Observations

Consider the case where \((C + x)\) has \(d\) digits. Then \(x\) must fall into the intersection of:

• \(1 \leq x \leq D\), and
• \(10^{d-1} - C \leq x \leq 10^d - 1 - C\).

Let \(L = \mathrm{max}(1, 10^{d-1} - C)\) and \(R = \mathrm{min}(D, 10^d - 1 - C)\). If \(L > R\), there is no such \(x\); if \(L \leq R\), it can take \(L \leq x \leq R\).

Since we assumed that \((C + x)\) has \(d\) digits, \(f(C, C+x) = C \times 10^d + (C+x)\). Thus, the answer for this range is the number of perfect squares between \(C \times 10^d + C + L\) and\(C \times 10^d + C + R\), inclusive.

As there are \(\lfloor \sqrt{k} \rfloor\) perfect squares less than or equal to \(k\), the sought count can be represented as \(\lfloor \sqrt{C \times 10^d + C + R} \rfloor - \lfloor \sqrt{C \times 10^d + C + L - 1} \rfloor\).

To find \(\lfloor \sqrt{k} \rfloor\), it is faster to use std::sqrt instead of performing a binary search.

The answer to the original problem can be found as the sum of the counts in the computation above for \(d = 1, \dots, \lfloor \log_{10} (C+D) \rfloor + 1\).

Assuming that square roots can be found in a constant time, the time complexity is \(O(\log(C+D))\), which is fast enough.

Sample code (C++)

#include <iostream>
using std::cin;
using std::cout;
using std::min;
using std::max;
#include <cmath>

typedef long long int ll;

// floor(sqrt(x))
ll f (ll x) {
	ll y = sqrt(x);
	while (y * y > x) y--;
	while ((y+1) * (y+1) <= x) y++;
	return y;
}
ll solve (ll c, ll d) {
	ll ans = 0;

	ll xmin = 1, xmax = 9, cshift = 10;
	while (xmin <= c + d) {
		ll l = max(xmin, c + 1);
		ll r = min(xmax, c + d);

		if (l <= r) {
			ll vl = c * cshift + l;
			ll vr = c * cshift + r;

			ans += (f(vr) - f(vl - 1));
		}

		xmin = xmin * 10;
		xmax = (xmax + 1) * 10 - 1;
		cshift *= 10;
	}

	return ans;
}

int main (void) {
	ll T;
	cin >> T;

	for (ll ti = 0; ti < T; ti++) {
		ll c, d;
		cin >> c >> d;

		ll ans = solve(c, d);
		cout << ans << "\n";
	}

	
	return 0;
}