集合问题

//下面函数从一个结合的全排列中取出第n个

void GetN(int N, char set1[], int n, char out1[])

{

      char * pset2=new char[N]; //拷贝原来的集合

       int pos;

      int j,k;

      for(pos=0; pos < N; pos++)

      {

          pset2[pos]=set1[pos];

      }

      for(pos =0; pos < N ; pos ++)

      {

             //选取一个元素

             k = n % (N-pos);

             out1[pos]= pset2[k];

             //把选出的元素删除

               for(j=k; j<N-1; j++)

             {

                   pset2[j]= pset2[j+1];

             } 

             n= n/(N-pos);

      }

      delete []pset2;

}

void GetN(

    int N, //每个集合中元素个数

    char set1[], char set2[], //两个集合

    int n, //第n中排列情况

    char set1o[], char set2o[]//输出的结果

)

{

     int N1 =1;

     int k;

     for(k=2; k< N ; k++) N1 *= k;

     GetN(N, set1, n%N1,set1o); 

     GetN(N, set2, n/N1,set2o); 

}

#include <stdio.h>



int  C=0;

char a[]={'1','2','3','4','5','6','7','8'};



void Permutation(char a[], int start, int end)

{

    int i, temp;    

	static char b[]={'A','B','C','D','E','F','G','H'};



    if(start == end)

    { 	

		++C;

		if( C<2 )

			Permutation( b, 0, sizeof(b)/sizeof(b[0])-1 );

		else{		

			for(i = 0; i <= end; i++)

				printf(" %c%c ",::a[i],b[i]);

			printf("\n");

		}

    }

    else

    {

        for(i = start; i <= end; i++)

        {

            temp=a[start]; a[start]=a[i]; a[i]=temp;

            Permutation(a, start+1, end);

            temp=a[start]; a[start]=a[i]; a[i]=temp;

        }

    }

}



int main()

{	

	Permutation( a, 0, sizeof(a)/sizeof(a[0])-1 );

	return 0;	

}