Educational Codeforces Round 38 C. Constructing Tests 思维

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 sizem × 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 of1'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

5 2
1 1
-1

题意：定义一种叫“m-free”的二进制矩阵，其大小为n*n，满足在其中任选一个m*m的区域内均含有至少一个0。现在给定一个数x，代表在某个“m-free”矩阵中最多的1的个数，求该矩阵的n与m的值。

思路：首先要思考如何可以使某个矩阵的1最多。因为在n*n中任选m*m的区域中有一个0即可，所以只要0的个数大于等于“(n/m)的平方”个，就必定满足题设。所以本题便变为求方程“n*n-(n/m)*(n/m)=x”的解。枚举n求解即可。

代码如下：

#include 
using namespace std;
typedef long long ll;

int main()
{
    ll x;
    int t;
    scanf("%d",&t);
    while (t--){
        scanf("%I64d",&x);
        if (x==0){
            printf("1 1\n");
            continue;
        }
        int f=0;
        for (ll n=1;n<=100000;n++){
            if (x>=n*n)
                continue;
            ll k=sqrt(n*n-x);
            ll m=n/k;
            if (k>0&&n*n-(n/m)*(n/m)==x){
                f=1;
                printf("%I64d %I64d\n",n,m);
                break;
            }
        }
        if (!f)
            printf("-1\n");
    }
    return 0;
}