Square

思路：枚举pow(10,k)，然后y = sqrt(x * pow(10,k))，取l = sqrt(y),r = sqrt(y) + 1即可，注意ll问题以及y的范围是

1~1e9

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int INF = 1e9;
typedef long long ll;
typedef pair<int,int> PII;

int b[15];

void solve(){
	int x;
	cin >> x;
	for(int i = 0; i < 15; i ++){
		int p = x * b[i];
		int l = sqrt(p);
		int r = l + 1;
		for(int j = l; j <= r; j ++){
			if(j * j / b[i] == x && j <= INF){
				cout << j << '\n';
				return;
			}
		}
	}
	cout << -1 << '\n';
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	int t;
	cin >> t;
	b[0] = 1;
	for(int i = 1; i < 15; i ++) b[i] = b[i - 1] * 10;
	while(t --){
		solve();
	}
	return 0;
}