o-p-p-o2023提前批0715-第一题 构造二阶行列式

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小欧希望你构造一个二阶行列式，满足行列式中每个数均为不超过20的正整数，且行列式的值恰好等于x。你能帮帮她吗?

提示:二阶行列式的计算方式:
$$
\begin{vmatrix}
a & b \
c & d
\end{vmatrix} = a * d - b * c
$$
输入描述

一个正整数x。-1000<=x<=1000

输出描述

如果无解，请输出-1。否则输出任意合法行列式即可(输出两行,每行输出两个不超过20的正整数)。

示例：

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输入：
2

输出：
3 2
5 4

分析：

难度：容易

1
由于结果要求每个数不超过20，因此直接使用四重循环暴力枚举，时间复杂度为20 * 20 * 20 * 20

代码

  • C++

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#include <iostream>
#include <cstring>
#include <string>
#include <climits>
#include <algorithm>

using namespace std;

int main() {
	int x;
	cin >> x;
	
	for (int a = 1; a <= 20; a ++)
		for (int b = 1; b <= 20; b ++)
			for (int c = 1; c <= 20; c ++)
				for (int d = 1; d <= 20; d ++) {
					if (a * d - b * c == x) {
						cout << a << " " << b << endl;
						cout << c << " " << d << endl;
						return 0;
					}
				}
				
	cout << -1 << endl;
	return 0;
}
  • java
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import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        int x = in.nextInt();

        for (int a = 1; a <= 20; a ++)
            for (int b = 1; b <= 20; b ++)
                for (int c = 1; c <= 20; c ++)
                    for (int d = 1; d <= 20; d ++) {
                        if (a * d - b * c == x) {
                            System.out.println(a + " " + b);
                            System.out.println(c + " " + d);
                            return ;
                        }
                    }

        System.out.println(-1);
    }
}
  • Python
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x = int(input())

def main():
    for a in range(1, 21):
        for b in range(1, 21):
            for c in range(1, 21):
                for d in range(1, 21):
                    if a * d - b * c == x:
                        print(str(a) + " " + str(b))
                        print(str(c) + " " + str(d))
                        return
   print(-1)

main()
  • Golang
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package main

import (
	"fmt"
)

func main() {
	var x int
	fmt.Scan(&x)

	for a := 1; a <= 20; a++ {
		for b := 1; b <= 20; b++ {
			for c := 1; c <= 20; c++ {
				for d := 1; d <= 20; d++ {
					if a*d-b*c == x {
						fmt.Println(a, b)
						fmt.Println(c, d)
						return
					}
				}
			}
		}
	}

	fmt.Println(-1)
}
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