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Discover product of GCDs of all attainable row-column pair of given Matrix


Given a matrix of dimension N*M, the duty is to seek out the product of all attainable pairs of (i, j) the place i and j are the row quantity and column quantity respectively.

Observe: Because the reply might be very massive output the reply modulo 1000000007.

Examples:

Enter: N = 5, M = 6
Output: 5760
Rationalization: The values of GCD of every attainable pair

1 1 1 1 1 1
1 2 1 2 1 2
1 3 1 1 3
1 2 1 4 1 2
1 1 1 1 5 1

The product of grid = 1*1*1*1*1*1*1*2*1*2*1*2*1*1*3*1*1*3*1*2*1*4*1*2*1*1*1*1*5*1 = 5760

Enter: N = 34, M = 46
Output: 397325354

Naive Strategy: To resolve the issue traverse all of the attainable pairs of row and column and discover the GCD of them and multiply them with the required reply.

Observe the steps talked about beneath to implement the concept:

  • Initialize a variable ans = 1 to retailer the product.
  • Iterate from i = 1 to N:
    • For every worth of i traverse from 1 to M.
    • Calculate the GCD of every pair.
    • Multiply this with ans.
  • Return the ultimate worth of ans because the required reply.

Under is the implementation of the above method.

C++

  

#embrace <bits/stdc++.h>

utilizing namespace std;

  

int M = 1e9 + 7;

  

int gridPower(int n, int m)

{

    lengthy lengthy ans = 1;

    for (int i = 1; i <= n; i++) {

        for (int j = 1; j <= m; j++) {

            ans = (ans * __gcd(i, j)) % M;

        }

    }

    return ans;

}

  

int important()

{

    int N = 5, M = 6;

  

    

    cout << gridPower(N, M) << endl;

    return 0;

}

Time Complexity: O(N*M*log(min(N, M)))
Auxiliary House: O(1)

Environment friendly Strategy: To resolve the issue observe the beneath thought:

It may be noticed that for each row, a sample is fashioned until the row quantity and after that, the identical sample repeats.

1 1 1 1 1 1
1 2 1 2 1 2
1 1 3 1 1 3
1 2 1 4 1 2
1 1 1 1 5 1

For instance within the above grid of 4 rows and 6 columns

In row 1, all of the values are 1
In row 2, until index 2 a sample is fashioned and after that very same sample repeats
In row 3, until index 3 a sample is fashioned and after that very same sample repeats

Comparable observations might be made for all different rows.

Therefore for each row, we solely want to seek out the sample as soon as and multiply that sample energy the variety of instances it happens. This may be accomplished utilizing Modular exponentiation methodology. And at last we have to multiply the remaining sample energy that is the same as Mpercenti for ith row.

Additionally, we are able to think about the row because the minimal of N and M to cut back time complexity additional.

Under is the implementation of the above method:

C++

  

#embrace <bits/stdc++.h>

utilizing namespace std;

  

int M = 1e9 + 7;

  

int fastpower(lengthy lengthy a, int p)

{

    lengthy lengthy res = 1;

    whereas (p > 0) {

        if (p % 2)

            res = (res * a) % M;

        p /= 2;

        a = (a * a) % M;

    }

    return res;

}

  

int gridPower(int n, int m)

{

    lengthy lengthy res = 1;

    for (int i = 1; i <= min(n, m); i++) {

        lengthy lengthy patternPower = 1;

  

        

        

        lengthy lengthy patternOccurence = max(n, m) / i;

  

        

        

        lengthy lengthy stays = max(n, m) % i;

  

        

        

        lengthy lengthy remainsPower = 1;

  

        

        

        for (int j = 1; j <= i; j++) {

            patternPower = (patternPower * __gcd(i, j)) % M;

  

            if (j == stays)

                remainsPower = patternPower;

        }

  

        res = (res

               * fastpower(patternPower, patternOccurence))

              % M;

        res = (res * remainsPower) % M;

    }

    return res;

}

  

int important()

{

    int N = 5, M = 6;

  

    

    cout << gridPower(N, M) << endl;

    return 0;

}

Time Complexity: min(N, M)*min(N, M)*log(min(N, M))
Auxiliary House: O(1)

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