模块化部门

给定三个正数a，b和m。在模m下计算a / b。任务基本上是找到一个数字c，使(b * c)％m = a％m。
例子：

Input  : a  = 8, b = 4, m = 5
Output : 2

Input  : a  = 8, b = 3, m = 5
Output : 1
Note that (1*3)%5 is same as 8%5

Input  : a  = 11, b = 4, m = 5
Output : 4
Note that (4*4)%5 is same as 11%5

以下文章是实现此目的的先决条件。
模乘逆
扩展的欧几里得算法
我们可以一直做模块划分吗?
答案是不”。首先，像普通算术一样，不定义除以0。例如，不允许使用4/0。在模块化算术中，不仅不允许4/0，而且也不允许模6下的4/12。原因是，模数为6时12等于0。
何时定义模块化部门?
当除数的模逆存在时，定义模除法。整数“ x”的倒数是另一个整数“ y”，使得(x * y)％m = 1，其中m是模量。
逆何时存在?如此处所讨论的，如果“ a”和“ m”是互质的，则模“ m”下存在数字“ a”的倒数，即它们的GCD为1。
如何找到模块化部门?

The task is to compute a/b under modulo m.
1) First check if inverse of b under modulo m exists or not. 
    a) If inverse doesn't exists (GCD of b and m is not 1), 
          print "Division not defined"
    b) Else return  "(inverse * a) % m" 

C

// C program to do modular division
#include 
 
// C function for extended Euclidean Algorithm
int gcdExtended(int a, int b, int *x, int *y);
 
// Function to find modulo inverse of b. It returns
// -1 when inverse doesn't
int modInverse(int b, int m)
{
    int x, y; // used in extended GCD algorithm
    int g = gcdExtended(b, m, &x, &y);
 
    // Return -1 if b and m are not co-prime
    if (g != 1)
        return -1;
 
    // m is added to handle negative x
    return (x%m + m) % m;
}
 
// Function to compute a/b under modlo m
void modDivide(int a, int b, int m)
{
    a = a % m;
    int inv = modInverse(b, m);
    if (inv == -1)
     printf ("Division not defined");
    else
    {
      int c = (inv * a) % m ;
       printf ("Result of division is %d", c) ;
    }
}
 
// C function for extended Euclidean Algorithm (used to
// find modular inverse.
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 Program
int main()
{
    int  a  = 8, b = 3, m = 5;
    modDivide(a, b, m);
    return 0;
}

Python 3

# Python3 program to do modular division
import math
 
# Function to find modulo inverse of b. It returns
# -1 when inverse doesn't
# modInverse works for prime m
def modInverse(b,m):
    g = math.gcd(b, m)
    if (g != 1):
        # print("Inverse doesn't exist")
        return -1
    else:
        # If b and m are relatively prime,
        # then modulo inverse is b^(m-2) mode m
        return pow(b, m - 2, m)
 
 
# Function to compute a/b under modulo m
def modDivide(a,b,m):
    a = a % m
    inv = modInverse(b,m)
    if(inv == -1):
        print("Division not defined")
    else:
        print("Result of Division is ",(inv*a) % m)
 
# Driver Program
a = 8
b = 3
m = 5
modDivide(a, b, m)
 
# This code is Contributed by HarendraSingh22

C++

// C++ program to do modular division
#include
using namespace std;
 
// C++ function for extended Euclidean Algorithm
int gcdExtended(int a, int b, int *x, int *y);
 
// Function to find modulo inverse of b. It returns
// -1 when inverse doesn't
int modInverse(int b, int m)
{
    int x, y; // used in extended GCD algorithm
    int g = gcdExtended(b, m, &x, &y);
 
    // Return -1 if b and m are not co-prime
    if (g != 1)
        return -1;
 
    // m is added to handle negative x
    return (x%m + m) % m;
}
 
// Function to compute a/b under modlo m
void modDivide(int a, int b, int m)
{
    a = a % m;
    int inv = modInverse(b, m);
    if (inv == -1)
       cout << "Division not defined";
    else
       cout << "Result of division is " << (inv * a) % m;
}
 
// C function for extended Euclidean Algorithm (used to
// find modular inverse.
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 Program
int main()
{
    int  a  = 8, b = 3, m = 5;
    modDivide(a, b, m);
    return 0;
}
 
//this code is contributed by khushboogoyal499

输出：

Result of division is 1

模除法不同于加法，减法和乘法。
一个区别是除法并不总是存在(如上所述)。以下是另一个区别。

Below equations are valid
(a * b) % m = ((a % m) * (b % m)) % m
(a + b) % m = ((a % m) + (b % m)) % m

// m is added to handle negative numbers
(a - b + m) % m = ((a % m) - (b % m) + m) % m 

But, 
(a / b) % m may NOT be same as ((a % m)/(b % m)) % m

For example, a = 10, b = 5, m = 5. 
   (a / b) % m is 2, but ((a % m) / (b % m)) % m 
                          is not defined.