Ka's retrospective programming exercise part6.5| | The permutation problem with repeating elements

#include <stdio.h>#defineBR printf ("\ n");intans=0;Charel[502]; intConfirmintIintK//to compare back if the element has the same skip this condition{    if(i>k) { while(i>k) {if(El[i]==el[k])return 0; K++; }    }    return 1; }voidOpintN) {ans++; intJ;  for(j=0; j<=n;j++) printf ("%c", El[j]); BR}voidPermintNintk)//N represents the number of ends of the current distance assignment (array pattern, -1) K represents the position of the current element{    intI,r; if(n==k) OP (n);//if two numbers have reached equal output    Else  for(i=k;i<=n;i++)//swap from the current array element to the end, one by one        if(Confirm (I,k))//Exchange based on the returned results{R=el[i];el[i]=el[k];el[k]=R; Perm (N,k+1);//one of the advancing Kr=el[i];el[i]=el[k];el[k]=R; } }intMain () {intN,i; scanf ("%d",&N);    GetChar ();  for(i=0; i<n;i++) scanf ("%c", &el[i]);//the above is the normal input linkPerm (n1,0); printf ("%d", ans); return 0;}

Forgive me for my bad taste hahaha.

The whole idea is this:

A A C c

0 1 2 3

First exchange with yourself

a a C c

a a C c rear letter disorderly order　

And then recursively go on, constantly nonalphanumeric each word back to the current position to find the next situation.

C a A C

If you're not swapping with yourself, check to see if you've swapped the same elements before. Like what:

A b c d　e F e H

a b c e D f E H rear letter disorderly order　

If the next and the next e exchange, then the effect and the first e the same, are in the following order regardless, ABCE down.

A b c D E F e h

a b c e E F D H　

Test to find out from the back, so that the first person can be found

A b c D E F e H

K I

A b c D E F e H

　　　　　　　　K I is equal, skipping this condition

Slowly realize it. It should be clear.

Ka's retrospective programming exercise part6.5| | The permutation problem with repeating elements