C language implementation of permutation/combination algorithm

/*
* FileName: PERMUTATION.C
* Use: Full array algorithm
* Programming Environment: WinXP SP2+CL 8.0
* Date of Completion: 2006.2 Ver 0.01
* Author: 88250
* Contact: e-mail:dl88250@gmail.com qq:845765
*/

#include < stdio.h >

#include < stdlib.h >



int count = 0;



void permutation (char per[], int m, int len)

{

int i;

char tmp;



if (M < len-1) {

Permutation (per, M + 1, len);

for (i = m + 1; i < Len; i + +) {

/* Select the time of the arrangement of "head" * *

TMP = Per[m];

PER[M] = Per[i];

Per[i] = tmp;

Permutation (per, M + 1, len);

/* Restore previous arrangement * *

TMP = Per[m];

PER[M] = Per[i];

Per[i] = tmp;

}

} else {

+ + count; /* To count the permutations * *

/* Adjust the format ^ ^ * *

printf ("%s", per);

}

return;

}

int main (void)

{

int i = time (NULL);

Char per[] = "ABCD";



Permutation (per, 0, strlen (per));

printf ("Total:%d", count);



return 0;

}

/*
* FileName: COMBINATION.C
* Use: Combinatorial algorithm
* Programming Environment: WinXP SP2+CL 8.0
* Date of Completion: 2006.2 Ver 0.01
* Author: 88250
* Contact: e-mail:dl88250@gmail.com qq:845765
*/

#include < stdio.h >



#define MAX_NUM 20



int Comb[max_num];



void combination (int m, int n)

{

int I, J;



for (i = m; i >= N; i--) {

Comb[n] = i; /* Select the current "header" element * *

if (n > 1) {

/* Go to the next smaller combination problem * *

Combination (i-1, n-1);

} else {

/* Full number of combinations required, output * *

for (j = comb[0]; j > 0; J--) {

printf ("%c", Comb[j] + 64);

}

printf ("");

}

}



return;

}



int main (int argc, char * argv[])

{



comb[0] = 2;

Combination (4, comb[0]); /* C (4, 2) */



return 0;

}