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”.
樣例輸入
3
2 2 2
2 2 7
4 7 47
樣例輸出
0101
Impossible
01010111011
看到這道題，我的第一思路就是排列組合，比如輸入n=2 ，m=3 ，那就是有2個0和3個1 ， 再對這5個數進行排列，用遞歸的思想。
第一位排0，則再對剩下的n-1和m排列
第一位排1，則再對剩下的n和m-1排列
利用這種思想也很容易得出總共的排列次數
我的代碼如下
#include<stdio.h>
#include<stdlib.h>
int total(int n,int m) //用來計算n和0和m個1排列的總個數
{
if(n==0 || m==0)
return 1;
else
return (total(n-1,m)+total(n,m-1));
}
void findthekth(int pailie[] ,int n,int m,int len ,int k,int index,int *count)
{
int i;
if( index >= len )
{
(*count)=(*count)+1;
if((*count) == k)
{
for(i=0;i<len;i++)
{
printf("%d",pailie[i]);
}
printf("\n");
}
return ;
}
if(n>0) //0的個數比0大
{
pailie[index] = 0;
findthekth(pailie,n-1,m,len,k,index+1,count);//剩下n-1個0和m個1去排
}
if(m>0) //1的個數比0大
{
pailie[index] = 1;
findthekth(pailie,n,m-1,len,k,index+1,count);//剩下n個0和m-1個1去排
}
}
int main()
{
int num;
int n ,m ,k,count ;
scanf("%d",&num);
int i,j;
int * pailie = (int *)malloc((n+m)*sizeof(int));
for(i=0;i<num;i++)
{
j=0;
count = 0;
scanf("%d",&n);
scanf("%d",&m);
scanf("%d",&k);
if(k<=total(n,m))
findthekth(pailie,n,m,n+m,k,0,&count);
else
printf("Impossible\n");
}
free(pailie);
}