微软2014年实习生在线机试第二题及解法

翻译我就不做了，看大家有什么好思路，

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

采用递归解法

#include<iostream>#include<string>using namespace std;//M代表0的个数int M=0;//N代表1的个数int N=0;//K代表第k个满足条件的数int K=0;/*a代表数组m代表0的个数n代表1的个数count代表满足要求的数的个数layer代表目前数的位数*/void findString(int layer, int &count,int m,int n,int a[]){if(m>M || n>N)//不满足条件，直接返回return;if(count>K)//只需要输出第K个，其它无需遍历了return;if( (layer+1)==(M+N) )//递归终止，位数到了{//这里(layer+1)是因为数组下标从0开始//最后一位放0，判断是否满足要求a[layer]=0;m=m+1;if(m==M && n==N)//判断是否满足条件:有m个0,n个1{count++;if(count==K){int i;for(i=0;i<=layer;i++)cout<<a[i];cout<<endl;}}--m;//最后一位放1，判断是否满足要求a[layer]=1;n=n+1;if(m==M && n==N)//判断是否满足条件:有m个0,n个1{count++;if(count==K){int i;for(i=0;i<=layer;i++)cout<<a[i];cout<<endl;}}return;}//当前位置先放0，使其满足字典排序a[layer]=0;findString(layer+1,count,m+1,n,a);//当前位置后放1a[layer]=1;findString(layer+1,count,m,n+1,a);}void main(){int a[35]={0};int count=0;K=4;M=2;N=2;//先判断K是否超出M和N组合的数的个数了findString(0, count,0,0,a);getchar();}