BOJ 15649 - N과 M (1)

문제

https://www.acmicpc.net/problem/15649

 

 

풀이

N과 M(3) 문제와는 달리 같은 숫자를 여러 번 사용할 수 없도록 조건식을 추가해야 한다

 

 

코드

python

def backtracking(depth):
    if depth == m:
        print(*answer)
        return
    for i in range(1, n + 1):
        if i in answer:
            continue
        answer.append(i)
        backtracking(depth + 1)
        answer.pop()
 
 
answer = []
 
n, m = map(int, input().split())
 
backtracking(0)
 

 

java

import java.util.Scanner;
 
public class Main {
 
    private static int n, m;
    private static int a[];
    private static boolean check[];
    private static StringBuilder sb = new StringBuilder();
 
    public static void main(String[] args) {
        input();
        solve();
    }
 
    private static void solve() {
        rec(0);
        System.out.println(sb.toString());
    }
 
    private static void rec(int k) {
        if (k == m) {
            for (int i = 0; i < m; i++) {
                sb.append(a[i]).append(" ");
            }
            sb.append("\n");
        } else {
            for (int i = 1; i <= n; i++) {
                if (check[i]) {
                    continue;
                }
                a[k] = i;
                check[i] = true;
 
                rec(k + 1);
 
                a[k] = 0;
                check[i] = false;
            }
        }
    }
 
    private static void input() {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        m = sc.nextInt();
        a = new int[m];
        check = new boolean[n + 1];
        sc.close();
    }
 
}
 

 

C++

#include <iostream>
 
using namespace std;
 
int N, M;
bool visit[9];
int arr[9];
 
void solve(int cnt)
{
    if (cnt == M)
    {
        for (int i=0; i<M; i++)
            cout << arr[i] << ' ';
        cout << '\n';
        return;
    }
    for (int i=1; i<=N; i++)
    {
        if (!visit[i])
        {
            visit[i] = true;
            arr[cnt] = i;
            solve(cnt+1);
            visit[i] = false;
        }
    }
}
 
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(0);
 
    cin >> N >> M;
 
    solve(0);
 
    return 0;
}