Take a range of 0 - 1000 Numbers and find the Prime numbers in that range. Store the prime numbers in a 2D Array, the first dimension represents the range 0-100, 100-200, and so on. While the second dimension represents the prime numbers in that range
Answers
Answer:
if its in java then
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class ExArrayPrimeNumberMatrix
{
// Function to check a number is prime or not
boolean isPrime(int n)
{
int c = 0;
for(int i = 1; i<=n; i++)
{
if(n%i == 0)
c++;
}
if(c == 2)
return true;
else
return false;
}
public static void main(String args[])throws IOException
{
// create object of Prime number matrix.
ExArrayPrimeNumberMatrix ob = new ExArrayPrimeNumberMatrix();
// create object of buffer stream.
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
// enter the number of rows and column.
System.out.print("Enter the number of rows : ");
int m=Integer.parseInt(br.readLine());
System.out.print("Enter the number of columns : ");
int n=Integer.parseInt(br.readLine());
// 2D array for storing 'm*n' prime numbers
int A[][]=new int[m][n];
// 1D array for storing 'm*n' prime numbers
int B[] = new int [m*n];
// For taking natural numbers
int i = 0, j;
int k = 1;
// for iD Array.
while(i < m*n)
{
if(ob.isPrime(k)==true)
{
B[i] = k;
i++;
}
k++;
}
// for 2D Array.
int x = 0;
for(i=0; i<m; i++)
{
for(j=0; j<n; j++)
{
A[i][j] = B[x];
x++;
}
}
// printing the result in 2D Array.
System.out.println("The Final Array is : ");
for(i=0; i<m; i++)
{
for(j=0; j<n; j++)
{
System.out.print(A[i][j]+"\t");
}
System.out.println();
}
}
}
for c++
// C++ program to print print all primes in a range
// using concept of Segmented Sieve
#include <iostream.h>
// This functions finds all primes smaller than limit
// using simple sieve of eratosthenes. It stores found
// primes in vector prime[]
void simpleSieve(int limit, vector<int>& prime)
{
bool mark[limit + 1];
memset(mark, false, sizeof(mark));
for (int i = 2; i <= limit; ++i) {
if (mark[i] == false) {
// If not marked yet, then its a prime
prime.push_back(i);
for (int j = i; j <= limit; j += i)
mark[j] = true;
}
}
}
// Finds all prime numbers in given range using
// segmented sieve
void primesInRange(int low, int high)
{
// Comput all primes smaller or equal to
// square root of high using simple sieve
int limit = floor(sqrt(high)) + 1;
vector<int> prime;
simpleSieve(limit, prime);
// Count of elements in given range
int n = high - low + 1;
// Declaring boolean only for [low, high]
bool mark[n + 1];
memset(mark, false, sizeof(mark));
// Use the found primes by simpleSieve() to find
// primes in given range
for (int i = 0; i < prime.size(); i++) {
// Find the minimum number in [low..high] that is
// a multiple of prime[i] (divisible by prime[i])
int loLim = floor(low / prime[i]) * prime[i];
if (loLim < low)
loLim += prime[i];
if(loLim==prime[i])
loLim += prime[i];
/* Mark multiples of prime[i] in [low..high]:
We are marking j - low for j, i.e. each number
in range [low, high] is mapped to [0, high - low]
so if range is [50, 100] marking 50 corresponds
to marking 0, marking 51 corresponds to 1 and
so on. In this way we need to allocate space only
for range */
for (int j = loLim; j <= high; j += prime[i])
mark[j - low] = true;
}
// Numbers which are not marked in range, are prime
for (int i = low; i <= high; i++)
if (!mark[i - low])
cout << i << " ";
}
// Driver program to test above function
int main()
{
int low = 10, high = 100;
primesInRange(low, high);
return 0;
}
Answer:
public class Prime2DArray {
// method to check number is prime number or not
public static boolean isPrime(int n) {
if (n == 0 || n == 1) {
return false;
}
int i = 2;
while (i <= n / 2) {
if (n % i == 0) {
return false;
}
i++;
}
return true;
}
public static void main(String[] args) {
int prime[][] = new int[11][2]; //creation of 2d array
int k = 0;
int count = 0; //to count how many prime number is there
for (int i = 0; i <= 1000; i++) {
boolean x = isPrime(i);
if (x) {
count++;
}
if (i % 100 == 0 && i != 0) {
prime[k][0] = i;
prime[k][1] = count;
count = 0;
k++;
}
}
int j = 0;
while (j < 2) {
for (int n = 0; n < 10; n++) {
System.out.print(prime[n][j] + " ");
}
System.out.println();
j++;
}
}
}
Explanation: