查找将X和Y除以产生相同余数的数字

给定两个整数X和Y ，任务是查找并打印将X和Y相除以产生相同余数的数字。
例子：

Input: X = 1, Y = 5 
Output: 1, 2, 4 
Explanation: 
Let the number be M. It can be any value in the range [1, 5]: 
If M = 1, 1 % 1 = 0 and 5 % 1 = 0 
If M = 2, 1 % 2 = 1 and 5 % 2 = 1 
If M = 3, 1 % 3 = 1 and 5 % 3 = 2 
If M = 4, 1 % 4 = 1 and 5 % 4 = 1 
If M = 5, 1 % 5 = 1 and 5 % 5 = 0 
Therefore, the possible M values are 1, 2, 4
Input: X = 8, Y = 10 
Output: 1, 2 
 

幼稚的方法：针对此问题的幼稚的方法是检查范围[1，max(X，Y)]内M的所有可能值的模值，并在条件满足时打印M的值。
下面是上述方法的实现：

C++

// C++ program to find numbers
// that divide X and Y
// to produce the same remainder
 
#include 
using namespace std;
 
// Function to find
// the required number as M
void printModulus(int X, int Y)
{
    // Finding the maximum
    // value among X and Y
    int n = max(X, Y);
 
    // Loop to iterate through
    // maximum value among X and Y.
    for (int i = 1; i <= n; i++) {
 
        // If the condition satisfies, then
        // print the value of M
        if (X % i == Y % i)
            cout << i << " ";
    }
}
 
// Driver code
int main()
{
 
    int X, Y;
    X = 10;
    Y = 20;
    printModulus(X, Y);
    return 0;
}

Java

// Java program to find numbers
// that divide X and Y
// to produce the same remainder
class GFG{
  
// Function to find
// the required number as M
static void printModulus(int X, int Y)
{
    // Finding the maximum
    // value among X and Y
    int n = Math.max(X, Y);
  
    // Loop to iterate through
    // maximum value among X and Y.
    for (int i = 1; i <= n; i++) {
  
        // If the condition satisfies, then
        // print the value of M
        if (X % i == Y % i)
            System.out.print(i + " ");
    }
}
  
// Driver code
public static void main(String[] args)
{
  
    int X, Y;
    X = 10;
    Y = 20;
    printModulus(X, Y);
}
}
 
// This code is contributed by Princi Singh

Python3

# Python program to find numbers
# that divide X and Y
# to produce the same remainder
 
# Function to find
# the required number as M
def printModulus( X, Y):
     
    # Finding the maximum
    # value among X and Y
    n = max(X, Y)
 
    # Loop to iterate through
    # maximum value among X and Y.
    for i in range(1, n + 1):
 
        # If the condition satisfies, then
        # print the value of M
        if (X % i == Y % i):
            print(i,end=" ")
 
# Driver code
X = 10
Y = 20
printModulus(X, Y)
 
# This code is contributed by Atul_kumar_Shrivastava

C#

// C# program to find numbers
// that divide X and Y
// to produce the same remainder
using System;
 
class GFG{
 
// Function to find
// the required number as M
static void printModulus(int X, int Y)
{
    // Finding the maximum
    // value among X and Y
    int n = Math.Max(X, Y);
 
    // Loop to iterate through
    // maximum value among X and Y.
    for (int i = 1; i <= n; i++) {
 
        // If the condition satisfies, then
        // print the value of M
        if (X % i == Y % i)
            Console.Write(i + " ");
    }
}
 
// Driver code
public static void Main()
{
    int X, Y;
    X = 10;
    Y = 20;
    printModulus(X, Y);
}
}
 
// This code is contributed by AbhiThakur

Javascript

 
// Javascript program to find numbers
// that divide X and Y
// to produce the same remainder
 
// Function to find
// the required number as M
function printModulus(X, Y)
{
    // Finding the maximum
    // value among X and Y
    var n = Math.max(X, Y);
 
    // Loop to iterate through
    // maximum value among X and Y.
    for (var i = 1; i <= n; i++) {
 
        // If the condition satisfies, then
        // print the value of M
        if (X % i == Y % i)
            document.write(i+" ");
    }
}
 
// Driver code
X = 10;
Y = 20;
printModulus(X, Y);
 
// This code is contributed by noob2000.

输出：

1 2 5 10

时间复杂度： O(max(X，Y))
高效方法：假设Y比X大D的差。

• 那么Y可以表示为

Y = X + D
and
Y % M = (X + D) % M
      = (X % M) + (D % M)

• 现在，条件变为X％M和X％M + D％M是否相等。
• 在此，由于X％M在两侧都是公共的，所以如果对于某些M， D％M ＝ 0 ，则M的值为真。
• 因此，M的要求值将成为D的因数。

下面是上述方法的实现：

CPP

// C++ program to find numbers
// that divide X and Y to
// produce the same remainder
 
#include 
using namespace std;
 
// Function to print all the possible values
// of M such that X % M = Y % M
void printModulus(int X, int Y)
{
    // Finding the absolute difference
    // of X and Y
    int d = abs(X - Y);
 
    // Iterating from 1
    int i = 1;
 
    // Loop to print all the factors of D
    while (i * i <= d) {
 
        // If i is a factor of d, then print i
        if (d % i == 0) {
            cout << i << " ";
 
            // If d / i is a factor of d,
            // then print d / i
            if (d / i != i)
                cout << d / i << " ";
        }
        i++;
    }
}
 
// Driver code
int main()
{
 
    int X = 10;
    int Y = 26;
 
    printModulus(X, Y);
    return 0;
}

Java

// Java program to find numbers
// that divide X and Y to
// produce the same remainder
import java.util.*;
 
class GFG{
  
// Function to print all the possible values
// of M such that X % M = Y % M
static void printModulus(int X, int Y)
{
    // Finding the absolute difference
    // of X and Y
    int d = Math.abs(X - Y);
  
    // Iterating from 1
    int i = 1;
  
    // Loop to print all the factors of D
    while (i * i <= d) {
  
        // If i is a factor of d, then print i
        if (d % i == 0) {
            System.out.print(i+ " ");
  
            // If d / i is a factor of d,
            // then print d / i
            if (d / i != i)
                System.out.print(d / i+ " ");
        }
        i++;
    }
}
  
// Driver code
public static void main(String[] args)
{
  
    int X = 10;
    int Y = 26;
  
    printModulus(X, Y);
}
}
 
// This code is contributed by Princi Singh

Python3

# Python program to find numbers
# that divide X and Y to
# produce the same remainder
 
# Function to prall the possible values
# of M such that X % M = Y % M
def printModulus(X, Y):
    # Finding the absolute difference
    # of X and Y
    d = abs(X - Y);
 
    # Iterating from 1
    i = 1;
 
    # Loop to prall the factors of D
    while (i * i <= d):
 
        # If i is a factor of d, then pri
        if (d % i == 0):
            print(i, end="");
 
            # If d / i is a factor of d,
            # then prd / i
            if (d // i != i):
                print(d // i, end=" ");
         
        i+=1;
     
 
 
# Driver code
if __name__ == '__main__':
 
    X = 10;
    Y = 26;
 
    printModulus(X, Y);
 
# This code contributed by Princi Singh

C＃

// C# program to find numbers
// that divide X and Y to
// produce the same remainder
using System;
 
public class GFG{
   
// Function to print all the possible values
// of M such that X % M = Y % M
static void printModulus(int X, int Y)
{
    // Finding the absolute difference
    // of X and Y
    int d = Math.Abs(X - Y);
   
    // Iterating from 1
    int i = 1;
   
    // Loop to print all the factors of D
    while (i * i <= d) {
   
        // If i is a factor of d, then print i
        if (d % i == 0) {
            Console.Write(i+ " ");
   
            // If d / i is a factor of d,
            // then print d / i
            if (d / i != i)
                Console.Write(d / i+ " ");
        }
        i++;
    }
}
   
// Driver code
public static void Main(String[] args)
{
   
    int X = 10;
    int Y = 26;
   
    printModulus(X, Y);
}
}
  
// This code contributed by Princi Singh

Java脚本

    // Javascript program to find numbers
    // that divide X and Y to
    // produce the same remainder
 
    // Function to print all the possible values
    // of M such that X % M = Y % M
    function printModulus(X, Y)
    {
     
      // Finding the absolute difference
      // of X and Y
      var d = Math.abs(X - Y);
 
      // Iterating from 1
      var i = 1;
 
      // Loop to print all the factors of D
      while (i * i <= d) {
 
        // If i is a factor of d, then print i
        if (d % i == 0) {
          document.write(i + " ");
 
          // If d / i is a factor of d,
          // then print d / i
          if (d / i != i)
            document.write(parseInt(d / i) + " ");
        }
        i++;
      }
    }
 
    // Driver code
    var X = 10;
    var Y = 26;
    printModulus(X, Y);
 
// This code is contributed by rrrtnx.
  

输出：

1 16 2 8 4

时间复杂度分析O(sqrt(D)) ，其中D是值X和Y之间的差。