题解 | #自守数#

#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;

    int ans = 0;
    int base = 10;
    for (int i = 0; i <= n; i++) {
        int m = i * i;
        if (i == base)
            base *= 10;
        if (m % base == i)
            ans++;
    }

    cout << ans;
}
// 64 位输出请用 printf("%lld")