Sum and sum of squares of digits -coprime - c++

Question

For the paragraph L, R
its sum of digits and the sum of its squares (in the decimal) is co-
prime.
Count how many numbers in the paragraph L, R meet the above conditions
I was stuck on the sub21
exceeds time limit when R = 10^8 and Max R = 10^18:
#include<bits/stdc++.h>
using namespace std;
class tinhtong
{
public:
long long getSum(long long n)
{
long long int sum = 0;
while (n != 0)
{
sum = sum + (n % 10);
n = n/10;
}
return sum;
}
};
class binhphuong
{
public:
long long getpow(long long n)
{
long long poww = 0;
while (n != 0)
{
poww = poww + (n % 10)*(n % 10);
n = n/10;
}
return poww;
}
};
int main()
{
tinhtong g;
binhphuong h;
long long TONG=0,k,l,ucln;
long long int m,n;
cin>>n>>m;
for(n;n<=m;n++)
{
ucln=0;
k=g.getSum(n);
l=h.getpow(n);
while(k!=0 && l!=0)
{
if(k>l)
k-=l;
else
l-=k;
}
if(k==0)
ucln=l;
else
ucln=k;
if (ucln==1)
TONG++;
}
cout<<TONG;
return 0;
}

Answer 1

As mentioned in comments, you can calculate the sum and the sum of squares in one loop, avoiding multiple calls of same % operation.
Moreover, to calculate that two numbers are coprime, it is better to use the version of Euclide's algorithm that uses modulo instead of substractions. The code is below.
In this code, I use the fact that the sum of squares is greater or equal the sum.
I also use the fact that if the sum is a multiple of 2, it is also the case for the sum of squares.
If this code is not fast enough, you can try the suggestion in a comment to first convert the number to a string.
Another possibility would be the check if both numbers are multiple of 3 or 5, before calculating the GCD, as GCD calculation can be longer.
#include <iostream>
#include <tuple>
std::pair<long long, long long> getSum_pow(long long n) {
long long int sum = 0;
long long int sumP = 0;
while (n != 0) {
long long r = n%10;
sum += r;
sumP += r*r;
n = n/10;
}
return {sumP, sum};
};
int main() {
long long TONG = 0, a, b;
long long int m, n;
std::cin >> n >> m;
for(; n <= m; n++) {
std::tie (a, b) = getSum_pow(n);
if (b%2 == 0) continue;
while (b != 0) {
a = a%b;
std::swap (a, b);
}
if (a == 1) {
TONG++;
}
}
std::cout << TONG;
return 0;
}