Educational Codeforces Round 38 (Rated for Div. 2) C. Constructing Tests_constructing tests codeforces

本文介绍了一个特定的算法问题：如何根据给定的最大1的数量构造出对应的n和m值，使得n*n的大正方形中所有m*m的子矩阵都至少包含一个0。通过逐步解析并提供示例代码，展示了枚举法在解决此类问题中的应用。

Let's denote a m-free matrix as a binary (that is, consisting of only 1's and 0's) matrix such that every square submatrix of size m × m of this matrix contains at least one zero.

Consider the following problem:

You are given two integers n and m. You have to construct an m-free square matrix of size n × n such that the number of 1's in this matrix is maximum possible. Print the maximum possible number of 1's in such matrix.

You don't have to solve this problem. Instead, you have to construct a few tests for it.

You will be given t numbers x1, x2, ..., xt. For every $\text{[math]}$ , find two integers ni and mi (ni ≥ mi) such that the answer for the aforementioned problem is exactly xi if we set n = ni and m = mi.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of tests you have to construct.

Then t lines follow, i-th line containing one integer xi (0 ≤ xi ≤ 109).

Note that in hacks you have to set t = 1.

Output

For each test you have to construct, output two positive numbers ni and mi (1 ≤ mi ≤ ni ≤ 109) such that the maximum number of 1's in a mi-free ni × ni matrix is exactly xi. If there are multiple solutions, you may output any of them; and if this is impossible to construct a test, output a single integer - 1.

Example
inputCopy
3
21
0
1
output
5 2
1 1
-1

题意：在一个由01组成的n*n的正方形内，找到全部m*m的小正方形，而且每个小正方形至少包含一个0，那么n*n大正方形对应一个最大的1的个数，现在给你最大1的个数，请构造出n和m。

思路：我们先找n，大正方形的方格数至少要大于x，否则一定没法求解，所以先枚举n，当有n*n>x时，开始求解m，由于是正方形，规律十分对称，所以我们只看一横行就可以了，那么一行中0的个数我们可以直接求出来，已知n，x，那么0的总数：n*n-x*x，开方得一行的0，现在已知一行的零k个，那么可以推得一行的小正方形个数，那么每个小正方形的边长m：n/k。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int main()
{
    ll x,t,m,tmp,flag;
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld",&x);
        flag=0;
        for(ll i=1;i<=40005;i++)   //枚举n
        {
            if(i*i>x)
            {
                tmp=sqrt(i*i-x);
                m=i/tmp;     //求解m
                if(i*i-(i/m)*(i/m)==x)
                {
                    flag=1;
                    printf("%lld %lld\n",i,m);
                    break;
                }
            }
        }
        if(flag==0)printf("-1\n");
    }
    return 0;
}