给定两个数字a和b，求出所有x使得a％x = b

给定两个数字a和b，求出所有x，使得％x = b。

例子：

Input : a = 21, b = 5
Output : 2
The answers of the Modular Equation are
8 and 16 since 21 % 8 = 21 % 16 = 5 .

这里出现3种情况：

1. 如果(a 则没有答案。
2. 如果(a = b)，那么所有大于a的数字都是答案，因此将有无限可能的解决方案。
3. 如果(a> b)假设x是我们方程的答案。然后x除以(a – b)。同样由于a％x = b，所以b O(sqrt(ab))中求解。这里只有一种情况出现，当(ab)是理想平方时，我们必须分开处理，然后我们将其平方根加两次，因此，如果出现这种情况，我们必须减去一次。

C++

// C++ program to find x such that a % x is equal
// to b.
#include 
using namespace std;
 
void modularEquation(int a, int b)
{
    // if a is less than b then no solution
    if (a < b) {
        cout << "No solution possible " << endl;
        return;
    }
 
    // if a is equal to b then every number
    // greater than a will be the solution
    // so its infinity
    if (a == b) {
        cout << "Infinite Solution possible " << endl;
        return;
    }
 
    // all resultant number should be greater than
    // b and (a-b) should be divisible by resultant
    // number
 
    // count variable store the number of values
    // possible
    int count = 0;
    int n = a - b;
    int y = sqrt(a - b);
    for (int i = 1; i <= y; ++i) {
        if (n % i == 0) {
 
            // checking for both divisor and quotient
            // whether they divide ( a-b ) completely
            // and greater than b .
            if (n / i > b)
                count++;
            if (i > b)
                count++;
        }
    }
 
    // Here y is added twice in the last iteration
    // so 1 y should be decremented to get correct
    // solution
    if (y * y == n && y > b)
        count--;
 
    cout << count << endl;
}
 
// Driver code
int main()
{
    int a = 21, b = 5;
    modularEquation(a, b);
    return 0;
}

Java

// Java program to find x such that
// a % x is equal to b.
import java.io.*;
class GFG {
     
static void modularEquation(int a, int b)
{
    // if a is less than b then no solution
    if (a < b) {
        System.out.println("No solution possible ");
        return;
    }
 
    // if a is equal to b then every number
    // greater than a will be the solution
    // so its infinity
    if (a == b) {
        System.out.println("Infinite Solution possible ");
        return;
    }
 
    // all resultant number should be greater
    // than b and (a-b) should be divisible
    // by resultant number
 
    // count variable store the number of
    // values possible
    int count = 0;
    int n = a - b;
    int y = (int)Math.sqrt(a - b);
    for (int i = 1; i <= y; ++i) {
        if (n % i == 0) {
 
            // checking for both divisor and
            // quotient whether they divide
            // ( a-b ) completely and
            // greater than b .
            if (n / i > b)
                count++;
            if (i > b)
                count++;
        }
    }
 
    // Here y is added twice in the last
    // iteration so 1 y should be decremented
    // to get correct solution
    if (y * y == n && y > b)
        count--;
 
    System.out.println(count);
}
 
// Driver code
public static void main(String[] args)
{
    int a = 21, b = 5;
    modularEquation(a, b);
}
}
 
// This code is contributed by Prerna Saini

Python3

# Python3 program to find x such
# that a % x is equal to b.
 
import math
def modularEquation(a, b) :
     
    # if a is less than b then no solution
    if (a < b) :
        print("No solution possible ")
        return
     
    # if a is equal to b then every number
    # greater than a will be the solution
    # so its infinity
    if (a == b) :
        print("Infinite Solution possible ")
        return
     
    # all resultant number should be
    # greater than b and (a-b) should
    # be divisible by resultant number
  
    # count variable store the number
    # of values possible
    count = 0
    n = a - b
    y = (int)(math.sqrt(a - b))
    for i in range(1, y+1) :
        if (n % i == 0) :
             
            # checking for both divisor
            # and quotient whether they
            # divide ( a-b ) completely
            # and greater than b .
            if (n / i > b) :
                count = count + 1
            if (i > b) :
                count = count  + 1
         
         
  
    # Here y is added twice in the
    # last iteration so 1 y should be
    # decremented to get correct
    # solution
    if (y * y == n and y > b) :
        count = count - 1
  
    print (count)
  
# Driver code
a = 21
b = 5
modularEquation(a, b)
 
# This code is contributed by Nikita Tiwari.

C#

// C# program to find x such that
// a % x is equal to b.
using System;
 
class GFG {
     
static void modularEquation(int a, int b)
{
    // if a is less than b then no solution
    if (a < b) {
        Console.WriteLine("No solution possible ");
        return;
    }
 
    // if a is equal to b then every number
    // greater than a will be the solution
    // so its infinity
    if (a == b) {
        Console.WriteLine("Infinite Solution possible ");
        return;
    }
 
    // all resultant number should be greater
    // than b and (a-b) should be divisible
    // by resultant number
 
    // count variable store the number of
    // values possible
    int count = 0;
    int n = a - b;
    int y = (int)Math.Sqrt(a - b);
    for (int i = 1; i <= y; ++i) {
        if (n % i == 0) {
 
            // checking for both divisor and
            // quotient whether they divide
            // ( a-b ) completely and
            // greater than b .
            if (n / i > b)
                count++;
            if (i > b)
                count++;
        }
    }
 
    // Here y is added twice in the last
    // iteration so 1 y should be decremented
    // to get correct solution
    if (y * y == n && y > b)
        count--;
 
    Console.WriteLine(count);
}
 
// Driver code
public static void Main()
{
    int a = 21, b = 5;
    modularEquation(a, b);
}
}
 
//This code is contributed by vt_m.

PHP

 $b)
                $count++;
            if ($i > $b)
                $count++;
        }
    }
 
    // Here y is added twice
    // in the last iteration
    // so 1 y should be
    // decremented to get correct
    // solution
    if ($y * $y == $n && $y > $b)
        $count--;
 
    echo $count ;
}
 
    // Driver Code
    $a = 21;
    $b = 5;
    modularEquation($a, $b);
 
// This code is contributed by anuj_67.
?>

Javascript

 
// Javascript program to find x
// such that a % x is equal
// to b.
function modularEquation(a, b)
{
     
    // If a is less than b
    // then no solution
    if (a < b)
    {
        document.write("No solution possible ");
        return;
    }
 
    // If a is equal to b then every
    // number greater than a will
    // be the solution so its infinity
    if (a == b)
    {
        document.write("Infinite Solution possible ");
        return;
    }
 
    // All resultant number should be
    // greater than b and (a-b) should be
    // divisible by resultant number
 
    // Count variable store the number
    // of values possible
    let count = 0;
    let n = a - b;
    let y = Math.sqrt(a - b);
    for(let i = 1; i <= y; ++i)
    {
        if (n % i == 0)
        {
             
            // checking for both
            // divisor and quotient
            // whether they divide
            // ( a-b ) completely
            // and greater than b .
            if (n / i > b)
                count++;
            if (i > b)
                count++;
        }
    }
 
    // Here y is added twice
    // in the last iteration
    // so 1 y should be
    // decremented to get correct
    // solution
    if (y * y == n && y > b)
        count--;
 
    document.write(count);
}
 
// Driver Code
let a = 21;
let b = 5;
 
modularEquation(a, b);
 
// This code is contributed by _saurabh_jaiswal.
 

输出：

2