微软机试题02

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本文介绍了一个算法问题，即如何找到由相同数量的0和1组成的字符串集中的第K个字典序字符串。文章详细解释了输入输出格式，并提供了一段C++代码示例来解决该问题。

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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 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”.

Sample In

3
2 2 2
2 2 7
4 7 47

Sample Out

0101
Impossible

01010111011

#include <iostream>
#include <cmath>

//统计整数x的二进制中值为1的位数
int count_1(int x)
{
	int count = 0;  
	while(x)  
	{  
		x = x & ( x - 1 );  
		count++;   
	}  
	return count;
}

int main()
{
	int nn;
	while(std::cin >> nn)
	{
		for(int kk = 0; kk < nn; ++kk)
		{
			int n, m, k;
			std::cin >> n >> m >> k;
			int obj = 0;
			obj = (int)pow((double)2, m) - 1;
			for(int ii = 0; ii < k;)
			{
				if(count_1(obj) == m)
				{
					++ii;
				}
				obj++;
			}
			obj = obj-1;

			if((obj) < (int)pow((double)2, m+n))
			{
				for(int ii = n+m-1; ii >= 0; ii--)
				{
					std::cout << ((obj >> ii) & 1);
				}
				std::cout << std::endl;
			}
			else
			{
				std::cout << "Impossible" << std::endl;
			}
			
		}
	}
	return 0;
}