计算模乘逆元原理上有四种方法：

1.暴力算法

2.扩展欧几里得算法

3.费尔马小定理

4.欧拉定理

模乘逆元定义：满足 ab≡1（mod m)，称b为a模乘逆元。以下是有关概念以及四种方法及程序。

文章出处：

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. This is equivalent to

1. Brute Force

We can calculate the inverse using a brute force approach where we multiply a with all possible valuesx and find ax such that Here’s a sample C++ code:

int modInverse(int a, int m) {    a %= m;    for(int x = 1; x < m; x++) {        if((a*x) % m == 1) return x;    }}

2. Using Extended Euclidean Algorithm

We have to find a number x such that a·x = 1 (mod m). This can be written as well as a·x = 1 + m·y, which rearranges into a·x – m·y = 1. Since x and y need not be positive, we can write it as well in the standard form, a·x + m·y = 1.

Iterative Method

/* This function return the gcd of a and b followed by    the pair x and y of equation ax + by = gcd(a,b)*/pair
    
      > extendedEuclid(int a, int b) {    int x = 1, y = 0;    int xLast = 0, yLast = 1;    int q, r, m, n;    while(a != 0) {        q = b / a;        r = b % a;        m = xLast - q * x;        n = yLast - q * y;        xLast = x, yLast = y;        x = m, y = n;        b = a, a = r;    }    return make_pair(b, make_pair(xLast, yLast));} int modInverse(int a, int m) {    return (extendedEuclid(a,m).second.first + m) % m;}
    

Recursive Method

/* This function return the gcd of a and b followed by    the pair x and y of equation ax + by = gcd(a,b)*/pair
    
      > extendedEuclid(int a, int b) {    if(a == 0) return make_pair(b, make_pair(0, 1));    pair
     
       > p;    p = extendedEuclid(b % a, a);    return make_pair(p.first, make_pair(p.second.second - p.second.first*(b/a), p.second.first));} int modInverse(int a, int m) {    return (extendedEuclid(a,m).second.first + m) % m;}
     
    

3. Using Fermat’s Little Theorem

Fermat’s little theorem states that if m is a prime and a is an integer co-prime to m, then

a^p − 1 will be evenly divisible by m. That is

or

Here’s a sample C++ code:

/* This function calculates (a^b)%MOD */int pow(int a, int b, int MOD) {int x = 1, y = a;    while(b > 0) {        if(b%2 == 1) {            x=(x*y);            if(x>MOD) x%=MOD;        }        y = (y*y);        if(y>MOD) y%=MOD;        b /= 2;    }    return x;} int modInverse(int a, int m) {    return pow(a,m-2,m);}

4. Using Euler’s Theorem

Fermat’s Little theorem can only be used if m is a prime. If m is not a prime we can use Euler’s Theorem, which is a generalization of Fermat’s Little theorem. According to Euler’s theorem, if a is coprime to m, that is, gcd(a, m) = 1, then

, where where φ(m) is Euler Totient Function. Therefore the modular multiplicative inverse can be found directly:

. The problem here is finding φ(m). If we know φ(m), then it is very similar to above method.

vector
     
       inverseArray(int n, int m) {    vector
      
        modInverse(n + 1,0);    modInverse[1] = 1;    for(int i = 2; i <= n; i++) {        modInverse[i] = (-(m/i) * modInverse[m % i]) % m + m;    }    return modInverse;}