Find product of GCDs of all possible 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: For the reason that reply could be very giant 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 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 unravel 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.

Comply with the steps talked about under to implement the thought:

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.

Beneath is the implementation of the above method.

C++

#embody <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 foremost()

{

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 unravel the issue comply with the under 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 could 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 completed utilizing Modular exponentiation technique. And eventually we have to multiply the remaining sample energy that is the same as Mpercenti for i^throw.

Additionally, we are able to take into account the row because the minimal of N and M to scale back time complexity additional.