Codeforces Educational 38 C. Constructing Tests （数学公式推导+暴力)

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本文介绍了一个特定的编程问题：如何构造有效的测试案例以验证一个算法，该算法的目标是找到一个m-free矩阵中1的最大数量。文章提供了输入输出示例及解析，并附带了一个C++代码示例来解决这一问题。

C. Constructing Tests
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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 , 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
Input
Copy

3
21
0
1

Output
Copy

5 2
1 1
-1

解析:画画图就可以发现发现n*n矩阵里面至少有n/m的平方个零,所以数学公式就是n*n-(n/m)^2=x.
对公式推导一下就能发现n*n最大为四分之三的1e9,所以设定上限为1e5即可,之后暴力枚举n.
还有一点就是存在一种情况比如7/5=1但是反过来7/1等于7得不到正确的结果,把这种情况考虑到就能a了

#include <bits/stdc++.h>
#define LL long long
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    int T;
    cin >> T;
    while(T--)
    {
        LL x;
        cin >> x;
        if(x == 0)
            cout << 1 << ' ' << 1 << endl;
        else
        {
            bool flag = false;
            for(LL i = 1; i <= 100000; i++)
            {
                if(x >= i*i) continue;
                LL k = sqrt(i*i-x);
                if(k*k == i*i-x)
                {
                    LL m = i/k;
                    if(i/m==k) // 避免例如7/5=1但是7/1=5的情况出现
                    {
                        cout << i << ' ' << m << endl;
                        flag = true; break;
                    }
                }
            }
            if(!flag) cout << -1 << endl;
        }
    }
}