题解 | #小红的取模构造#

我们发现如果a和b都是正数，相同的时候无解，不同的时候只需要给小的数加上大的数即可，然后特判一下和0有关的情况

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 5;
int __t = 1, n, a, b;
void solve() {
    cin >> a >> b;
    if (a == 0 && b == 0) {
        cout << "1 1\n";
        return;
    }
    if (a == 0) {
        cout << b + b << " " << b << "\n";
        return;
    }
    if (b == 0) {
        cout << a << " " << a + a << "\n";
        return;
    }
    if (a > b) {
        cout << a << " " << b + a << "\n";
        return;
    }
    if (b > a) {
        cout << a + b << " " << b << "\n";
        return;
    }
    cout << "-1 -1\n";
    return;
}
int32_t main() {
#ifdef ONLINE_JUDGE
    ios::sync_with_stdio(false);
    cin.tie(0);
#endif
    cin >> __t;
    while (__t--)
        solve();
    return 0;
}