Code for sorting and combination

1. Generate class cyclic Arrangement
Input example
3 2
Output example
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1

# Include <iostream> # include <ctime> using namespace std; int A [100]; void Print_permutation (int n, int * A, int cur) {int I, j; if (cur = n) {/* for (I = 0; I <n; ++ I) {cout <A [I] <"";} cout <endl; */} else {for (I = 1; I <= n; I ++) {bool flag = true; for (j = 0; j <cur; ++ j) {if (A [j] = I) {flag = false; break ;}} if (flag) {A [cur] = I; print_permutation (n, A, cur + 1) ;}}int main () {int n = 0; clock_t start, end; cin> n; start = clock (); Print_permutation (n, A, 0); end = clock (); cout <endl <(double) (end-start) /CLOCKS_PER_SEC <"s" <endl; return 0;} In fact, This method uses recursion to implement multiple loops. This recursive program is equivalent to n reloops, each
When the length of the recycle is m, the output contains a total of m ^ n rows.

 

 

2. Do not repeat:

The number of n is input, and the number of n is arranged. Repeated items are not allowed.
Input example
3
1 1 2
Output example
1 1 2
1 2 1
2 1 1

Version One :　

LRJ code:

#include <iostream> using namespace std ; int P[100] , A[100] ; void Print_permutation(int n , int *P , int *A , int cur) { if (cur == n) { for (int i = 0 ; i < n ; ++ i) { cout << A[i] << " " ; } cout << endl ; } else { for (int i = 0 ; i < n ; ++ i) { if (!i || P[i] != P[i-1]) { int c1 = 0 , c2 = 0 ; for (int j = 0 ; j < cur ; ++ j) { if (A[j] == P[i]) { c1 ++ ; } } for (int j = 0 ; j < n ; ++ j) { if (P[i] == P[j]) { c2 ++ ; } } if (c1 < c2) { A[cur] = P[i] ; Print_permutation(n,P,A,cur+1) ; } } } } } int main() { int n ; cin >> n ;  for (int i = 0 ; i < n ; ++ i) { cin >> P[i] ; } Print_permutation(n,P,A,0) ; return 0 ; } 

 

Version Two :

#include <iostream> using namespace std ; const int MAXN = 100 ; int A[MAXN] , P[MAXN] , used[MAXN] , tmp[MAXN] ;  void Print_permutation(int n , int m , int *A , int *P , int cur) {if (cur == n) { for (int i = 0 ; i < n ; ++ i) { cout << A[i] << " " ; } cout << endl ; } else { for (int i = 0 ; i < m ; ++ i) { if (used[i] > -1) { used[i] -- ; A[cur] = P[i] ; Print_permutation(n,m,A,P,cur+1) ; used[i] ++ ; } } } } 　　　int main() { int n , m ; cin >> n ; int count = 0 ; memset(P,0,sizeof(P)) ; for(int i = 0 ; i < n ; ++ i) {cin >> tmp[i] ; P[count++] = tmp[i] ; for (int j = 0 ; j < i ; ++ j) { if (tmp[j] == tmp[i]) { used[j] ++ ; count -- ; break ; } } } for (int i = 0 ; i < count ; ++ i) { cout << P[i] << " " ; } cout << endl ; Print_permutation(n,count,A,P,0) ; return 0 ; } 

Version Three:

(Core Implementation of next_permutation in STL)

template<class _BidIt> inline
 bool next_permutation(_BidIt _First, _BidIt _Last)
 { // permute and test for pure ascending, using operator<
   _DEBUG_RANGE(_First, _Last);
   return (_Next_permutation(_Unchecked(_First), _Unchecked(_Last)));
 }

template<class _BidIt> inlinebool _Next_permutation(_BidIt _First, _BidIt _Last){// permute and test for pure ascending, using operator<_BidIt _Next = _Last;if (_First == _Last || _First == --_Next)return (false);for (; ; ){// find rightmost element smaller than successor_BidIt _Next1 = _Next;if (_DEBUG_LT(*--_Next, *_Next1)){// swap with rightmost element that's smaller, flip suffix_BidIt _Mid = _Last;for (; !_DEBUG_LT(*_Next, *--_Mid); );_STD iter_swap(_Next, _Mid);_STD reverse(_Next1, _Last);return (true);}if (_Next == _First){// pure descending, flip all_STD reverse(_First, _Last);return (false);}}}

3. Full arrangement: (input without repeated elements)

#include <iostream> using namespace std ; const int MAXN = 100 ; int used[MAXN] , A[MAXN] , P[MAXN] ; void Print_permutation(int n , int *A , int *P , int cur) { if (cur == n) { for (int i = 0 ; i < n ; ++ i) { cout << A[i] << " " ; } cout << endl ; } else { for (int i = 0 ; i < n ; ++ i) { if (!used[i]) { used[i] = 1 ; A[cur] = P[i] ; Print_permutation(n,A,P,cur+1) ; used[i] = 0 ; } } }} int main() { int n = 0 ; cin >> n ; for (int i = 0 ; i < n ; ++ i) { cin >> P[i] ; } memset(used,0,sizeof(used)) ; Print_permutation(n,A,P,0) ; return 0 ; }