E1. Divisible Numbers (easy version)

E1. Divisible Numbers (easy version)

This is an easy version of the problem. The only difference between an easy and a hard version is the constraints on $a, b, c$ and $d$.

You are given $4$ positive integers $a, b, c, d$ with $a<c$ and $b<d$. Find any pair of numbers $x$ and $y$ that satisfies the following conditions:

• $a < x \leq c$ , $b < y \leq d$,
• $x \cdot y$ is divisible by $a \cdot b$.

Note that required $x$ and $y$ may not exist.

Input

The first line of the input contains a single integer $t$ $(1 \leq t \leq 10)$, the number of test cases.

The descriptions of the test cases follow.

The only line of each test case contains four integers $a, b, c$ and $d$ $(1 \leq a< c \leq {10}^{5}, 1 \leq b < d \leq {10}^{5})$.

Output

For each test case print a pair of numbers $a < x \leq c$ and $b < y \leq d$ such that $x \cdot y$ is divisible by $a \cdot b$. If there are multiple answers, print any of them. If there is no such pair of numbers, then print $\text{-1 -1}$.

Example

input

5
1 1 2 2
3 4 5 7
8 9 15 18
12 21 14 24
36 60 48 66

output

2 2
4 6
12 12
-1 -1
-1 -1

 

解题思路

　　容易想到的暴力做法是双重循环枚举$x$和$y$，这样做肯定会超时，因此可以想一下能否只枚举$x$，$y$则根据$ab \mid xy$这个条件直接求出来。

　　因为$xy$被$ab$整除，因此$y$必然被$\frac{a \cdot b}{\gcd(a \cdot b, x)}$整除。令$t = \frac{a \cdot b}{\gcd(a \cdot b, x)}$，因此当我们枚举到某个$x$，看一下在区间$[b+1, d]$内是否存在一个$t$的倍数，设这个数为$y$，如果存在则表示$x, y$就是一组解。

　　如何判断在$[b+1, d]$内是否存在一个$t$的倍数呢？只要看一下严格大于$b$的最小的$t$的倍数是否不超过$d$就可以了。

• 如果$t \mid b$，那么很明显严格大于$b$的最小的$t$的倍数是$(\frac{b}{t} + 1) \times t$。
• 如果$t \nmid b$，那么严格大于$b$的最小的$t$的倍数是$\left\lceil {\frac{b}{t}} \right\rceil \times t$。

　　因此可以统一表示成$\left( {\left\lfloor {\frac{b}{t}} \right\rfloor + 1} \right) \times t$。

　　AC代码如下：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

int gcd(LL a, LL b) {
    return b ? gcd(b, a % b) : a;
}

void solve() {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    LL k = 1ll * a * b;
    for (int i = a + 1; i <= c; i++) {    // 枚举x 
        LL t = k / gcd(k, i);    // t|y，t是y的因子 
        if ((b / t + 1) * t <= d) {    // 大于b的最小的t的倍数不超过d 
            printf("%d %d\n", i, d / t * t);
            return;
        }
    }
    printf("-1 -1\n");
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

 

参考资料

　　Codeforces Round #828 (Div. 3) Editorial：https://codeforces.com/blog/entry/108101