Rezolvare BAC Informatica 2009 – Varianta 51 – Subiectul al III-lea problema 3
Mai jos puteti gasi rezolvarea problemei 3 de la subiectul III din varianta 51 pentru examenul de bacalaureat din anul 2009
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| #include <iostream> | |
| using namespace std; | |
| int dist2(int xa, int ya, int xb, int yb) { | |
| return (xa - xb) * (xa - xb) + (ya - yb) * (ya - yb); | |
| } | |
| typedef struct{ | |
| int x; | |
| int y; | |
| }TPunct; | |
| int main() { | |
| TPunct v[4]; | |
| int i; | |
| for (i = 0; i < 4; i++) { | |
| cin>>v[i].x>>v[i].y; | |
| } | |
| int latura = -1,distantaMinima = 0,ePatrat = 1; | |
| for (i = 0; i < 4 && ePatrat == 1; i++) { | |
| int j,distantaMinima = 0; | |
| for (j = 0; j < 4; j++) { | |
| if (i != j) { | |
| if (distantaMinima == 0) | |
| distantaMinima = dist2(v[i].x,v[i].y,v[j].x,v[j].y); | |
| else if (dist2(v[i].x, v[i].y,v[j].x,v[j].y) < distantaMinima) | |
| distantaMinima = dist2(v[i].x,v[i].y,v[j].x,v[j].y); | |
| } | |
| } | |
| if (latura == -1) | |
| latura = distantaMinima; | |
| else if (distantaMinima != latura) | |
| ePatrat = 0; | |
| } | |
| if (ePatrat) | |
| cout<<"DA"; | |
| else | |
| cout<<"NU"; | |
| return 0; | |
| } |
by Spor la lucru! Daca aveti intrebari nu ezitati sa le lasati in comentarii, va vom raspunde cat de repede putem 
Exista o problema cu acest algoritm. In cazul in care valorile introduse corespund unor puncte coliniare (de ex 1,1 / 2,2, / 3/3 , 4/4) ploblema retunreaza “DA”, cand ar trebui sa returneze nu. Acesta este algoritmul la care m-am gandit eu.
#include
using namespace std;
int x[4],y[4],i;
void interschimbare(int a, int b)
{
int aux=x[a];
x[a]=x[b];
x[b]=aux;
aux=y[a];
y[a]=y[b];
y[b]=aux;
}
void ordonare()
{
for(i=1;i<4;i++)
if(x[i]<x[0] || y[i]<y[0])
interschimbare(0,i);
for(i=0;ix[2] || y[i]>y[2])
interschimbare(2,i);
}
void afisare()
{
for(i=0;i<4;i++)
cout<<x[i]<<' '<<y[i]<<endl;
}
int dist(int xa, int ya, int xb, int yb)
{
return (xb-xa)*(xb-xa)+(yb-ya)*(yb-ya);
}
int main()
{
int ok=1;
for(i=0;i>x[i]>>y[i];
cout<<endl;
ordonare();
afisare();
int pd=dist(x[0],y[0],x[1],y[1]);
for(i=0;i<=2;i++)
if(pd!=dist(x[i],y[i],x[i+1],y[i+1])) {ok=0;}
if(ok==1 && dist(x[0],y[0],x[2],y[2])==2*pd && dist(x[1],y[1],x[3],y[3])==2*pd )
cout<<"DA";
else cout<<"NU";
}