模乘逆
给定两个整数‘a’和‘m’，求模‘m’下‘a’的模乘逆。 模乘逆是一个整数‘x’，这样。
a x ≅ 1 (mod m)
x 的值应该在{ 1，2，… m-1}内，即在整数模 m 的范围内(注意 x 不能为 0 作为*0 mod m 永远不会为 1 ) 当且仅当 a 和 m 相对素数(即 gcd(a，m) = 1)时，“a 模 m”的乘法逆存在。
示例:
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。
下面是这个方法的实现。
C++
// C++ program to find modular
// inverse of a under modulo m
#include <iostream>
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 语言(一种计算机语言，尤用于创建网站)
// 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.*/
Python 3
# 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.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<≅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.
≅>
java 描述语言
<script>
// Javascript 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.
</script>
Output
4
时间复杂度: O(m)。
方法 2(当 m 和 a 是互素时有效) 想法是使用 扩展欧几里德算法 ，该算法采用两个整数‘a’和‘b’，找到它们的 gcd，还找到‘x’和‘y ’,这样
ax + by = gcd(a, b)
为了在“m”下找到“a”的乘法逆，我们在上面的公式中放入了 b = m。既然我们知道 a 和 m 是相对素数，我们可以把 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)
我们可以去掉左边的第二项，因为“my (mod m)”对于整数 y 总是 0。
ax ≅ 1 (mod m)
所以我们可以用扩展欧几里德算法找到的“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 <iostream>
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 <stdio.h>
// 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;
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<≅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
≅>
Output
Modular multiplicative inverse is 4
迭代实现:
C++
// Iterative C++ program to find modular
// inverse using extended Euclid algorithm
#include <bits/stdc++.h>
using namespace std;
// 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
cout << "Modular multiplicative inverse is "<< modInverse(a, m);
return 0;
}
// this code is contributed by shivanisinghss2110
C
// Iterative C program to find modular
// inverse using extended Euclid algorithm
#include <stdio.h>
// 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 语言(一种计算机语言，尤用于创建网站)
// 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.*/
Python 3
# 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.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<≅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 描述语言
<script>
// 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
</script>
Output
Modular multiplicative inverse is 4
时间复杂度: O(Log m)
方法 3(m 为素数时有效) 如果知道 m 为素数，那么也可以用 费马的小定理 求逆。
am-1 ≅ 1 (mod m)
如果我们将两边乘以一个 -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 <iostream>
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 语言(一种计算机语言，尤用于创建网站)
// 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.
Python 3
# 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.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<≅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 描述语言
<script>
// 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.
</script>
Output
Modular multiplicative inverse is 4
时间复杂度: O(Log m)
我们讨论了三种求乘法逆模 m 的方法。 1)天真法，O(m) 2)扩展欧拉 GCD 算法，O(Log m)【当 a 和 m 是互质时起作用】 3)费马小定理，O(Log m)【当‘m’是素数时起作用】
应用: 计算模乘逆是 RSA 公钥加密方法中必不可少的一步。
参考文献: https://en . Wikipedia . org/wiki/Modular _ 乘法 _ 逆 http://e-maxx.ru/algo/reverse_element 本文由 Ankur 供稿。如果您发现任何不正确的地方，或者您想分享更多关于上面讨论的主题的信息，请写评论
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