Program to print numbers such that no two consecutive numbers are co-prime and every three consecutive numbers are co-prime Using C++

C++Server Side ProgrammingProgramming

In this tutorial, we will be discussing a program to print numbers such that no two consecutive numbers are co-prime and every three consecutive numbers are coprime.

In this, we will be given an integer N, we have to print N integers less than 109 such that no two consecutive numbers are coprime but a pair of 3 consecutive integers must be coprime.

For example, let us say we have the integer 4. Then the numbers that follow both of the above-provided conditions are

6 15 35 14

Example

#include <bits/stdc++.h>
using namespace std;
#define limit 1000000000
#define MAX_PRIME 2000000
#define MAX 1000000
#define I_MAX 50000
map<int, int> map1;
int b[MAX];
int p[MAX];
int j = 0;
bool prime[MAX_PRIME + 1];
void sieve(int n){
   memset(prime, true, sizeof(prime));
   for (int p = 2; p * p <= n; p++){
      if (prime[p] == true){
         for (int i = p * p; i <= n; i += p)
            prime[i] = false;
      }
   }
   for (int p = 2; p <= n; p++){
      if (prime[p]) {
         b[j++] = p;
      }
   }
}
int gcdiv(int a, int b){
   if (b == 0)
      return a;
   return gcdiv(b, a % b);
}
//printing the required series
void print_elements(int n){
   sieve(MAX_PRIME);
   int i, g, k, l, m, d;
   int ar[I_MAX + 2];
   for (i = 0; i < j; i++){
      if ((b[i] * b[i + 1]) > limit)
         break;
      p[i] = b[i];
      map1[b[i] * b[i + 1]] = 1;
   }
   d = 550;
   bool flag = false;
   for (k = 2; (k < d - 1) && !flag; k++){
      for (m = 2; (m < d) && !flag; m++){
         for (l = m + k; l < d; l += k){
            if (((b[l] * b[l + k]) < limit)
               && (l + k) < d && p[i - 1] != b[l + k]
               && p[i - 1] != b[l] && map1[b[l] * b[l + k]] != 1){
                  if (map1[p[i - 1] * b[l]] != 1){
                     p[i] = b[l];
                     map1[p[i - 1] * b[l]] = 1;
                     i++;
                  }
            }
            if (i >= I_MAX) {
               flag = true;
               break;
            }
         }
      }
   }
   for (i = 0; i < n; i++)
      ar[i] = p[i] * p[i + 1];
   for (i = 0; i < n - 1; i++)
      cout << ar[i] << " ";
   g = gcdiv(ar[n - 1], ar[n - 2]);
   cout << g * 2 << endl;
}
int main(){
   int n = 4;
   print_elements(n);
   return 0;
}

Output

6 15 35 14
raja
Published on 01-Nov-2019 10:51:09
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