微软2014实习生在线测试之K-th string

最新推荐文章于 2025-09-13 12:59:37 发布

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本博客探讨了如何在指定数量的0和1组成的字符串集中，找到第K个字符串，并提供了相应的算法实现。

//Time Limit: 10000ms
//Case Time Limit: 1000ms
//Memory Limit: 256MB</p><p>//Description</p><p>//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.</p><p>//Input</p><p>//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.</p><p>//Output</p><p>//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</p><p>//Sample Out
//0101
//Impossible
//01010111011

//下面为正规解法,最坏情况下:O(C(m+n,m))

#include <iostream>
using namespace std;
int m, n, k;
int element[2] = {0,1};
int map[2] = {0};
int count = 0; //统计字符串个数
bool hasFound = false; //hasFound为true表示，找到了，否则表示没找到.主要是在找到时，能够立刻退出backtrack。
void backtrack(int t, short *x)
{
	if (t >= m+n)
	{
		count += 1;
		if (count == k)
		{
			for (int i=0; i<t; ++i)
			{
				cout << element[x[i]] ;
// 				cout << x[i]; //因为此题element[i]==i，故能够用此代替上面的
			}
			cout << endl;
			hasFound = true;
			return ;
		}

	}
	else
		for (short i=0; i<2; ++i)
		{
			if (!hasFound)
			{
				if (map[i] > 0)
				{
					--map[i];
					x[t] = i;
					backtrack(t+1, x);
					++map[i];
				}
			}
			else
				return;
		}
}//end func backtrack
int main()
{
	cout << "please input the value of m, n k,such as 2 2 3:";
	while (cin>>m>>n>>k)
	{
		if (!m && !n)
		{
			break;
		}
		short* x = new short[m+n];
		map[0] = m; //0的个数
		map[1] = n; //1的个数
		backtrack(0, x);
		if (count < k) //总的个数小于k
		{
			cout << "Impossible" << endl;
		}
		delete [] x;
		cout << "please input the value of m, n k,such as 2 2 3:";
		
	}//end while
	return 0;
}//end main