Scheme logice-programare referat

Sa se realizeze schema logica pentru determinarea maximului unui sir X de n elemente si a pozitiei lui in cadrul sirului.

START

CITESTE

I=1

I=I+1

I<=N

CITESTE

V[I]

MAX=V[1]

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

POZ=1

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

I=2

MAX<

V[i]

I<=N

41832htk53lxv9q

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA

MAX=V[i]

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

AFISEAZA

MAX, POZ

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU

POZ=i

STOP

I=I+1

Sa se realizeze schema logica pentru calculul sumei a n numere reale(X₁,X₂,X₃,………..,X_N) citite de la tastatura.

START

S=0

CITESTE

I=1

I<=N

CITESTE

X[I]

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA

AFISEAZA

S=S+X[I]

I=I+1

STOP

Sa se realizeze schema logica pentru calculul produsului a n numere reale(X₁,X₂,X₃,………..,X_N) citite de la tastatura.

START

P=1

CITESTE

I=1

I<=N

CITESTE

X[I]

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA

AFISEAZA

P=P*X[I]

I=I+1

STOP

4.Sa se realizeze schema logica pentru calculul valorii functiei f(x)=xⁿ, cu x si n dati de la tastatura.

START

CITESTE

X,N

P=1

I=1

I>N

P=P*X

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU

I=I+1

AFISEAZA

STOP

Sa se realizeze schema logica pentru calculul expresiei:

START

CITESTE

S=0

P=1

I=1

I>N

CITESTE X

E=S+P

S=S+X

AFISEAZA

P=P*X

STOP

I=I+1

6. Sa se realizeze schema logica pentru rezolvarea urmatoarei probleme x este suma ce o are de incasat o persoana pentru munca prestata si y suma care trebuie sa-I fie retinuta. Evidentiati toate situatiile ce se pot ivi atunci cand trebuie determinata suma ramasa de plata. Afisati rezultatul.

START

CITESTE

X,Y

X>Y

S=X-Y

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q DA

X=Y

41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

S=0

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

S=Y-X

AFISEAZA

STOP

7. Fiind date trei valori reale A,B,C sa se determine si sa se afiseze cea mai mare dintre ele. Aceeasi problema pentru cea mai mica valoare.

START

CITESTE A,B,C

A>B

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA 41832htk53lxv9q

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

A>C

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA

A<C

NU DA

MIN=B

MAX=A

41832htk53lxv9q 41832htk53lxv9q

MAX=C

B>C

NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA

MIN=C

MIN=A

MIN=B

MIN=C

MAX=B

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

B>C

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA

41832htk53lxv9q

MAX=C

MAX=B

AFISEAZA MIN,MAX

STOP

8. Sa se realizeze schema logica pentru calculul sumei elementelor unui vector de 5 elemente.

Generalizarea pentru n elemente

START

S=0

I=1

I>5

CITESTE

V[I]

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

AFISEAZA

S=S+v[I]

I=I+1

STOP

Pe caz general

START

CITESTE

S=0

I=1

I>N

CITESTE

V[I]

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

AFISEAZA

S=S+v[I]

I=I+1

STOP

9. Sa se realizeze schema logica pentru evaluarea expresiei:

START

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

CITESTE

A,B

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

A-B²=0

DA NU

AFISEAZA

“Expesia nu are valoare”

E=((A*A)+B)/(A-(B*B))

AFISEAZA

STOP

10. Sa se realizeze schema logica pentru evaluarea expresiei:

START

CITESTE A,B,C

C<0

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

C=0

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA

E=A²-B

A²-B>=0

AFISEAZA

41832htk53lxv9q

41832htk53lxv9q 41832htk53lxv9q

Afiseaza

“Expesia nu are sens”

AFISEAZA

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

A=0

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA

Afiseaza

“Expesia nu are sens”

41832htk53lxv9q

AFISEAZA

STOP

Fie a,b,c variabile ale caror valori sunt numere reale distincte, iar x o variabila booleana(0 sau 1). Sa se determine valoarea variabilei v

START

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

CITESTE

A,B,C

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

X=1

NU DA

V=A+B+C

C=0

41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA

AFISEAZA

“Expesia nu are valoare”

V=(A+B)/C

AFISEAZA

STOP

12. Sa se realizeze schema logica pentru determinarea mediei de varsta a unei colectivitati, cunoscand varsta fiecarui individ, pentru cazul in care se stie N numarul de membri.

START

S=0

CITESTE

I=1

I<=N

CITESTE

M=S/N

AFISEAZA

S=S+v

I=I+1

STOP

- pentru cazul in care nu se stie cati membri sunt

START

S=0

N=0

CITESTE

S=S+v

N=N+1

MAI SUNT?

NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA

M=S/N

AFISEAZA

STOP

13. Sa se realizeze schema logica pentru calculul numarului de elemente negative, pozitive si nule ale unui vector de n elemente.

START

NEG=0

POZ=0

NULE=0

CITESTE N

I=1

I<=N

CITESTE V[I]

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

AFISEAZA

POZ,NEG,NULE

V[I]>0

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q NU 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q DA 41832htk53lxv9q

V[I]=0

POZ=POZ+1

STOP

41832htk53lxv9q 41832htk53lxv9q

NEG=NEG+1

NULE=NULE+1

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q

I=I+1

41832htk53lxv9q

program p1;

type vect=array[1..100] of integer;

var i,n,poz,max:integer;

41832htk53lxv9q v:vect;

begin

write('n=');

readln(n);

for i:=1 to n do

readln(v[i]);

max:=v[1];

poz:=1;

i:=2;

for i:=1 to n do

if max<v[i] then begin

41832htk53lxv9q max:=v[i];

41832htk53lxv9q poz:=i;

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q end;

write(max,poz);

end.

program p2;

type vect=array[1..100] of real;

var i,n:integer;

41832htk53lxv9q s:real;

41832htk53lxv9q x:vect;

begin

s:=0;

write('n=');

readln(n);

for i:=1 to n do begin

readln(x[i]);

s:=s+x[i];

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q end;

write(s);

end.

program p3;

type vect=array[1..100] of real;

var i,n:integer;

41832htk53lxv9q p:real;

41832htk53lxv9q x:vect;

begin

p:=1;

write('n=');

readln(n);

for i:=1 to n do begin

readln(x[i]);

p:=p*x[i];

41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q 41832htk53lxv9q end;

write(p);

end.

program p4;

var x,n,i,p:integer;

begin

write('n=');

readln(n);

write('x=