原题链接：

Problem - E1 - Codeforces

题目描述：

This is an easy version of the problem. The only difference between an easy and a hard version is the constraints on aa, bb, cc and dd.

You are given 44 positive integers aa, bb, cc, dd with a<ca<c and b<db<d. Find any pair of numbers xx and yy that satisfies the following conditions:

a<x≤ca<x≤c, b<y≤db<y≤d,
x⋅yx⋅y is divisible by a⋅ba⋅b.

Note that required xx and yy may not exist.

Input

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

The descriptions of the test cases follow.

The only line of each test case contains four integers aa, bb, cc and dd (1≤a<c≤1051≤a<c≤105, 1≤b<d≤1051≤b<d≤105).

Output

For each test case print a pair of numbers a<x≤ca<x≤c and b<y≤db<y≤d such that x⋅yx⋅y is divisible by a⋅ba⋅b. If there are multiple answers, print any of them. If there is no such pair of numbers, then print -1 -1.

题目大意：

给定a,b,c,d，问有没有x,y满足a < x <= c, b < y <= d且x * y能被a * b整除，如果有多组答案，则输出任意一组x，y即可，如果不存在答案，则输出-1。

解题思路：

数据不大，1e5，O(n)是完全可承受的，我们可以枚举x，设e = a * b / gcd(a * b, x)，则任何一个大于b且小于等于d的e的倍数都可以满足题目的需求。我们让y = (b + e) / e * e即可得到最小的大于b的e的倍数，只要他同时满足小于等于d，那么这个x和y就可以是答案。如果枚举完所有的x都找不到合适的y，即无解。

代码（CPP）：

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long
typedef unsigned long long ull;
const int maxn = 1e3 + 10;
const int INF = 0x3fffffff;signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout << fixed;cout.precision(18);int t;cin >> t;while(t--){int a, b, c, d;cin >> a >> b >> c >> d;bool flag = false;for (int x = a + 1; x <= c; x++){int e = a * b / __gcd(a * b , x);int y = (e + b) / e * e;if(y > b && y <= d){cout << x << " " << y << endl;flag = true;break;}}if(!flag)cout << "-1 -1\n";}return 0;
}

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