查找具有两个元素之差的m个元素的集合可被k整除

给定n个正整数和正整数k的数组，找到一组正好m个元素，以使任意两个元素的差等于k。
例子 ：

Input : arr[] = {4, 7, 10, 6, 9},
        k = 3, m = 3
Output : Yes 
         4 7 10

Input : arr[] = {4, 7, 10, 6, 9}, 
        k = 12, m = 4
Output : No

Input : arr[] = {4, 7, 10, 6, 9}, 
        k = 3, m = 4
Output : No

方法：要解决此问题，只需将元素除以k即可记录余数。创建一个大小为k的多维数组restder_set [] []，其索引显示余数，并且该数组的元素将除以k时将按照其对应的余数进行元素化。例如，如果arr [i]％k = 3，则arr [i]将是restder_set [3]的元素，以此类推。现在，遍历余数集，就可以轻松获得一个大小大于或等于所需大小m的集合(如果存在)。并且可以肯定的是，该集合中任何元素的差都可以被k整除。

C++

// C++ program for finding reemainder set
#include 
using namespace std;
 
// function to find remainder set
void findSet(int arr[], int n, int k, int m) {
 
  vector remainder_set[k];
 
  // calculate remainder set array
  // and push element as per their remainder
  for (int i = 0; i < n; i++) {
    int rem = arr[i] % k;
    remainder_set[rem].push_back(arr[i]);
  }
 
  // check whether sizeof any remainder set
  // is equal or greater than m
  for (int i = 0; i < k; i++) {
    if (remainder_set[i].size() >= m) {
      cout << "Yes \n";
      for (int j = 0; j < m; j++)
        cout << remainder_set[i][j] << " ";     
      return;
    }
  }
 
  cout << "No";
}
 
// driver program
int main() {
  int arr[] = {5, 8, 9, 12, 13, 7, 11, 15};
  int k = 4;
  int m = 3;
  int n = sizeof(arr) / sizeof(arr[0]);
  findSet(arr, n, k, m);
}

Java

// Java program for finding reemainder set
import java.util.*;
class GFG
{
 
// function to find remainder set
static void findSet(int arr[], int n,
                    int k, int m)
{
    Vector []remainder_set = new Vector[k];
    for (int i = 0; i < k; i++)
    {
        remainder_set[i] = new Vector();
    }
     
    // calculate remainder set array
    // and push element as per their remainder
    for (int i = 0; i < n; i++)
    {
        int rem = arr[i] % k;
        remainder_set[rem].add(arr[i]);
    }
     
    // check whether sizeof any remainder set
    // is equal or greater than m
    for (int i = 0; i < k; i++)
    {
        if (remainder_set[i].size() >= m)
        {
            System.out.println("Yes");
            for (int j = 0; j < m; j++)
                    System.out.print(remainder_set[i].get(j) + " ");    
            return;
        }
    }
    System.out.print("No");
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = {5, 8, 9, 12, 13, 7, 11, 15};
    int k = 4;
    int m = 3;
    int n = arr.length;
    findSet(arr, n, k, m);
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to find exacktly m-element set
# where differencec of any two is divisible by k
def findSet( arr, k, m):
 
    arr_size = len(arr);
    remainder_set=[0]*k;
 
    # initialize remainder set with blank array
    for i in range(k):
        remainder_set[i] = [];
 
    # calculate remainder set array
    # and push element as per their reainder
    for i in range(arr_size):
        rem = arr[i] % k;
        remainder_set[rem].append(arr[i]);
 
    # check whether sizeof any remainder set
    # is equal or greater than m
    for i in range(k):
        # if size exist then print yes and all elements
        if(len(remainder_set[i]) >= m):
            print("Yes");
            for j in range(m):
                print(remainder_set[i][j],end="");
                print(" ",end="");
 
            # return if remainder set found
            return;
 
    # print no if no remiander set found
    print("No");
 
arr = [5, 8, 9, 12, 13, 7, 11, 15];
k = 4; # constant k
m = 3; # size of set required
findSet(arr, k, m);
 
# This code is contributed by mits.

C#

// C# program for finding
// remainder set
using System;
using System.Collections.Generic;
 
class GFG
{
// function to find
// remainder set
static void findSet(int []arr, int n,
                    int k, int m)
{
    List[] remainder_set =
                       new List[k];
    for(int i = 0; i < k; i++)
        remainder_set[i] =
                       new List();
     
    // calculate remainder set
    // array and push element
    // as per their remainder
    for (int i = 0; i < n; i++)
    {
        int rem = arr[i] % k;
        remainder_set[rem].Add(arr[i]);
    }
     
    // check whether sizeof
    // any remainder set is
    // equal or greater than m
    for (int i = 0; i < k; i++)
    {
        if (remainder_set[i].Count >= m)
        {
        Console.Write("Yes \n");
        for (int j = 0; j < m; j++)
            Console.Write(remainder_set[i][j] +
                                          " ");    
        return;
        }
    }
     
    Console.Write("No");
    }
     
// Driver Code
static void Main()
{
    int []arr = new int[]{5, 8, 9, 12,
                          13, 7, 11, 15};
    int k = 4;
    int m = 3;
    int n = arr.Length;
    findSet(arr, n, k, m);
}
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

PHP

= $m)
        {
            print("Yes\n");
            for($j = 0; $j < $m; $j++)
            {
                print($remainder_set[$i][$j]);
                print(" ");
            }
 
            // return if remainder set found
            return;
        }
    }
 
    // print no if no remiander set found
    print("No");
}
 
$arr = array(5, 8, 9, 12, 13, 7, 11, 15);
$k = 4; // constant k
$m = 3; // size of set required
findset($arr, $k, $m);
?>

Javascript

 
// Javascript program to find exacktly m-element set
// where differencec of any two is divisible by k
 
 
function findSet(arr, k, m)
{
    let arr_size = arr.length;
    let remainder_set = []
    // initialize remainder set with blank array
    for (let i = 0; i < k; i++)
    {
        remainder_set[i] = new Array();
    }
 
    // calculate remainder set array
    // and push element as per their reainder
    for (let i = 0; i < arr_size; i++)
    {
        let rem = arr[i] % k;
        remainder_set[rem].push(arr[i]);
    }
 
    // check whether sizeof any remainder set
    // is equal or greater than m
    for(let i = 0; i < k; i++)
    {
        // if size exist then print yes and all elements
        if(remainder_set[i].length >= m)
        {
            document.write("Yes<br>");
            for(let j = 0; j < m; j++)
            {
                document.write(remainder_set[i][j]);
                document.write(" ");
            }
 
            // return if remainder set found
            return;
        }
    }
 
    // print no if no remiander set found
    document.write("No");
}
 
let arr = [5, 8, 9, 12, 13, 7, 11, 15];
let k = 4; // constant k
let m = 3; // size of set required
findSet(arr, k, m);
 
 
// This code is contributed by _saurabh_jaiswal
 

输出：

Yes 
5 9 13