微软2014实习生及校招秋令营技术类职位在线测试：2.K-th string

Time Limit: 10000ms

Case Time Limit: 1000ms

Memory Limit: 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 s​uch 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”.

Sample In

3

2 2 2

2 2 7

4 7 47

Sample Out

0101

Impossible
01010111011

谈谈自己对本题的分析思路：

我们首先谈一般的情况，就是排除了所有特殊情况的地方，加入输入为2个 0 ，2 个 1，那么它的所有情况就是这样的，{0011, 0101, 0110, 1001, 1010, 1100}，实际上这些序列的个数就是将2个 1  插入到2 个 0中所有的情况个数，实际上就是组合的个数的C(2,4)，而最后一个参数给出的K就是距离第一个组合数的个数（包括第一个组合数），

我们可以利用下面的公式：

K=K-C(i,j) ，C(i,j)<=K<=C(i+1,j)其中i代表0的个数，j代表1的个数C(i,j)的意义是(i+j)!/(i! * j!)，这时候K值总是不小于0，这样便可以确定最高位的1所在的位置，然后依次下去，确定次高位的1位置，知道所有的1都确定位置。当然我们也要考虑到程序可能出现的特殊情况，这样特殊情况比较多，下面还是对着程序来讨论吧。

package com.microsoft;

import java.util.Scanner;
import java.util.StringTokenizer;

/*
* 微软校招笔试第二题
*/
public class K_th {
private int Combe(int n, int m) {
return fMuti(m + n) / (fMuti(m) * fMuti(n));
}

private int fMuti(int n) {
int muti = 1;
for (int i = 1; i <= n; ++i)
muti *= i;
return muti;
}

public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
K_th kth = new K_th();
int i = 0;
int a[] = new int[3];
String s;
while (sc.hasNext()) {
i = 0;
s = sc.nextLine();
StringTokenizer st = new StringTokenizer(s);
while (st.hasMoreTokens()) {
a[i++] = Integer.parseInt(st.nextToken());
}
if (i == 3) { // 表明输入3个数的时候才可以打印出来
if ((a[0] + a[1] >= 2 && a[0] + a[1] <= 33) && a[0] >= 0
&& a[1] >= 0 && (a[2] >= 1 && a[2] <= 1000000000)) { // 这是判定条件，题目本身给的
int thres = kth.Combe(a[0], a[1]);
if (a[2] > thres) {
System.out.println("Impossible");
continue;
}
}
int j;
int[] result = new int[a[0] + a[1]]; // 这是最终盛放结果的数组
// 下面是本题的核心源代码
while (a[1] > 0) {
if (a[2] == 1) { //如果最后K变成1，则只要列出最小的组合数就行了
if (a[1] > 1) {
for (j = 0; j < a[1]; ++j)
result[j] = 1;
}
result[0] = 1;
} else {
for (j = 0; j <= (a[0] - 1); ++j) {
if (kth.Combe(j, a[1]) <= a[2]
&& kth.Combe(j + 1, a[1]) >= a[2]) {
result[a[1] + j] = 1;
a[2] -= kth.Combe(j, a[1]); //这里就是递归的式子
break;
}
}
}
a[1]--;
}
for (j = result.length - 1; j >= 0; --j)
System.out.print(result[j]);
System.out.println();
}
}
}
}
运行结果截图如下：