Number of distinct rectangles in which diagonal is passing in N squares - algorithm

Question

I'm solving CS problem and I need little help. I have number N, and I need to count the number of distinct rectangles in which diagonal is passing in N squares if the rectangle is splited on rectangles with size 1x1. This picture will help you understand.
This picture is showing all 4 combinations if N = 4, actually the rectangles in which the diagonal is passing in 4 squares are with sizes 1x4, 2x3, 4x2 and 4x4.
I found the formula if we have given the two sizes of the rectangles it is:
A + B - gcd(A,B)
since N<=10^6, i go up to 10^6 and check for each N the divisors of N, complexity of that is O(Nsqrt(N)), since the divisors of A is gcd(A,B)i solve the system of equations
q is divisor of A and q is gcd(A,B)
A+B-q=N and gcd(A,B)=q
I solved this in O(Nsqrt(N)*log(N))
where i assume that log(N) is the time to find gcd of two numbers.
Because the time limit is 3 seconds it fails on time. I need help on optimizing the solution.
Update: Here is my code:
#include <bits/stdc++.h>
#define ll long long
using namespace std;
int a;
int gcd(int a, int b) {
if(b>a) swap(a,b);
if(b==0) return a;
return gcd(b, a%b);
}
bool valid(int n, int m, int gc, int a) {
if(n+m-gc==a) return true;
return false;
}
int main() {
cin>>a;
int counter=0;
for(int i=1;i<=a/2;i++) {
for(ll j=1;j<=sqrt(i);j++) {
if(i%j==0) {
if(j!=i/j) {
int m1 = a+j-i;
int div=i/j;
int m2 = a+div-i;
if(valid(i, m1, j, a)) {
if(gcd(i, m1)==j)
counter++;
}
if(valid(i, m2, i/j, a)) {
if(gcd(i,m2)==i/j)
counter++;
}
}
else {
int m1 = a+j-i;
if(valid(i, m1, j, a)) {
if(gcd(i, m1)==j)
counter++;
}
}
}
}
}
cout<<counter+1;
return 0;
}
Thanks in advance.

Answer 1

Although O(n*sqrt(n)*log(n)) sounds a bit much for n <= 10^6, and you likely need a slightly better algorithm, your code supports some optimizations:
int gcd(int a, int b) {
if(b>a) swap(a,b);
if(b==0) return a;
return gcd(b, a%b);
}
Get rid of the swap, it will work just fine without it.
While you're at it, get rid of the recursion too:
int gcd(int a, int b) {
while (b) {
int r = a % b;
a = b;
b = r;
}
return a;
}
Next:
for(int i=1;i<=a/2;i++) {
for(ll j=1;j<=sqrt(i);j++) {
Compute a/2 and sqrt(i) outside of their respective loops. There is no need to compute it at each iteration. The compiler may or may not be smart enough (or set up) to do this itself, but you shouldn't rely on it, especially in an online judge setting.
You can also precompute i / j further down so as to not recompute it every time. A lot of divisions can be slow.
Next, do you really need long long for j? i is an int, and j goes up to its square root. So you don't need long long for j, use int.