E Bogo Sort

置換求環 + 大整數 + lcm
就是求出這個置換有多少環，然後對這些環的大小進行求lcm
對於置換求環

#include <iostream>
#include <cstdio>
using namespace std;
const int N = 1e5 + 5;
int p[N], ans[N], cnt;
bool vis[N];
int main(){
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
        cin >> p[i]; // 置換
    for(int i = 1; i <= n; i++) {
        if(vis[i]) continue;
        int tmp = i, tot = 0;
        while(vis[tmp] == 0) { //沿著置換遍歷 
            vis[tmp] = 1;
            tmp = p[tmp];
            tot++;
        }
        ans[++cnt] = tot; // 第cnt個環的大小
    }
    return 0;
}
 

好像在java裡，new一個BIgInteger比較慢，但是直接呼叫one，ten的大整數快

import java.math.BigInteger;
import java.util.*;
public class Main {
	static int p[] = new int[100010];
	static int num[] = new int[100010];
	static boolean vis[] = new boolean[100010];
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt(), cnt = 0;
		for(int i = 1; i <= n; i++) {
			p[i] = in.nextInt();
		}
		BigInteger mod = BigInteger.ONE;
		for(int i = 1; i <= n; i++) {
			mod = mod.multiply(BigInteger.TEN);
			if(vis[i]) continue;
			int tmp = i, tot = 0;
			while(!vis[tmp]) {
				vis[tmp] = true;
				tmp = p[tmp];
				tot++;
			}
			num[++cnt] = tot;
		}
		BigInteger ans = new BigInteger(String.valueOf(num[1]));
		for(int i = 1; i <= cnt; i++) {
			BigInteger g = ans.gcd(new BigInteger(String.valueOf(num[i])));
			ans = ans.multiply(new BigInteger(String.valueOf(num[i])));
			ans = ans.divide(g);
		}
		System.out.println(ans.mod(mod));
	    in.close();
	}
}