微软2014实习生在线测试之K-th string

问题描述：

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

基本思路：

简单点就是不断查找第K个数，但时间复杂度比较高。O((M+N)*K)

这里使用排列组合思想。时间复杂度O((M+N)*log(M))

首先分析共有多少可能，即排除Impossible。一共M+N位数字，M个'1'，则共有种可能，凡是大于此数的K均为不合法。

然后对结果从高位到低位分析，若第1位是'0'，剩下的数字中就有M个'1'，N-1个0，共有种可能。 

• 若K<k_max，则第1位是'0'
• 若K>k_max，则第1位是'1'，K = K - k_max
• 若K==k_max，则前M位均为'1'，剩下的N位为'0'

简单思路的程序

 

排列组合的程序

#include <iostream>
#include <vector>
using std::cout;
using std::cin;
using std::endl;
using std::vector;

int cc(int s, int x)
{
	int result = 0;
	double temp = 1.0;
	for (int i = 0; i < x; i++)
	{
		temp *= (double)(s - i) / (double)(x - i);
	}
	result = (int)(temp + 0.0001);
	return result;
}
int main(int argc, char **argv)
{
	vector<vector<int>> input;
	int total = 0;
	cin >> total;
	while (total--)
	{
		int N, M, K;
		cin >> N >> M >> K;
		vector<int> temp;
		temp.push_back(N);
		temp.push_back(M);
		temp.push_back(K);
		input.push_back(temp);
	}
	
	for (int i = 0; i < input.size( ); i++)
	{
		int N, M, K;
		N = input[i][0];
		M = input[i][1];
		K = input[i][2];
		//cout << N << ", " << M << ", " << K << endl;
		
		int max = cc(M + N, N);
		if (max < K)
		{
			cout << "Imposibile" << endl;
		}

		int n, m;
		int _k = K;
		char *digits = new char[M + N + 1];
		int idx = 0;
		memset(digits, 0, M + N + 1);
		m = M; n = N;
		while (idx < M + N)
		{
			// if highest is 0;
			int t = cc(m + n - 1, n - 1);
			if (t > _k)
			{
				digits[idx++] = '0';
				m = m;
				n = n - 1;
			}
			else if (t < _k)
			{
				digits[idx++] = '1';
				m = m - 1;
				n = n;
				_k = _k - t;
			}
			else if (t == _k)
			{
				digits[idx++] = '0';
				n--;
				while (m--)
				{
					digits[idx++] = '1';
				}
				while (n--)
				{
					digits[idx++] = '0';
				}
			}
		}
		cout << digits << endl;
		delete[] digits;
	}

}