Rezumat problemă

PermPF

#202

Se citeşte un număr natural nenul n. Să se afişeze, în ordine lexicografică, permutările fără puncte fixe ale mulţimii {1,2,..,n}.

Fișiere Candale Silviu (silviu) medie

Clasa 11 Backtracking Elemente de combinatorică

Rezolvă

Detalii

Problema	PermPF	Operații I/O	permpf.in/permpf.out
Limita timp	0.5 secunde	Limita memorie	Total: 64 MB / Stivă 8 MB
Id soluție	#64268034	Utilizator	Raileanu Sofia (RaileanuSofia)
Fișier	permpf.cpp	Dimensiune	679 B
Data încărcării	27 Aprilie 2026, 11:35	Scor/rezultat	Eroare de compilare

Evaluare

Mesaj compilare

permpf.cpp:4:1: error: expected ‘,’ or ‘;’ before ‘ofstream’
    4 | ofstream fout("permpf.out")
      | ^~~~~~~~
permpf.cpp: In function ‘int valid(int)’:
permpf.cpp:9:8: error: ‘v’ was not declared in this scope
    9 |     if(v[k]==k)return 0;
      |        ^
permpf.cpp:11:12: error: ‘v’ was not declared in this scope
   11 |         if(v[i]==v[k])return 0;return 1;
      |            ^
permpf.cpp:11:9: warning: this ‘if’ clause does not guard... [-Wmisleading-indentation]
   11 |         if(v[i]==v[k])return 0;return 1;
      |         ^~
permpf.cpp:11:32: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the ‘if’
   11 |         if(v[i]==v[k])return 0;return 1;
      |                                ^~~~~~
permpf.cpp:10:5: warning: this ‘for’ clause does not guard... [-Wmisleading-indentation]
   10 |     for(i=1;i<=k-1;i++)
      |     ^~~
permpf.cpp:11:32: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the ‘for’
   11 |         if(v[i]==v[k])return 0;return 1;
      |                                ^~~~~~
permpf.cpp: In function ‘int solutie(int)’:
permpf.cpp:15:11: error: ‘n’ was not declared in this scope
   15 |     if(k==n)
      |           ^
permpf.cpp:15:5: warning: this ‘if’ clause does not guard... [-Wmisleading-indentation]
   15 |     if(k==n)
      |     ^~
permpf.cpp:16:18: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the ‘if’
   16 |         return 1;return 0;
      |                  ^~~~~~
permpf.cpp: In function ‘void afisare(int)’:
permpf.cpp:21:22: error: ‘fout’ was not declared in this scope
   21 |     for(i=1;i<=k;i++)fout<<v[i]<<' ';fout<<'\n';
      |                      ^~~~
permpf.cpp:21:28: error: ‘v’ was not declared in this scope
   21 |     for(i=1;i<=k;i++)fout<<v[i]<<' ';fout<<'\n';
      |                            ^
permpf.cpp:21:5: warning: this ‘for’ clause does not guard... [-Wmisleading-indentation]
   21 |     for(i=1;i<=k;i++)fout<<v[i]<<' ';fout<<'\n';
      |     ^~~
permpf.cpp:21:38: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the ‘for’
   21 |     for(i=1;i<=k;i++)fout<<v[i]<<' ';fout<<'\n';
      |                                      ^~~~
permpf.cpp:21:38: error: ‘fout’ was not declared in this scope
permpf.cpp: In function ‘void backt(int)’:
permpf.cpp:27:16: error: ‘n’ was not declared in this scope
   27 |     for(i=1;i<=n;i++)
      |                ^
permpf.cpp:29:9: error: ‘v’ was not declared in this scope
   29 |         v[k]=i;
      |         ^
permpf.cpp:30:11: warning: suggest explicit braces to avoid ambiguous ‘else’ [-Wdangling-else]
   30 |         if(valid(k)==1)
      |           ^
permpf.cpp: In function ‘int main()’:
permpf.cpp:37:8: error: ‘n’ was not declared in this scope
   37 |   fin>>n;backt(1);
      |        ^
permpf.cpp:39:3: error: ‘fout’ was not declared in this scope
   39 |   fout.close():
      |   ^~~~

Cum funcționează evaluarea?