Greedy from small to large.

We can delete the numbers from small to large. Thus, previously removed numbers will not affect the future choices (if $$$x \lt y$$$, then $$$x$$$ cannot be a multiple of $$$y$$$). So an integer $$$x$$$ ($$$l\le x\le r$$$) can be removed if and only if $$$k\cdot x\le r$$$, that is, $$$x\le \left\lfloor\frac{r}{k}\right\rfloor$$$. The answer is $$$\max\left(\left\lfloor\frac{r}{k}\right\rfloor-l+1,0\right)$$$.

Time complexity: $$$\mathcal{O}(1)$$$ per test case.

#include <bits/stdc++.h>
#define all(s) s.begin(), s.end()
using namespace std;
using ll = long long;
using ull = unsigned long long;
 
const int _N = 1e5 + 5;
 
int T;
 
void solve() {
	int l, r, k; cin >> l >> r >> k;
	cout << max(r / k - l + 1, 0) << endl;
	return;
}
 
int main() {
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> T;
	while (T--) {
		solve();
	}
}

for _ in range(int(input())):
    l, r, k = map(int, input().split())
    print(max(r // k - l + 1, 0))

($$$\texttt{01}$$$ or $$$\texttt{10}$$$ exists) $$$\Longleftrightarrow$$$ (both $$$\texttt{0}$$$ and $$$\texttt{1}$$$ exist).

Each time we do an operation, if $$$s$$$ consists only of $$$\texttt{0}$$$ or $$$\texttt{1}$$$, we surely cannot find any valid indices. Otherwise, we can always perform the operation successfully. In the $$$i$$$-th operation, if $$$t_i=\texttt{0}$$$, we actually decrease the number of $$$\texttt{1}$$$-s by $$$1$$$, and vice versa. Thus, we only need to maintain the number of $$$\texttt{0}$$$-s and $$$\texttt{1}$$$-s in $$$s$$$. If any of them falls to $$$0$$$ before the last operation, the answer is NO, otherwise, the answer is YES.

Time complexity: $$$\mathcal{O}(n)$$$ per test case.

#include <bits/stdc++.h>
#define all(s) s.begin(), s.end()

using namespace std;
using ll = long long;

const int _N = 1e5 + 5;

void solve() {
	int n; cin >> n;
	string s, t; cin >> s >> t;
	int cnt0 = count(all(s), '0'), cnt1 = n - cnt0;
	for (int i = 0; i < n - 1; i++) {
		if (cnt0 == 0 || cnt1 == 0) {
			cout << "NO" << '\n';
			return;
		}
		if (t[i] == '1') cnt0--;
		else cnt1--;
	}
	cout << "YES" << '\n';
}

int main() {
	int T; cin >> T;
	while (T--) {
		solve();
	}
}

for _ in range(int(input())):
    n = int(input())
    s = input()
    one = s.count("1")
    zero = s.count("0")
    ans = "YES"
    for ti in input():
        if one == 0 or zero == 0:
            ans = "NO"
            break
        one -= 1
        zero -= 1
        if ti == "1":
            one += 1
        else:
            zero += 1
    print(ans)