Uva 10825 Anagram and Multiplication dfs 爆搜排列

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题目链接：点击打开链接

题意:

给定m( 3<=m<=6), n ( 4<=n<=400)

构造一个m位n进制的数。

使得这个数*2, *3, *i ( 2<=i<=m)的结果是 这个数的排列组合。

如：

m = 6, n = 10;

则 这个数可能为 142857

2*142857 = 285714

3*142857 = 428571

4*142857 = 571428

5*142857 = 714285

6*142857 = 857142

输出：

题目保证至多只有一个解，若有这样的解则输出这个数（每位数间有一个空格）若无输出"Not found."

思路:

枚举个位，然后不断用 (2~m)去乘个位得到其他位，则这样得到的结果的某种排列才有可能是答案，

判断所有排列是否可行。

import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.TreeSet;
import java.util.Queue;

public class Main {
	int m, n;
	boolean ok;
	int[] ans = new int[7], b = new int[500], a = new int[500];
	boolean check(int x, int[] y){
		for(int i = 0; i < n; i++)a[i] = b[i];
		int tmp = 0;
		for(int i = m; i >= 1; i--){
			tmp = tmp + x*y[i];
			int k = tmp %n;
			if(--a[k] < 0)return false;
			tmp /= n;
		}
		return tmp == 0;
	}
	boolean[] use = new boolean[7];
	int[] v = new int[7];
	boolean dfs(int pos){
		if(pos > m){
			for(int i = 2; i <= m; i++)
				if(false == check(i,v))return false;
			for(int i = 1; i <= m; i++)ans[i] = v[i];
			return true;
		}
		else {
			for(int i = 1; i <= m; i++){
				if(use[i])continue;
				use[i] = true;
				v[pos] = ans[i];
				if(true == dfs(pos+1))return true;
				use[i] = false;
			}
			return false;
		}
	}
	boolean judge(int x){
		int tmp = 0;
		for(int i = 0; i < n; i++)b[i] = 0;
		for(int i = 1; i <= m; i++){
			tmp = (x + tmp)%n;
			ans[i] = tmp;
			b[ans[i]]++;
			use[i] = false;
		}
		return dfs(1);
	}
	void put(int x[]){
		out.print(x[1]);
		for(int i = 2; i <= m; i++)out.print(" "+x[i]);
		out.println();
	}
	void work() {
		while(true){
		  	m = cin.nextInt(); n = cin.nextInt();
			if(m == 0 && n == 0)break;
			ok = false;
			for (int i = 1; i < n && ok == false; i++) 
				if (judge(i)) 
					ok = true;
			if(ok)
				put(ans);
			else 
				out.println("Not found.");
		}
	}
	Main() {
		cin = new Scanner(System.in);
		out = new PrintWriter(System.out);
	}

	public static void main(String[] args) {
		Main e = new Main();
		e.work();
		out.close();
	}

	public Scanner cin;
	public static PrintWriter out;
}

Uva 10825 Anagram and Multiplication dfs 爆搜排列

标签：

原文地址：http://blog.csdn.net/qq574857122/article/details/42970477