【微软2014实习生及秋令营技术类职位在线測试】题目2 : K-th string

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

Description

Consider a string set that each of them consists of {0, 1} only. All strings in the set have the same number of 0s and 1s. Write a program to find and output the K-th string according to the dictionary order. If such a string doesn’t exist, or the input
is not valid, please output “Impossible”. For example, if we have two ‘0’s and two ‘1’s, we will have a set with 6 different strings, {0011, 0101, 0110, 1001, 1010, 1100}, and the 4th string is 1001.

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10000), the number of test cases, followed by the input data for each test case.
Each test case is 3 integers separated by blank space: N, M(2 <= N + M <= 33 and N , M >= 0), K(1 <= K <= 1000000000). N stands for the number of ‘0’s, M stands for the number of ‘1’s, and K stands for the K-th of string in the set that needs to be printed
as output.

Output

For each case, print exactly one line. If the string exists, please print it, otherwise print “Impossible”.

例子输入

3
2 2 2
2 2 7
4 7 47

例子输出

0101
Impossible
01010111011

思路：带有反复字符的全排列问题

import java.lang.String;
import java.lang.StringBuilder;
import java.net.Inet4Address;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;

public class Permutation {
	static int sum ;

	public static void main(String[] args)
	{
		int T ;
		Scanner jin = new Scanner(System.in);
		T = jin.nextInt();
		while(T-- > 0)
		{
			int N, M, K;
			StringBuilder stringBuilder = new StringBuilder();
			N = jin.nextInt();M = jin.nextInt();K = jin.nextInt();

			if (!((M+N) >= 2 && (N+M) <= 33)) {
				System.out.println("Impossible");
				continue;
			}

			if(N < 0 || M < 0)
			{
				System.out.println("Impossible");
				continue;
			}

			for (int i = 0; i < N; i++) {
				stringBuilder.append(‘0‘);
			}
			for (int i = 0; i < M; i++) {
				stringBuilder.append(‘1‘);
			}
			sum = 0;
			permutation(stringBuilder.toString().toCharArray(), 0, stringBuilder.length()-1, K);
			if (sum < K) {
				System.out.println("Impossible");
			}
		}
	}
	public static void permutation(char[] str, int start, int end, int K) {
		if (start == end) {
			sum++;
			//System.out.print(sum);
			//System.out.print("	");
			//System.out.print(new String(str));
			//System.out.print("	");
			if (sum == K) {
				System.out.println(new String(str));
				return ;
			}

		}
		else {
			for (int i = start; i <= end; i++) {
				if(!Isduplicated(str, start, i))
				{
					char tmp = str[start];
					str[start] = str[i];
					str[i] = tmp;

					permutation(str, start+1, end, K);

					tmp = str[start];
					str[start] = str[i];
					str[i] = tmp;
				}

			}
		}

	}
	public static boolean Isduplicated(char[] ch, int start, int end) {
		for (int i = start; i < end; i++) {
			if (ch[i] == ch[end]) {
				return true;
			}
		}
		return false;
	}

}