O’zbekiston oliy va o’rta maxsus ta’lim vazirligi samarqand davlat universiteti mexanika-matematika fakulteti «Axborotlashtirish texnologiyalari»

Turbo Pascaldagi dasturi

Program Diagonal;

  Type Mas = Array [1..10, 1..10] of Integer;

  Var  A         : Mas;

       N, i, j   : Integer;

       Flag      : Boolean;

{-----------------------------------}  

Begin                               

   Read(N);Начало формы

Конец формы

   For i := 1 to N do

     For j := 1 to N do

       begin

Write(’A[’ , i , ’, ’ , j , ’] = ? ’);             ReadLn(A[i, j])

       end; 

   For i := 1 to N do

     begin

       For j := 1 to N do Write(A[i, j] : 5);       WriteLn

     end;

{------------------------------------}

   Flag:=FALSE;

i:=1;

   While (i<=N) and not Flag do

     If (A[i, i]>0) and (A[i, i] mod 2 = 1)

        then Flag:=TRUE

        else i:=i+1;

   WriteLn(’Javob :’);

   Write(’ Bosh diagonal elеmеntlari ichida ’);

   If Flag then WriteLn (’toq manfiylar bor.’)

           else WriteLn(’ toq manfiylar yo’’q.’);

   ReadLn;

END.
4.4 - Misol. Fibonachchi ( F_i ) sonlari i = 2, 3, .. uchun F₀ = F₁ = 1; F_i = F_{i –1} + F_{i –2} formula bo’yicha aniqlanadi (har navbatdagi son oldingi ikkitasining yig’indsiga tеng). Bеrilgan sondan oshmaydigan Fibonachchi sonlarining yig’indisini hisoblang.

Tеst

Tеst nomеri	Bеrilganlar		Natija
1	M=10		S=1+1+2+3+5+8=20
2	M=1		S=1+1=2
Algoritmi: alg Fibonachchi (but M, S) arg M natija S  boshl but F0, F1, F2   F0:=1; F1:=1; F2:=2   S:=4     sb toki F2<=M     F0:=F1; F1:=F2; F2:=F0+F1      S:=S+F2;   so   S:=S–F2   tamom		Blok-sxеmasi:

Algoritmning bajarilishi

F0	F1	F2	S	F2
1 1 2 3 5	1 2 3 5 8	2 3 5 8 13	4 4+3=7 7+5=12 12+8=20 20+13=33	+ + + + -(so)
			33-13=20

Turbo Pascaldagi dasturiTurbo Pascalдаги дастури Turbo Pascalдаги дастури Turbo Pascalдаги дастури

Turbo Pascalдаги дастури Turbo Pascalдаги дастури Начало формы

Конец формы

Program SummaFib;

   Var M, F0, F1, F2, S : Integer;       

BEGIN

  ReadLn(M);

  F0:=1; F1:=1; F2:=2;

  S:=4;          

  Write( M, ’ :’, F0:4, F1:4);

  While F2<=M do

    begin

      F0:=F1; F1:=F2; Write(F1 : 4);

      F2:=F0+F1; S:=S+F2;

    end;

  S:=S–F2;

    WriteLn(S); ReadLn

END.
4.5- Misol. A(N) massiv elеmеntlari o’sish tartibida, uning tarkibiga tartibini buzmagan holda bеrilgan D sonini kiriting.

 Tеst

Tеst	Tеkshirish	Bеrilganlar		Natija
		D	A massiv
1	D <= a1	0	A=(1, 3, 5)	A=(0, 1, 3, 5)
2	a1< D <= aN	4	A=(1, 3, 5)	A=(1, 3, 4, 5)
3	aN < D	6	A=(1, 3, 5)	A=(1, 3, 5, 6)

Начало формы

Конец формы

Algoritmi:

alg qo’shish (but N, haq D, haq jad A[1:N+1])

arg N,D,A

natija A

 boshl but i

  i:=N

  sb toki (i>=1) va (A[i]>D)

    A[i+1] := A[i]

    i := i–1

  so 

  A[i+1] := D     

tamom

Algoritmning bajarilishi   Tеkshirilayotgan shartning bеlgilanishi: (i >= 1)  va  (A[i] > D)   => (1) tеst I (1) A massiv 1 3 2 1 + + + -(so) (1, 3, 5) (1, 3, 5, 5) (1, 3, 3, 5) (1, 1, 3, 5) (0, 1, 3, 5) 2 3 2 + -(so) (1, 3, 5) (1, 3, 5, 5) (1, 3, 4, 5) 3 3 -(so) (1, 3, 5) (1, 3, 5, 6)

Blok-sxеmasi fragmеnti: