Question

A set of a set of points over a straight line is defined as correlative to some k if the absoute difference between any two point is multiple of k given n (2<=n=100000) points and some integer k (1<=k=1000) your task to find the largest set exists. n and k will be in first line of the input n line will follow each one single integer representing the location of the one point which is correlated to k in the first line of input remaing line will contain the points of the set one per line in increasing order

Answer 1

Answer:

set of points over a straight line is defined as correlative to some k if the absoute difference between any two point is multiple of k given n (2<=n=100000) points and some integer k (1<=k=1000) your task to find the largest set exists. n and k will be in first line of the input n line will follow each one single integer representing the location of the one point which is correlated to k in the first line of input remaing line will contain the points of the set one per line in increasing orderset of points over a straight line is defined as correlative to some k if the absoute difference between any two point is multiple of k given n (2<=n=100000) points and some integer k (1<=k=1000) your task to find the largest set exists. n and k will be in first line of the input n line will follow each one single integer representing the location of the one point which is correlated to k in the first line of input remaing line will contain the points of the set one per line in increasing order.

Answer 2

Answer:

#include <bits/stdc++.h>

using namespace std;

// function to find remainder set

int findSet(int arr[], int n, int k, int m) {

vector<int> 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 <<m<< "\n";

for (int j = 0; j < m; j++){

 cout << remainder_set[i][j] << "\n";  

     

}

return 1;

}

}

return 0;

}

// driver program

int main() {

   int n,k;

   cin>>n>>k;

int arr[n];

for(int i=0;i<n;i++)

   cin>>arr[i];

int z;

int m = sizeof(arr)/sizeof(int);

for(int i=m;i>0;i--)

{

   z=findSet(arr, n, k, i);

   if(z==1)

   break;

}

}

Explanation: