Given an array a of size N. The task is to find the number of pairs such that gcd(a[i], a[j]) is equal to 1, where 1 ? i < j ? N.
Examples:
Input : a[] = {1, 2, 4, 6}
Output : 3
{1, 2}, {1, 4}, {1, 6} are such pairsInput : a[] = {1, 2, 3, 4, 5, 6}
Output : 11
Approach :
The answer is to sum of ?(X) * C(D(X), 2) overall integer X. Where, ?(X) is Mobius function, C(N, K) is the selection of K things from N and D(X) is the number of integers in the given sequence that are divisible by X.
The correctness of the solution follows from the fact that we can do an inclusion-exclusion principle solution and to show that it is, in fact, equal to our answer. That means that we will add to the answer the number of pairs that are divisible by some intermediate (in the IEP) product D if D is formed by multiplication of even number of prime numbers and subtract this number of pairs otherwise.
So, we get:
- 1 for addition, because that is Möbius function for square-free numbers with even number of prime divisors.
- -1 for subtraction, that is Mobius function for square-free numbers with an odd number of prime divisors.
- 0 for square-free numbers. By the definition, they can’t occur in our IEP solution.
Below is the implementation of the above approach :
// CPP program to find the number of pairs // such that gcd equals to 1 #include <bits/stdc++.h> using namespace std;
#define N 100050 int lpf[N], mobius[N];
// Function to calculate least // prime factor of each number void least_prime_factor()
{ for ( int i = 2; i < N; i++)
// If it is a prime number
if (!lpf[i])
for ( int j = i; j < N; j += i)
// For all multiples which are not
// visited yet.
if (!lpf[j])
lpf[j] = i;
} // Function to find the value of Mobius function // for all the numbers from 1 to n void Mobius()
{ for ( int i = 1; i < N; i++) {
// If number is one
if (i == 1)
mobius[i] = 1;
else {
// If number has a squared prime factor
if (lpf[i / lpf[i]] == lpf[i])
mobius[i] = 0;
// Multiply -1 with the previous number
else
mobius[i] = -1 * mobius[i / lpf[i]];
}
}
} // Function to find the number of pairs // such that gcd equals to 1 int gcd_pairs( int a[], int n)
{ // To store maximum number
int maxi = 0;
// To store frequency of each number
int fre[N] = { 0 };
// Find frequency and maximum number
for ( int i = 0; i < n; i++) {
fre[a[i]]++;
maxi = max(a[i], maxi);
}
least_prime_factor();
Mobius();
// To store number of pairs with gcd equals to 1
int ans = 0;
// Traverse through the all possible elements
for ( int i = 1; i <= maxi; i++) {
if (!mobius[i])
continue ;
int temp = 0;
for ( int j = i; j <= maxi; j += i)
temp += fre[j];
ans += temp * (temp - 1) / 2 * mobius[i];
}
// Return the number of pairs
return ans;
} // Driver code int main()
{ int a[] = { 1, 2, 3, 4, 5, 6 };
int n = sizeof (a) / sizeof (a[0]);
// Function call
cout << gcd_pairs(a, n);
return 0;
} |
// Java program to find the number of pairs // such that gcd equals to 1 class GFG
{ static int N = 100050 ;
static int []lpf = new int [N];
static int []mobius = new int [N];
// Function to calculate least // prime factor of each number static void least_prime_factor()
{ for ( int i = 2 ; i < N; i++)
// If it is a prime number
if (lpf[i] == 0 )
for ( int j = i; j < N; j += i)
// For all multiples which are not
// visited yet.
if (lpf[j] == 0 )
lpf[j] = i;
} // Function to find the value of Mobius function // for all the numbers from 1 to n static void Mobius()
{ for ( int i = 1 ; i < N; i++)
{
// If number is one
if (i == 1 )
mobius[i] = 1 ;
else
{
// If number has a squared prime factor
if (lpf[i / lpf[i]] == lpf[i])
mobius[i] = 0 ;
// Multiply -1 with the previous number
else
mobius[i] = - 1 * mobius[i / lpf[i]];
}
}
} // Function to find the number of pairs // such that gcd equals to 1 static int gcd_pairs( int a[], int n)
{ // To store maximum number
int maxi = 0 ;
// To store frequency of each number
int []fre = new int [N];
// Find frequency and maximum number
for ( int i = 0 ; i < n; i++)
{
fre[a[i]]++;
maxi = Math.max(a[i], maxi);
}
least_prime_factor();
Mobius();
// To store number of pairs with gcd equals to 1
int ans = 0 ;
// Traverse through the all possible elements
for ( int i = 1 ; i <= maxi; i++)
{
if (mobius[i] == 0 )
continue ;
int temp = 0 ;
for ( int j = i; j <= maxi; j += i)
temp += fre[j];
ans += temp * (temp - 1 ) / 2 * mobius[i];
}
// Return the number of pairs
return ans;
} // Driver code public static void main (String[] args)
{ int a[] = { 1 , 2 , 3 , 4 , 5 , 6 };
int n = a.length;
// Function call
System.out.print(gcd_pairs(a, n));
} } // This code is contributed by PrinciRaj1992 |
# Python3 program to find the number of pairs # such that gcd equals to 1 N = 100050
lpf = [ 0 for i in range (N)]
mobius = [ 0 for i in range (N)]
# Function to calculate least # prime factor of each number def least_prime_factor():
for i in range ( 2 , N):
# If it is a prime number
if (lpf[i] = = 0 ):
for j in range (i, N, i):
# For all multiples which are not
# visited yet.
if (lpf[j] = = 0 ):
lpf[j] = i
# Function to find the value of Mobius function # for all the numbers from 1 to n def Mobius():
for i in range ( 1 , N):
# If number is one
if (i = = 1 ):
mobius[i] = 1
else :
# If number has a squared prime factor
if (lpf[ (i / / lpf[i]) ] = = lpf[i]):
mobius[i] = 0
# Multiply -1 with the previous number
else :
mobius[i] = - 1 * mobius[i / / lpf[i]]
# Function to find the number of pairs # such that gcd equals to 1 def gcd_pairs(a, n):
# To store maximum number
maxi = 0
# To store frequency of each number
fre = [ 0 for i in range (N)]
# Find frequency and maximum number
for i in range (n):
fre[a[i]] + = 1
maxi = max (a[i], maxi)
least_prime_factor()
Mobius()
# To store number of pairs with gcd equals to 1
ans = 0
# Traverse through the all possible elements
for i in range ( 1 , maxi + 1 ):
if (mobius[i] = = 0 ):
continue
temp = 0
for j in range (i, maxi + 1 , i):
temp + = fre[j]
ans + = temp * (temp - 1 ) / / 2 * mobius[i]
# Return the number of pairs
return ans
# Driver code a = [ 1 , 2 , 3 , 4 , 5 , 6 ]
n = len (a)
# Function call print (gcd_pairs(a, n))
# This code is contributed by Mohit Kumar |
// C# program to find the number of pairs // such that gcd equals to 1 using System;
class GFG
{ static int N = 100050;
static int []lpf = new int [N];
static int []mobius = new int [N];
// Function to calculate least // prime factor of each number static void least_prime_factor()
{ for ( int i = 2; i < N; i++)
// If it is a prime number
if (lpf[i] == 0)
for ( int j = i; j < N; j += i)
// For all multiples which are not
// visited yet.
if (lpf[j] == 0)
lpf[j] = i;
} // Function to find the value of Mobius function // for all the numbers from 1 to n static void Mobius()
{ for ( int i = 1; i < N; i++)
{
// If number is one
if (i == 1)
mobius[i] = 1;
else
{
// If number has a squared prime factor
if (lpf[i / lpf[i]] == lpf[i])
mobius[i] = 0;
// Multiply -1 with the previous number
else
mobius[i] = -1 * mobius[i / lpf[i]];
}
}
} // Function to find the number of pairs // such that gcd equals to 1 static int gcd_pairs( int []a, int n)
{ // To store maximum number
int maxi = 0;
// To store frequency of each number
int []fre = new int [N];
// Find frequency and maximum number
for ( int i = 0; i < n; i++)
{
fre[a[i]]++;
maxi = Math.Max(a[i], maxi);
}
least_prime_factor();
Mobius();
// To store number of pairs with gcd equals to 1
int ans = 0;
// Traverse through the all possible elements
for ( int i = 1; i <= maxi; i++)
{
if (mobius[i] == 0)
continue ;
int temp = 0;
for ( int j = i; j <= maxi; j += i)
temp += fre[j];
ans += temp * (temp - 1) / 2 * mobius[i];
}
// Return the number of pairs
return ans;
} // Driver code public static void Main (String[] args)
{ int []a = { 1, 2, 3, 4, 5, 6 };
int n = a.Length;
// Function call
Console.Write(gcd_pairs(a, n));
} } // This code is contributed by Rajput-Ji |
<script> // Javascript program to find the number of pairs // such that gcd equals to 1 var N = 100050;
var lpf = Array(N).fill(0);
var mobius = Array(N).fill(0);
// Function to calculate least
// prime factor of each number
function least_prime_factor() {
for (i = 2; i < N; i++)
// If it is a prime number
if (lpf[i] == 0)
for (j = i; j < N; j += i)
// For all multiples which are not
// visited yet.
if (lpf[j] == 0)
lpf[j] = i;
}
// Function to find the value of Mobius function
// for all the numbers from 1 to n
function Mobius() {
for (i = 1; i < N; i++) {
// If number is one
if (i == 1)
mobius[i] = 1;
else {
// If number has a squared prime factor
if (lpf[i / lpf[i]] == lpf[i])
mobius[i] = 0;
// Multiply -1 with the previous number
else
mobius[i] = -1 * mobius[i / lpf[i]];
}
}
}
// Function to find the number of pairs
// such that gcd equals to 1
function gcd_pairs(a , n) {
// To store maximum number
var maxi = 0;
// To store frequency of each number
var fre = Array(n+1).fill(0);
// Find frequency and maximum number
for (i = 0; i < n; i++) {
fre[a[i]]++;
maxi = Math.max(a[i], maxi);
}
least_prime_factor();
Mobius();
// To store number of pairs with gcd equals to 1
var ans = 0;
// Traverse through the all possible elements
for (i = 1; i <= maxi; i++) {
if (mobius[i] == 0)
continue ;
var temp = 0;
for (j = i; j <= maxi; j += i)
temp = parseInt(temp+fre[j]);
ans += parseInt(temp * (temp - 1) / 2 * mobius[i]);
}
// Return the number of pairs
return ans;
}
// Driver code
var a = [ 1, 2, 3, 4, 5, 6 ];
var n = a.length;
// Function call
document.write(gcd_pairs(a, n));
// This code contributed by Rajput-Ji </script> |
11
Time Complexity: O(N2)
Auxiliary Space: O(N)