Modular multiplicative inverse

In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that

That is, it is the multiplicative inverse in the ring of integers modulo m, denoted . Once defined, x may be noted , where the fact that the inversion is m-modular is implicit. The multiplicative inverse of a modulo m exists if and only if a and m are coprime (i.e., if gcd(a, m) = 1).[1] If the modular multiplicative inverse of a modulo m exists, the operation of divisionby a modulo m can be defined as multiplying by the inverse, which is in essence the same concept as division in the field of reals. wiki

example : http://codeforces.com/contest/560/problem/E

test code : are each number of factorial series and 10e9+7 coprime ?

#include <cstdio>
using namespace std;
const int MOD = 1000000007;
int fact[100001]; // 10e5
void make_factorial_series() {
    fact[0] = fact[1] = 1;
    for (int i = 2; i <= 100000; i++) {
        fact[i] = (int)((long long)fact[i-1] * i % MOD);
    }
}
int gcd(int a, int b) {
    return b ? gcd(b, a%b) : a;
}
int main() {
    make_factorial_series();
    for (int i = 2; i <= 100000; i++) {
        int _gcd = gcd(fact[i], MOD);
        printf("%d %d ==> gcd(%d) - %s\n", fact[i], MOD, _gcd, (_gcd == 1) ? "o" : "X");
    }
    return 0;
}

extended Euclidean algorithm

The modular multiplicative inverse of a modulo m can be found with the extended Euclidean algorithm. The algorithm finds solutions to Bézout’s identity

한국어: https://medium.com/@jongukim/extended-euclidean-algorithm-8101bafb9164

test codes :

#include <iostream>
using namespace std;
void cal(int quotient , int &r, int &new_r) {
    int temp(r);
    r = new_r;
    new_r = temp - quotient * new_r;
}
int inverse(int a, int m) {
    int t(0), new_t(1), r(m), new_r(a);
    while (new_r) {
        int quotient = r / new_r;
        cal(quotient, t, new_t);
        cal(quotient, r, new_r);
    }
    if (r > 1) return -1; // something wrong
    if (t < 0) t = t + m;
    return t;
}
int main() {
    while (true) {
        int a, b;
        cin >> a >> b;
        if (a == 0) break;
        cout << inverse(a, b) << endl;
    }
    return 0;
}

Written on August 21, 2015