模乘逆

给定两个整数“ a”和“ m”，在模“ m”下找到“ a”的模乘逆。
模乘逆是这样的整数“ x”。

a x ≅ 1 (mod m)

x的值应在{1，2，…m-1}中，即在整数模m的范围内。 (请注意，x不能为0，因为a * 0 mod m永远不会为1)
当且仅当a和m相对质数(即gcd(a，m)= 1)时，才存在“模m”的乘法逆。

例子：

Input:  a = 3, m = 11
Output: 4
Since (4*3) mod 11 = 1, 4 is modulo inverse of 3(under 11).
One might think, 15 also as a valid output as "(15*3) mod 11" 
is also 1, but 15 is not in ring {1, 2, ... 10}, so not 
valid.

Input:  a = 10, m = 17
Output: 12
Since (10*12) mod 17 = 1, 12 is modulo inverse of 10(under 17).

方法1(天真)
天真的方法是尝试从1到m的所有数字。对于每个数字x，检查(a * x)％m是否为1。

下面是此方法的实现。

C++

// C++ program to find modular
// inverse of a under modulo m
#include 
using namespace std;
 
// A naive method to find modular
// multiplicative inverse of 'a'
// under modulo 'm'
int modInverse(int a, int m)
{
    for (int x = 1; x < m; x++)
        if (((a%m) * (x%m)) % m == 1)
            return x;
}
 
// Driver code
int main()
{
    int a = 3, m = 11;
   
    // Function call
    cout << modInverse(a, m);
    return 0;
}

Java

// Java program to find modular inverse
// of a under modulo m
import java.io.*;
 
class GFG {
 
    // A naive method to find modulor
    // multiplicative inverse of 'a'
    // under modulo 'm'
    static int modInverse(int a, int m)
    {
      
        for (int x = 1; x < m; x++)
            if (((a%m) * (x%m)) % m == 1)
                return x;
        return 1;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int a = 3, m = 11;
       
        // Function call
        System.out.println(modInverse(a, m));
    }
}
 
/*This code is contributed by Nikita Tiwari.*/

Python3

# Python3 program to find modular
# inverse of a under modulo m
 
# A naive method to find modulor
# multiplicative inverse of 'a'
# under modulo 'm'
 
 
def modInverse(a, m):
     
    for x in range(1, m):
        if (((a%m) * (x%m)) % m == 1):
            return x
    return -1
 
 
# Driver Code
a = 3
m = 11
 
# Function call
print(modInverse(a, m))
 
# This code is contributed by Nikita Tiwari.

C#

// C# program to find modular inverse
// of a under modulo m
using System;
 
class GFG {
 
    // A naive method to find modulor
    // multiplicative inverse of 'a'
    // under modulo 'm'
    static int modInverse(int a, int m)
    {
         
        for (int x = 1; x < m; x++)
            if (((a%m) * (x%m)) % m == 1)
                return x;
        return 1;
    }
 
    // Driver Code
    public static void Main()
    {
        int a = 3, m = 11;
       
        // Function call
        Console.WriteLine(modInverse(a, m));
    }
}
 
// This code is contributed by anuj_67.

PHP

<≅php
// PHP program to find modular
// inverse of a under modulo m
 
// A naive method to find modulor
// multiplicative inverse of
// 'a' under modulo 'm'
function modInverse( $a, $m)
{
     
    for ($x = 1; $x < $m; $x++)
        if ((($a%$m) * ($x%$m)) % $m == 1)
            return $x;
}
 
    // Driver Code
    $a = 3;
    $m = 11;
 
    // Function call
    echo modInverse($a, $m);
 
// This code is contributed by anuj_67.
≅>

Javascript

 
// Javascirpt program to find modular
// inverse of a under modulo m
 
// A naive method to find modulor
// multiplicative inverse of
// 'a' under modulo 'm'
function modInverse(a, m)
{
    for(let x = 1; x < m; x++)
        if (((a % m) * (x % m)) % m == 1)
            return x;
}
 
// Driver Code
let a = 3;
let m = 11;
 
// Function call
document.write(modInverse(a, m));
 
// This code is contributed by _saurabh_jaiswal.
 

输出

4

时间复杂度： O(m)。

方法2(当m和a为互质时有效)
这个想法是使用扩展的欧几里得算法，它采用两个整数“ a”和“ b”，找到它们的gcd，还找到“ x”和“ y”，使得

ax + by = gcd(a, b)

为了在“ m”下找到“ a”的乘法逆，我们将b = m放在上式中。由于我们知道a和m是相对质数，因此我们可以将gcd的值设为1。

ax + my = 1

如果我们在两边取模m，我们得到

ax + my ≅ 1 (mod m)

我们可以删除左侧的第二项，因为对于整数y，“ my(mod m)”始终为0。

ax  ≅ 1 (mod m)

因此，我们可以使用扩展Euclid算法找到的’x’是’a’的乘法逆。

下面是上述算法的实现。

C++

// C++ program to find multiplicative modulo
// inverse using Extended Euclid algorithm.
#include 
using namespace std;
 
// Function for extended Euclidean Algorithm
int gcdExtended(int a, int b, int* x, int* y);
 
// Function to find modulo inverse of a
void modInverse(int a, int m)
{
    int x, y;
    int g = gcdExtended(a, m, &x, &y);
    if (g != 1)
        cout << "Inverse doesn't exist";
    else
    {
         
        // m is added to handle negative x
        int res = (x % m + m) % m;
        cout << "Modular multiplicative inverse is " << res;
    }
}
 
// Function for extended Euclidean Algorithm
int gcdExtended(int a, int b, int* x, int* y)
{
     
    // Base Case
    if (a == 0)
    {
        *x = 0, *y = 1;
        return b;
    }
     
    // To store results of recursive call
    int x1, y1;
    int gcd = gcdExtended(b % a, a, &x1, &y1);
 
    // Update x and y using results of recursive
    // call
    *x = y1 - (b / a) * x1;
    *y = x1;
 
    return gcd;
}
 
// Driver Code
int main()
{
    int a = 3, m = 11;
   
    // Function call
    modInverse(a, m);
    return 0;
}
 
// This code is contributed by khushboogoyal499

C

// C program to find multiplicative modulo inverse using
// Extended Euclid algorithm.
#include 
 
// C function for extended Euclidean Algorithm
int gcdExtended(int a, int b, int* x, int* y);
 
// Function to find modulo inverse of a
void modInverse(int a, int m)
{
    int x, y;
    int g = gcdExtended(a, m, &x, &y);
    if (g != 1)
        printf("Inverse doesn't exist");
    else
    {
        // m is added to handle negative x
        int res = (x % m + m) % m;
        printf("Modular multiplicative inverse is %d", res);
    }
}
 
// C function for extended Euclidean Algorithm
int gcdExtended(int a, int b, int* x, int* y)
{
    // Base Case
    if (a == 0)
    {
        *x = 0, *y = 1;
        return b;
    }
 
    int x1, y1; // To store results of recursive call
    int gcd = gcdExtended(b % a, a, &x1, &y1);
 
    // Update x and y using results of recursive
    // call
    *x = y1 - (b / a) * x1;
    *y = x1;
 
    return gcd;
}
 
// Driver Code
int main()
{
    int a = 3, m = 11;
   
    // Function call
    modInverse(a, m);
    return 0;
}

的PHP

<≅php
// PHP program to find multiplicative modulo
// inverse using Extended Euclid algorithm.
// Function to find modulo inverse of a
function modInverse($a, $m)
{
    $x = 0;
    $y = 0;
    $g = gcdExtended($a, $m, $x, $y);
    if ($g != 1)
        echo "Inverse doesn't exist";
    else
    {
        // m is added to handle negative x
        $res = ($x % $m + $m) % $m;
        echo "Modular multiplicative " .
                   "inverse is " . $res;
    }
}
 
// function for extended Euclidean Algorithm
function gcdExtended($a, $b, &$x, &$y)
{
    // Base Case
    if ($a == 0)
    {
        $x = 0;
        $y = 1;
        return $b;
    }
 
    $x1;
    $y1; // To store results of recursive call
    $gcd = gcdExtended($b%$a, $a, $x1, $y1);
 
    // Update x and y using results of
    // recursive call
    $x = $y1 - (int)($b/$a) * $x1;
    $y = $x1;
 
    return $gcd;
}
 
// Driver Code
$a = 3;
$m = 11;
 
// Function call
modInverse($a, $m);
 
// This code is contributed by chandan_jnu
≅>

输出

Modular multiplicative inverse is 4

迭代实现：

C++

// Iterative C++ program to find modular
// inverse using extended Euclid algorithm
#include 
 
// Returns modulo inverse of a with respect
// to m using extended Euclid Algorithm
// Assumption: a and m are coprimes, i.e.,
// gcd(a, m) = 1
int modInverse(int a, int m)
{
    int m0 = m;
    int y = 0, x = 1;
 
    if (m == 1)
        return 0;
 
    while (a > 1) {
        // q is quotient
        int q = a / m;
        int t = m;
 
        // m is remainder now, process same as
        // Euclid's algo
        m = a % m, a = t;
        t = y;
 
        // Update y and x
        y = x - q * y;
        x = t;
    }
 
    // Make x positive
    if (x < 0)
        x += m0;
 
    return x;
}
 
// Driver Code
int main()
{
    int a = 3, m = 11;
 
    // Function call
    printf("Modular multiplicative inverse is %d\n",
           modInverse(a, m));
    return 0;
}

Java

// Iterative Java program to find modular
// inverse using extended Euclid algorithm
 
class GFG {
 
    // Returns modulo inverse of a with
    // respect to m using extended Euclid
    // Algorithm Assumption: a and m are
    // coprimes, i.e., gcd(a, m) = 1
    static int modInverse(int a, int m)
    {
        int m0 = m;
        int y = 0, x = 1;
 
        if (m == 1)
            return 0;
 
        while (a > 1) {
            // q is quotient
            int q = a / m;
 
            int t = m;
 
            // m is remainder now, process
            // same as Euclid's algo
            m = a % m;
            a = t;
            t = y;
 
            // Update x and y
            y = x - q * y;
            x = t;
        }
 
        // Make x positive
        if (x < 0)
            x += m0;
 
        return x;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int a = 3, m = 11;
         
        // Function call
        System.out.println("Modular multiplicative "
                           + "inverse is "
                           + modInverse(a, m));
    }
}
 
/*This code is contributed by Nikita Tiwari.*/

Python3

# Iterative Python 3 program to find
# modular inverse using extended
# Euclid algorithm
 
# Returns modulo inverse of a with
# respect to m using extended Euclid
# Algorithm Assumption: a and m are
# coprimes, i.e., gcd(a, m) = 1
 
 
def modInverse(a, m):
    m0 = m
    y = 0
    x = 1
 
    if (m == 1):
        return 0
 
    while (a > 1):
 
        # q is quotient
        q = a // m
 
        t = m
 
        # m is remainder now, process
        # same as Euclid's algo
        m = a % m
        a = t
        t = y
 
        # Update x and y
        y = x - q * y
        x = t
 
    # Make x positive
    if (x < 0):
        x = x + m0
 
    return x
 
 
# Driver code
a = 3
m = 11
 
# Function call
print("Modular multiplicative inverse is",
      modInverse(a, m))
 
# This code is contributed by Nikita tiwari.

C＃

// Iterative C# program to find modular
// inverse using extended Euclid algorithm
using System;
class GFG {
 
    // Returns modulo inverse of a with
    // respect to m using extended Euclid
    // Algorithm Assumption: a and m are
    // coprimes, i.e., gcd(a, m) = 1
    static int modInverse(int a, int m)
    {
        int m0 = m;
        int y = 0, x = 1;
 
        if (m == 1)
            return 0;
 
        while (a > 1) {
            // q is quotient
            int q = a / m;
 
            int t = m;
 
            // m is remainder now, process
            // same as Euclid's algo
            m = a % m;
            a = t;
            t = y;
 
            // Update x and y
            y = x - q * y;
            x = t;
        }
 
        // Make x positive
        if (x < 0)
            x += m0;
 
        return x;
    }
 
    // Driver Code
    public static void Main()
    {
        int a = 3, m = 11;
       
        // Function call
        Console.WriteLine("Modular multiplicative "
                          + "inverse is "
                          + modInverse(a, m));
    }
}
 
// This code is contributed by anuj_67.

的PHP

<≅php
// Iterative PHP program to find modular
// inverse using extended Euclid algorithm
 
// Returns modulo inverse of a with respect
// to m using extended Euclid Algorithm
// Assumption: a and m are coprimes, i.e.,
// gcd(a, m) = 1
function modInverse($a, $m)
{
    $m0 = $m;
    $y = 0;
    $x = 1;
 
    if ($m == 1)
    return 0;
 
    while ($a > 1)
    {
         
        // q is quotient
        $q = (int) ($a / $m);
        $t = $m;
 
        // m is remainder now,
        // process same as
        // Euclid's algo
        $m = $a % $m;
        $a = $t;
        $t = $y;
 
        // Update y and x
        $y = $x - $q * $y;
        $x = $t;
    }
 
    // Make x positive
    if ($x < 0)
    $x += $m0;
 
    return $x;
}
 
    // Driver Code
    $a = 3;
    $m = 11;
 
    // Function call
    echo "Modular multiplicative inverse is\n",
                            modInverse($a, $m);
         
// This code is contributed by Anuj_67
≅>

Java脚本

 
// Iterative Javascript program to find modular
// inverse using extended Euclid algorithm
 
// Returns modulo inverse of a with respect
// to m using extended Euclid Algorithm
// Assumption: a and m are coprimes, i.e.,
// gcd(a, m) = 1
function modInverse(a, m)
{
    let m0 = m;
    let y = 0;
    let x = 1;
 
    if (m == 1)
        return 0;
 
    while (a > 1)
    {
         
        // q is quotient
        let q = parseInt(a / m);
        let t = m;
 
        // m is remainder now,
        // process same as
        // Euclid's algo
        m = a % m;
        a = t;
        t = y;
 
        // Update y and x
        y = x - q * y;
        x = t;
    }
 
    // Make x positive
    if (x < 0)
        x += m0;
 
    return x;
}
 
// Driver Code
let a = 3;
let m = 11;
 
// Function call
document.write(`Modular multiplicative inverse is ${modInverse(a, m)}`);
     
// This code is contributed by _saurabh_jaiswal
 

输出

Modular multiplicative inverse is 4

时间复杂度： O(Log m)方法3(当m为素数时有效)
如果我们知道m是素数，那么我们也可以使用Fermats的小定理来找到逆定理。

am-1 ≅ 1 (mod m)

如果将两边都乘以^-1 ，我们得到

a-1 ≅ a m-2 (mod m)

以下是上述想法的实现。

C++

// C++ program to find modular inverse of a under modulo m
// This program works only if m is prime.
#include 
using namespace std;
 
// To find GCD of a and b
int gcd(int a, int b);
 
// To compute x raised to power y under modulo m
int power(int x, unsigned int y, unsigned int m);
 
// Function to find modular inverse of a under modulo m
// Assumption: m is prime
void modInverse(int a, int m)
{
    int g = gcd(a, m);
    if (g != 1)
        cout << "Inverse doesn't exist";
    else
    {
        // If a and m are relatively prime, then modulo
        // inverse is a^(m-2) mode m
        cout << "Modular multiplicative inverse is "
             << power(a, m - 2, m);
    }
}
 
// To compute x^y under modulo m
int power(int x, unsigned int y, unsigned int m)
{
    if (y == 0)
        return 1;
    int p = power(x, y / 2, m) % m;
    p = (p * p) % m;
 
    return (y % 2 == 0) ? p : (x * p) % m;
}
 
// Function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Driver code
int main()
{
    int a = 3, m = 11;
 
    // Function call
    modInverse(a, m);
    return 0;
}

Java

// Java program to find modular
// inverse of a under modulo m
// This program works only if
// m is prime.
import java.io.*;
 
class GFG {
 
    // Function to find modular inverse of a
    // under modulo m Assumption: m is prime
    static void modInverse(int a, int m)
    {
        int g = gcd(a, m);
        if (g != 1)
            System.out.println("Inverse doesn't exist");
        else
        {
            // If a and m are relatively prime, then modulo
            // inverse is a^(m-2) mode m
            System.out.println(
                "Modular multiplicative inverse is "
                + power(a, m - 2, m));
        }
    }
   
      static int power(int x, int y, int m)
    {
        if (y == 0)
            return 1;
        int p = power(x, y / 2, m) % m;
        p = (int)((p * (long)p) % m);
        if (y % 2 == 0)
            return p;
        else
            return (int)((x * (long)p) % m);
    }
 
    // Function to return gcd of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int a = 3, m = 11;
        
        // Function call
        modInverse(a, m);
    }
}
 
// This code is contributed by Nikita Tiwari.

Python3

# Python3 program to find modular
# inverse of a under modulo m
 
# This program works only if m is prime.
 
# Function to find modular
# inverse of a under modulo m
# Assumption: m is prime
 
 
def modInverse(a, m):
 
    g = gcd(a, m)
 
    if (g != 1):
        print("Inverse doesn't exist")
 
    else:
 
        # If a and m are relatively prime,
        # then modulo inverse is a^(m-2) mode m
        print("Modular multiplicative inverse is ",
              power(a, m - 2, m))
 
# To compute x^y under modulo m
 
 
def power(x, y, m):
 
    if (y == 0):
        return 1
 
    p = power(x, y // 2, m) % m
    p = (p * p) % m
 
    if(y % 2 == 0):
        return p
    else:
        return ((x * p) % m)
 
# Function to return gcd of a and b
 
 
def gcd(a, b):
    if (a == 0):
        return b
 
    return gcd(b % a, a)
 
 
# Driver Code
a = 3
m = 11
 
# Function call
modInverse(a, m)
 
 
# This code is contributed by Nikita Tiwari.

C＃

// C# program to find modular
// inverse of a under modulo m
// This program works only if
// m is prime.
using System;
class GFG {
 
    // Function to find modular
    // inverse of a under modulo
    // m Assumption: m is prime
    static void modInverse(int a, int m)
    {
        int g = gcd(a, m);
        if (g != 1)
            Console.Write("Inverse doesn't exist");
        else {
            // If a and m are relatively
            // prime, then modulo inverse
            // is a^(m-2) mode m
            Console.Write(
                "Modular multiplicative inverse is "
                + power(a, m - 2, m));
        }
    }
 
    // To compute x^y under
    // modulo m
    static int power(int x, int y, int m)
    {
        if (y == 0)
            return 1;
 
        int p = power(x, y / 2, m) % m;
        p = (p * p) % m;
 
        if (y % 2 == 0)
            return p;
        else
            return (x * p) % m;
    }
 
    // Function to return
    // gcd of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
 
    // Driver Code
    public static void Main()
    {
        int a = 3, m = 11;
       
        // Function call
        modInverse(a, m);
    }
}
 
// This code is contributed by nitin mittal.

的PHP

<≅php
// PHP program to find modular
// inverse of a under modulo m
// This program works only if m
// is prime.
 
// Function to find modular inverse
// of a under modulo m
// Assumption: m is prime
function modInverse( $a, $m)
{
    $g = gcd($a, $m);
    if ($g != 1)
        echo "Inverse doesn't exist";
    else
    {
         
        // If a and m are relatively
        // prime, then modulo inverse
        // is a^(m-2) mode m
        echo "Modular multiplicative inverse is "
                        , power($a, $m - 2, $m);
    }
}
 
// To compute x^y under modulo m
function power( $x, $y, $m)
{
    if ($y == 0)
        return 1;
    $p = power($x, $y / 2, $m) % $m;
    $p = ($p * $p) % $m;
 
    return ($y % 2 == 0)? $p : ($x * $p) % $m;
}
 
// Function to return gcd of a and b
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}
 
// Driver Code
$a = 3;
$m = 11;
 
// Function call
modInverse($a, $m);
     
// This code is contributed by anuj_67.
≅>

Java脚本

// Javascript program to find modular inverse of a under modulo m
// This program works only if m is prime.
 
// Function to find modular inverse of a under modulo m
// Assumption: m is prime
function modInverse(a, m)
{
    let g = gcd(a, m);
    if (g != 1)
        document.write("Inverse doesn't exist");
    else
    {
        // If a and m are relatively prime, then modulo
        // inverse is a^(m-2) mode m
        document.write("Modular multiplicative inverse is "
             + power(a, m - 2, m));
    }
}
 
// To compute x^y under modulo m
function power(x, y, m)
{
    if (y == 0)
        return 1;
    let p = power(x, parseInt(y / 2), m) % m;
    p = (p * p) % m;
 
    return (y % 2 == 0) ? p : (x * p) % m;
}
 
// Function to return gcd of a and b
function gcd(a, b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Driver code
let a = 3, m = 11;
 
// Function call
modInverse(a, m);
 
// This code is contributed by subham348.

输出

Modular multiplicative inverse is 4

时间复杂度： O(Log m)

我们讨论了三种找到乘法逆模m的方法。
1)天真的方法，O(m)
2)扩展的欧拉GCD算法O(Log m)[当a和m为互素时有效]
3)费马小定理，O(Log m)[当’m’为素数时有效]

应用范围：
模数乘法逆的计算是RSA公钥加密方法中必不可少的步骤。