < b)
        swap(a, b);
    if (b == 0)
        return a;
    else
        return gcd(b, a%b);
}

/// Summation A+B
int sum(int a, int b, int m) {
    return (a%m + b%m)%m;
}

/// Multiplication A*B
int mult(int a, int b, int m) {
    return (a%m * b%m)%m;
}

/// Exponentiation A^B
int pow(int a, int b, int m) {
    int answer = 1;
    for (int i=0; i<b; i++)
        answer = mult(answer, a, m);
    return answer;
}

/// Fast Exponentiation A^B
int pow_fast(int a, int b, int m) {
    if (b == 1)
        return a%m;

    int temp = pow_fast(a, b/2, m);
    if (b%2 == 0)
        return mult(temp, temp, m);
    else
        return mult(a, mult(temp, temp, m), m);
}

/// Euler's Function phi(n)
/// Counts the positive numbers <= n that are co-prime to n
int phi(int n) {
    int answer = 0;
    for (int i=1; i<n; i++)
        if (gcd(n, i) == 1)
            answer++;
    return answer;
}

/// Opposite element
/// a*x = 1 (mod m)
/// a^(phi(m)) = 1 (mod m) for co-prime numbers a and m <- Euler's Theorem
/// So x = a^(phi(m) - 1) (mod m)
int opp(int a, int m) {
    if (gcd(a, m) != 1) {
        cout << "There isn't opposite element";
        return m;
    }
    return pow(a, phi(m)-1, m);
}

/// Division is multiplication with an opposite element
int div(int a, int b, int m) {
    return mult(a, opp(b, m), m);
}

int main()
{
    int a, b, m;
    cin >> a >> b >> m;

    printf("GCD(%d, %d) = %d\n", a, b, gcd(a, b));
    printf("%d + %d mod %d = %d\n", a, b, m, sum(a, b, m));
    printf("%d * %d mod %d = %d\n", a, b, m, mult(a, b, m));
    printf("%d ^ %d mod %d = %d\n", a, b, m, pow(a, b, m));
    printf("%d ^ %d mod %d = %d (fast)\n", a, b, m, pow_fast(a, b, m));
    printf("%d / %d mod %d = %d\n", a, b, m, div(a, b, m));
    printf("phi(%d) = %d\n", a, phi(a));
    printf("phi(%d) = %d\n", b, phi(b));
    printf("phi(%d) = %d\n", m, phi(m));
    printf("opposite element for %d is %d (mod %d)\n", a, opp(a, m), m);
    printf("opposite element for %d is %d (mod %d)\n", b, opp(b, m), m);

    return 0;
}