"Algorithm series learning One" full-array generation algorithm

The last section of the algorithm class refers to the full array of generation problems, today I found some information on the Internet, summed up a few ways:

A. Recursive collation algorithm.

Two. Dictionary ordering method.

Three. Increment the carry numeral method.

Four. Descending carry numeral method.

Five. Adjacent position Exchange method.

Six. N binary method.

Here's a look at some of these algorithms.

A. Recursive collation algorithm.

Recursive collation algorithm is relatively concise, the implementation of a variety of methods.

1. Recursive algorithm (non-dictionary order)

1#include <iostream>2#include <algorithm>3 4 5 using namespacestd;6 intsum=0;7 inta[Ten];8 intN;9 //consider the full arrangement of [l,n-1] intervalsTen voidPerm (intl) One {  A     //output, total number of permutations plus one -     if(l==n-1) -     { the          for(intI=0; i<n;i++) -         { -printf"%d", A[i]); -         } +printf"\ n"); -sum++; +      }  A      Else at      { -          //consider the arrangement of [l+1,n-1] -Perm (L +1); -           for(inti=l+1; i<n;i++) -          { -              //The current bit (l) is exchanged separately from each subsequent bit (l+i).  in swap (a[l],a[i]); -              //The current bit (l) has been determined, now consider the [l+1,n-1] arrangement.  toPerm (L +1); +              //Restore - swap (a[l],a[i]); the          } *      } $ }Panax Notoginseng intMain () - { theFreopen ("Data.out","W", stdout); +scanf"%d",&n); A      for(intI=0; i<n;i++) the     { +a[i]=i+1; -     } $Perm (0); $printf"A total of%d permutations. \ n", sum); -     return 0; -}

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2. Backtracking method.

Two. Dictionary ordering method.

1#include <iostream>2#include <algorithm>3 4 using namespacestd;5 intN;6 Const intmaxn=1e2+Ten;7 intA[MAXN];8 Const intinf=0x3f3f3f;9 BOOLnext_perm ()Ten { One     intk=-1; A //for (int i=0;i<n-1;i++) - //    { - //if (a[i]<a[i+1]) the //        { - //k=i; - //        } - //    } +     //from right to left to meet a[i]<a[i+1] The most right I, arranged from this position began to change -      for(inti=n-2; i>=0; i--) +     { A         if(a[i]<a[i+1]) at         { -k=i; -              Break; -         } -     } -     //indicates that this is already the largest dictionary order. in     if(k==-1) -     { to         return false;  +     } -     //next find the smallest value larger than a[k in [k+1,n-1], so that it is the smallest permutation of the ordered dictionary the     intans=INF; *     intindex=0; $      for(inti=k+1; i<n;i++)Panax Notoginseng     { -         if(a[i]>A[k]) the         { +             if(a[i]<ans) A             { theindex=i; +ans=A[i];  -             } $         } $     } - swap (A[k],a[index]); -     //next the number of [k+1,n-1] is inverted the     intI=1; -      while(1)Wuyi     { the         if(k+i>=n-i) -         { Wu              Break; -         } AboutSwap (a[k+i],a[n-i]); $i++; -     } -     return true;  - } A intMain () + { thescanf"%d",&n); -      for(intI=0; i<n;i++) $     { thescanf"%d",&a[i]); the     } the next_perm (); the      for(intI=0; i<n;i++) -     { inprintf"%d", A[i]);  the     } theprintf"\ n"); About     return 0; the}

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"Algorithm series learning One" full-array generation algorithm