Bazele programării în C

Problema 1

Rezolvarea ecuatiei ax2 + bx + c = 0, a,b,c Î R.

Pseudocod

{ real a, b, c, x, x1, x2, im, d, r;

scrie (”coef ecuatiei: ”); citeste (a,b,c);

if (a) then { d=b*b – 4*a*c;

if (d>=0) then

{ if(d>0) then

{ x1= (real) (-b+ sqrt(d)) /2/a;

x2= (real) (-b+ sqrt (d))/2/a;

scrie (”x1= ”, x1, ”x2= ”, x2);}

else { x=-b/2/a;

scrie (”x1=x2=”,x);}

}

else { r=-b/2/a;

im=(real) abs( sqrt(-d)/2/a);

scrie(”x1= ”,r,im,”x2= ”,r,im);}

else

if (b!=0) then scrie (x=-c/b);

else if (c!=0) then scrie (”ecuatie imposibila!”);

else scrie (”ecuatie nedeterminata!”);

}

Valori numerice

a=2;

b=4;

c=2;

d=16--4*2*2=16-16=0;

x=-4/2*2=-1;

x1=x2=-1

Codificare

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{clrscr();

float a,b,c,x,x1,x2,im,d, r;

printf("coeficientii ecuatiei: ");

scanf("%f %f %f", &a, &b, &c);

if (a )

{ d=b*b-4*a*c;

if(d >=0)

{ if (d> 0)

{ x1=(float)(-b+sqrt(d))/2/a;

x2=(float)(-b-sqrt(d))/2/a;

printf ("x1 = %5.2f , x2=%5.2f", x1, x2);

}

else { x=-b/2/a;

printf("x1 = x2 = %5.2f", x);}

}

else { r = -b/2/a;

im=(float)fabs(sqrt(-d)/2/a);

printf("x1=%5.2f+%5.2fi;x2=%5.2f-%5.f2fi",r,im,r,im);

} }

else

if(b!=0) printf("x= %5.2f", -c/b);

else if(c!=0) printf("ecuatie imposibila!");

else printf("ecuatie nedeterminata!");

}

Rezultatele rularii

Varianta 2

Codificare

include<process.h>

#include<stdio.h>

#include<math.h>

#include<float.h>

void main()

{

double a, b,c ,x1,x2,delta;

printf("n introduceti a:"); scanf("%lf",&a);

printf("n introduceti b:"); scanf("%lf",&b);

printf("n introduceti c:"); scanf("%lf",&c);

if (a==0&&b==0&&c==0)

printf("n coeficientii sunt nuli");

if (a==0&&b==0&&c!=0)

printf("n ecuatie imposibila: a=0,b=0,c diferit de 0");

if (a==0&&b!=0)

printf("n ecuatia are gradul unu,x=%f",-c/b);

if (a!=0){

delta=b*b-4*a*c;

if (delta==0)

printf("n ecuatia are radacinile egale x1=x2=%f",-b/(2*a));

else

if (delta<0)

{ printf("n ecuatia are radacini complexe");

delta=-delta;

delta=sqrt(delta);

printf("n x1=%f+%fi",(-b)/(2*a),delta/(2*a));

printf("n x2=%f-%fi",-b/(2*a),delta/(2*a));

}

else

{delta =sqrt(delta);

printf("n ecuatie are radacini reale distincte");

printf("n x1=%f",(-b+delta)/(2*a));

printf("n x2=%f",(-b-delta)/(2*a));

}

}

}

Rezultatele rularii