Question

Write a Java program to store the modulus of every element of array A [ ] with 7 in array B[ ]. Print the elements of array B[ ]. Also display the sum of those elements divisible by 7. (Array A [ ] consists of 5 elements).
Sample input - 35 53 70 40 14 Output - 0 4 0 5 0
Sum of elements divisible by 7 : 119

Answer 1

Answer:

Step by step descriptive logic to count duplicate elements in array.

Input size and elements in array from user. ...

Initialize another variable count with 0 to store duplicate count.

To count total duplicate elements in given array we need two loops. ...

Run another inner loop to find first duplicate of current array element.

Explanation:

// C++ implementation of finding all k

// such that arr[i]%k is same for each i

#include<bits/stdc++.h>

using namespace std;

// Prints all k such that arr[i]%k is same for all i

void printEqualModNumbers (int arr[], int n)

{

// sort the numbers

sort(arr, arr + n);

// max difference will be the difference between

// first and last element of sorted array

int d = arr[n-1] - arr[0];

// Case when all the array elements are same

if(d==0){

cout<<"Infinite solution";

return;

}

// Find all divisors of d and store in

// a vector v[]

vector <int> v;

for (int i=1; i*i<=d; i++)

{

if (d%i == 0)

{

v.push_back(i);

if (i != d/i)

v.push_back(d/i);

}

}

// check for each v[i] if its modulus with

// each array element is same or not

for (int i=0; i<v.size(); i++)

{

int temp = arr[0]%v[i];

// checking for each array element if

// its modulus with k is equal to k or not

int j;

for (j=1; j<n; j++)

if (arr[j] % v[i] != temp)

break;

// if check is true print v[i]

if (j == n)

cout << v[i] <<" ";

}

}

// Driver function

int main()

{

int arr[] = {38, 6, 34};

int n = sizeof(arr)/sizeof(arr[0]);

printEqualModNumbers(arr, n);

return 0;

}