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sâmbătă, 3 noiembrie 2012
Blog-uri
Asemanatoare cu problema nr 8
Folosind metoda backtracking sa se descompuna in toate modurile un numar natural n ca suma de numere prime distincte ordonate crescator.
#include<iostream.h>
int n,x[100],v[100],m;
int prim(int n)
{int i;
if(n==0 || n==1) return 0;
else if(n==2 || n==3) return 1;
else if(n%2==0) return 0;
for(i=3;i*i<=n;i+=2)
if(n%i==0) return 0;
return 1;
}
void afis(int n)
{ int i;
for(i=1;i<=n;i++)
cout<<x[i]<<" ";
cout<<endl;
}
void back(int k,int sp)
{ int i;
for(i=1;i<=m;i++)
{ x[k]=v[i];
sp=sp+x[k];
if(sp<=n && x[k]>x[k-1]) if(sp==n) afis(k);
else back(k+1,sp);
sp=sp-x[k];
}
}
void main()
{ int i;
cin>>n;
m=0;
for(i=2;i<=n;i++) if(prim(i)) { m++;
v[m]=i;
}
back(1,0);
}
Problema 19 (nu e rezolvata in clasa,oricum e o idee)
n camile sunt asezate in sir indian. Sa se afiseze toate posibilitatile de rearanjare a acestora astfel incat o camila sa aiba in fata ei o camila diferita de cea din configuratia initiala.
#include<iostream>
using namespace std;
int n,x[10],nrsol;
int afisare()
{int i;
nrsol++;
for(i=1;i<=n;i++)
cout<<x[i]<<" ";
cout<<endl;
}
int verif(int i)
{int j;
for(j=1;j<i;j++)
if(x[j]==x[i] || x[i]-x[i-1]==1 && i>1)
return 0;
return 1;
}
void back(int i)
{int j;
for(j=1;j<=n;j++)
{x[i]=j;
if(verif(i))
if(i==n)
afisare();
else back(i+1);
}
}
int main()
{int i;
cin>>n;
back(1);
cout<<endl<<"Nr. sol= "<<nrsol;
}
#include<iostream>
using namespace std;
int n,x[10],nrsol;
int afisare()
{int i;
nrsol++;
for(i=1;i<=n;i++)
cout<<x[i]<<" ";
cout<<endl;
}
int verif(int i)
{int j;
for(j=1;j<i;j++)
if(x[j]==x[i] || x[i]-x[i-1]==1 && i>1)
return 0;
return 1;
}
void back(int i)
{int j;
for(j=1;j<=n;j++)
{x[i]=j;
if(verif(i))
if(i==n)
afisare();
else back(i+1);
}
}
int main()
{int i;
cin>>n;
back(1);
cout<<endl<<"Nr. sol= "<<nrsol;
}
Se citesc de la tastatura numele a n copii. Sa se afiseze toate posibilitatile de aranjare a acestora pe n scaune.
#include<iostream>
using namespace std;
char a[10][10];
int n,x[10],nrsol;
int afisare()
{int i;
nrsol++;
for(i=1;i<=n;i++)
cout<<a[x[i]]<<" ";
cout<<endl;
}
int verif(int i)
{int j;
for(j=1;j<i;j++)
if(x[j]==x[i])
return 0;
return 1;
}
void back(int i)
{int j;
for(j=1;j<=n;j++)
{x[i]=j;
if(verif(i))
if(i==n)
afisare();
else back(i+1);
}
}
int main()
{int i;
cin>>n;
for(i=1;i<=n;i++)
cin>>a[i];
back(1);
cout<<endl<<"Nr. sol= "<<nrsol;
}
using namespace std;
char a[10][10];
int n,x[10],nrsol;
int afisare()
{int i;
nrsol++;
for(i=1;i<=n;i++)
cout<<a[x[i]]<<" ";
cout<<endl;
}
int verif(int i)
{int j;
for(j=1;j<i;j++)
if(x[j]==x[i])
return 0;
return 1;
}
void back(int i)
{int j;
for(j=1;j<=n;j++)
{x[i]=j;
if(verif(i))
if(i==n)
afisare();
else back(i+1);
}
}
int main()
{int i;
cin>>n;
for(i=1;i<=n;i++)
cin>>a[i];
back(1);
cout<<endl<<"Nr. sol= "<<nrsol;
}
Problema 44
Se citesc n culori. Sa se formeze toate drapelele de 3 culori astfel incat oricare doua culori alaturate sa fie distincte.
#include<iostream>
using namespace std;
char a[10][10];
int n,x[10],nrsol,k;
int afisare()
{int i;
nrsol++;
for(i=1;i<=k;i++)
cout<<a[x[i]]<<" ";
cout<<endl;
}
int verif(int i)
{int j;
for(j=1;j<i;j++)
if(x[j]==x[i] || x[i]==x[i-1] && i>1)
return 0;
return 1;
}
void back(int i)
{int j;
for(j=1;j<=n;j++)
{x[i]=j;
if(verif(i))
if(i==k)
afisare();
else back(i+1);
}
}
int main()
{int i;
cout<<"n=";cin>>n;
k=3;
for(i=1;i<=n;i++)
cin>>a[i];
back(1);
cout<<endl<<"Nr. sol= "<<nrsol;
}
Problema 61
Să se genereze toate codurile de lungime n morse formate din '.' şi '-' astfel încât să nu existe două puncte alăturate.
#include<iostream>
using namespace std;
char x[100];
int n,nr=0;
int verif(int i)
{if(i>1 && x[i]=='.' && x[i-1]=='.')
return 0;
else return 1;
}
void back(int i)
{char j;
for(j='-';j<='.';j=j+'.'-'-')
{x[i]=j;
if(verif(i))
if(i==n)
{ cout<<x+1<<endl;nr++;}
else back(i+1);
}
}
int main()
{cin>>n;
x[n+1]=NULL;
back(1);
cout<<endl<<nr<<" solutii";
}
#include<iostream>
using namespace std;
char x[100];
int n,nr=0;
int verif(int i)
{if(i>1 && x[i]=='.' && x[i-1]=='.')
return 0;
else return 1;
}
void back(int i)
{char j;
for(j='-';j<='.';j=j+'.'-'-')
{x[i]=j;
if(verif(i))
if(i==n)
{ cout<<x+1<<endl;nr++;}
else back(i+1);
}
}
int main()
{cin>>n;
x[n+1]=NULL;
back(1);
cout<<endl<<nr<<" solutii";
}
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