3-laboratoriya topshiriqlari 7-variant (1-jadval)

Muhammad al-Xorazmiy nomidagi Toshkent axborot texnologiyalari universiteti 413-20-guruh talabasi Davronov Diyorbekning Dasturlash_1 fanidan bajargan 3-laboratoriya ishi



3-laboratoriya topshiriqlari

7-variant (1-jadval)

7	;	1	b=4 c=-2 d=8	X=0.750
		2	b=4 c=-3 d=7	X=0.250

Dastur kodi

#include

#include

#include

using namespace std;

int main(){

float a, b, c, d, X;

cout << "b = "; cin >> b;

cout << "c = "; cin >> c;

cout << "d = "; cin >> d;

a = (b*b - c*c) / d;

if (a>0) {

X = pow(a,3) / (3 + a);

cout << "X = " << setprecision(3) << fixed << X << endl; }

else {

X = sqrt(fabs( (a*a+2) / (1+a) ));

cout << "X = " << setprecision(3) << fixed << X << endl; }

return 0;

}



Natijalar





7-masala (2-jadval)

7.	OX va OY o‘qlarida yotmaydigan nuqta koordinatalari bilan berilgan. Bu nuqta joylashgan koordinata choragi aniqlansin.

Dastur kodi

#include

using namespace std;

int main()

{

float x, y;

cout << "A nuqtani koordinatasini kiriting: \n";

cout << "x = "; cin >> x; cout << "y = "; cin >> y;

if (x>0 and y>0)

cout << "A nuqta 1-chorakda joylashgan!" << endl;

else

if (x<0 and y>0)

cout << "A nuqta 2-chorakda joylashgan!" << endl;

else

if (x<0 and y<0)

cout << "A nuqta 3-chorakda joylashgan!" << endl;

else

if (x>0 and y<0)

cout << "A nuqta 4-chorakda joylashgan!" << endl;

main();

return 0;

}

Natija

9-masala (2-jadval)

9.	Haqiqiy x,y va z sonlari berilgan bo‘lib, x

Dastur kodi

#include

#include

using namespace std;

int main()

{

float x, y, z;

cout << "x= "; cin >> x;

cout << "y= "; cin >> y;

cout << "z= "; cin >> z;

if ( xx) {

cout << "Ikkilangan qiymatlar:\n";

cout << x*2 << endl;

cout << y*2 << endl;

cout << z*2 << endl; }

else {

x = fabs(x); y = fabs(y); z = fabs(z);

cout << x << endl;

cout << y << endl;

cout << z << endl; }

main();

return 0;

}

Natija

11-masala (2-jadval)

11.	Sonlar o‘qida uchta A, B va C nuqtalar joylashgan. B va C nuqtalardan qaysi biri A nuqtaga yaqin masofada joylashgan bo`lsa, shu masofa chop etilsin.

Dastur kodi

#include

#include

using namespace std;

int main()

{

float A, B, C;

cout << "A = "; cin >> A;

cout << "B = "; cin >> B;

cout << "C = "; cin >> C;

if ( fabs(A-B) > fabs(A-C) )

cout << "|AB| = " << fabs(A-B) << endl;

else

if ( fabs(A-B) < fabs(A-C) )

cout << "|AC| = " << fabs(A-C) << endl;

else

cout << "|AB| = |AC| = " << fabs(A-B) << endl;

return 0;

}



Natija

12-masala (2-jadval)

12.	Berilgan uch xonali son raqamlari orasida bir xillari bor yoki yo‘qligi aniqlansin?

Dastur kodi

#include

#include

using namespace std;

int main()

{

int a, x, y;

boshi:

cout << "3 xonali butun son kiriting: "; cin >> a;

x = a%10;

y = (a/10) % 10;

a = a/100;

if (((a>=100 and a<=999) or (a>=-999 and a<=-100)) or (x==y or x==a or a==y)) {

cout << "Berilgan sonda raqamlar takrorlangan!" << endl; }

else

if ((a>999 or a<100) or (a>-100 or a<-1000)) {

cout << "! Siz 3 xonali butun son kiritmadingiz!!!\n" << endl;

goto boshi; }

else

cout << "Raqamlar takrorlanmagan!" << endl;

return 0;

}

Natija

Uyga vazifa

18-masala: 100-999 oraliqdagi sonlarni so`zda ifodalovchi programma tuzilsin.

Dastur kodi

#include

using namespace std;

int main()

{

int n;

start:cout << "N = "; cin >> n; //123

if(n>=100 and n<=999) {

if (n/100 == 1)

cout << "Bir yuz " << endl;

else

if (n/100 == 2)

cout << "Ikki yuz " << endl;

else

if (n/100 == 3)

cout << "Uch yuz " << endl;

else

if (n/100 == 4)

cout << "To`rt yuz " << endl;

else

if (n/100 == 5)

cout << "Besh yuz " << endl;

else

if (n/100 == 6)

cout << "Olti yuz " << endl;

else

if (n/100 == 7)

cout << "Yetti yuz " << endl;

else

if (n/100 == 8)

cout << "Sakkiz yuz " << endl;

else

if (n/100 == 9)

cout << "To`qqiz yuz " << endl;

if(n/10%10 == 0)

cout << "_" << endl;

if(n/10%10 == 1)

cout << "o`n " << endl;

if(n/10%10 == 2)

cout << "yigirma " << endl;

if(n/10%10 == 3)

cout << "o`ttiz " << endl;

if(n/10%10 == 4)

cout << "qirq " << endl;

if(n/10%10 == 5)

cout << "ellik " << endl;

if(n/10%10 == 6)

cout << "oltmish " << endl;

if(n/10%10 == 7)

cout << "yetmish " << endl;

if(n/10%10 == 8)

cout << "sakson " << endl;

if(n/10%10 == 9)

cout << "to`qson " << endl;

if (n%10 == 0)

cout << "_" << endl;

if (n%10 == 1)

cout << "bir " << endl;

if (n%10 == 2)

cout << "ikki " << endl;

if (n%10 == 3)

cout << "uch " << endl;

if (n%10 == 4)

cout << "to`rt " << endl;

if (n%10 == 5)

cout << "besh " << endl;

if (n%10 == 6)

cout << "olti " << endl;

if (n%10 == 7)

cout << "yetti " << endl;

if (n%10 == 8)

cout << "sakkiz " << endl;

if (n%10 == 9)

cout << "to`qqiz " << endl;

}

else {

cout << "[100;999] oraliqda son kiriting!\n";

goto start; }

return 0;

}

15-masala (2-jadval)

15.	tenglamaning haqiqiy ildizlarini topadigan yoki ildizi yo‘qligi haqida ma’lumot beradigan dastur tuzilsin.

Dastur kodi

#include

#include

#include

using namespace std;

int main()

{

float y1, y2, a, b, c, D;

cout << "Bikvadrat tenglama koeffitsientlarini kiriting:\n";

cout << "a = "; cin >> a;

cout << "b = "; cin >> b;

cout << "c = "; cin >> c;

// x*x=y

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

if (D >= 0) {

y1 = (-b + D)/(2*a);

y2 = (-b - D)/(2*a); }

else

cout << "Bikvadrat tenglama yechimga ega emas1" << endl;

if (y1 >= 0 and y2 >= 0)

{

cout << "Bikvadrat tenglama 4ta yechimga ega: " << endl;

cout << "x = " << -sqrt(y1) << " , " << sqrt(y1) << endl;

cout << "x = " << -sqrt(y2) << " , " << sqrt(y2) << endl;

}

else

if (y1 >= 0 and y2 < 0)

cout << "2ta yechimga ega: x= " << -sqrt(y1) << " , " << sqrt(y1) << endl;

else

if (y2 >= 0 and y1 < 0)

cout << "2ta yechimga ega: x= " << -sqrt(y2) << " , " << sqrt(y2) << endl;

else

if (y1 < 0 and y2 < 0)

cout << "Bikvadrat tenglama yechimga ega emas2" << endl;

main();

system("pause");

}

Natija

20-masala 2-jadval

Natural n soni berilgan. Sonni to‘rt xonali deb hisobga olgan holda ushbu sonning palindrom ekanligi aniqlansin (chapdan va o‘ngdan bir xil o‘qiladigan sonlar, masalan, 1221, 5555, 440 sonlari palindrom sonlar hisoblanadi).

Dastur kodi

#include

using namespace std;

int main(){

int n, m, k, t, y, x;

cout << "4 xonali Natural son kiriting: "; cin >> n; //9543

if (n>0) {

m = n/1000; //9

k = n%1000; // 543

t = k/100; //5

y = (k/10)%10; //4

x = k%10; //3

if (m==x and t==y)

cout << "Kiritilgan son palindrom!\n";

else

cout << "Palindrom son emas!\n";

}

else

cout << "Siz 4xonali Natural son kiritmadingiz!!!" << endl;

main();

return 0;

}

Natija

3-jadval 7-masala

Case7. Og’irlik birliklari quyidagi tartibda berilgan. 1-kilogramm, 2-milligramm, 3-gramm, 4-tonna, 5-sentner. Og’irlik birligini bildiruvchi son berilgan (1-5 oraliqda) va shu birlikdagi og’irlik qiymati berilgan (haqiqiy son). Og’irlikni kilogrammda ifodalovchi dastur tuzilsin.

Dastur kodi

#include

using namespace std;

int main(){

int a;

float m, kg;

cout << "Og`irlik birligini kiriting: 1:kilogram, 2:milligram, 3:gram, 4:tonna, 5:sentner \n";

cin >> a;

switch(a){

case 1: cout << "Og`irlikni Kilogramda kiriting: "; cin >> m;

kg = m; cout << kg << " kg" << endl; break;

case 2: cout << "Og`irlikni milligramda kiriting: "; cin >> m;

kg = m/1000000; cout << kg << " kg" << endl; break;

case 3: cout << "Og`irlikni gramda kiriting: "; cin >> m;

kg = m/1000; cout << kg << " kg" << endl; break;

case 4: cout << "Og`irlikni tonnada kiriting: "; cin >> m;

kg = m*1000; cout << kg << " kg" << endl; break;

case 5: cout << "Og`irlikni sentnerda kiriting: "; cin >> m;

kg = m*100; cout << kg << " kg" << endl; break;

default: cout << "Xatolik! Qaytadan kiriting!\n"; }

main();

return 0; }

Natija

Case13-masala

Case13. Teng yonli uchburchakning elementlari quyidagi tartibda nomerlangan: 1-katet a, 2-gipotenuza , 3-gipotenuzaga tushirilgan balandlik 4-yuzasi Shu elementlardan bittasi berilgan qolganlarini topuvchi dastur tuzilsin.

Dastur kodi

#include

#include

using namespace std;

int main(){

float S, a, c, h;

int k;

cout << "Kiriting 1: katet, 2: gipotenuza, 3: balandlik, 4: yuzasi\n";

cin >> k;

switch(k) {

case 1: cout << "Katetni kiriting: "; cin >> a;

c=sqrt(2)*a; cout << "Gipotenuza: " << c << endl;

h=c/2; cout << "Balandlik: " << h << endl;

S=c*h/2; cout << "Yuza: " << S << endl; break;

case 2: cout << "Gipotenuzani kiriting: "; cin >> c;

a=c/sqrt(2); cout << "Katet: " << a << endl;

h=c/2; cout << "Balandlik: " << h << endl;

S=c*h/2; cout << "Yuza: " << S << endl; break;

case 3: cout << "Balandlikni kiriting: "; cin >> h;

c=2*h; cout << "Gipotenuza: " << c << endl;

a=c/sqrt(2); cout << "Katet: " << a << endl;

S=c*h/2; cout << "Yuza: " << S << endl; break;

case 4: cout << "Yuzani kiriting: "; cin >> S;

c=sqrt(4*S); cout << "Gipotenuza: " << c << endl;

h=c/2; cout << "Balandlik: " << h << endl;

a=c/sqrt(2); cout << "Katet: " << a << endl; break;

default: cout << "Xatolik! Qayta kiriting" << endl; }

main(); return 0; }

Natija

Nazorat savollariga javob

While takrorlash operatori

while takrorlash operatori, operator yoki blokni tarorlash sharti yolg’on (false yoki 0) bo’lguncha takror bajaradi. U quyidagi sintaksisga ega:

while () ;

Agar rost qiymatli o’zgarmas ifoda bo’lsa, takrorlash cheksiz bo’ladi. Huddi shunday, takrorlash boshlanishida rost bo’lib, uning qiymatiga takrorlash tanasidagi hisoblash ta’sir etmasa, ya’ni uning qiymati o’zgarmasa, takrorlash cheksiz bo’ladi.

Xossalari:

Sikl sharti har safar tekshiriladi.

Agar sikl sharti yolg’on bo’lsa, sikl umuman bajarilmaydi.

Agar sikl sharti yolg’on bo’lmasa, dastur takrorlanaveradi.

Do-while so`ng shartli takrorlash operatori

do-while takrorlash operatori while operatoridan farqli ravishda oldin operator yoki blokni bajaradi, keyin takrorlash shartini tekshiradi. Bu qurilma takrorlash tanasini kamida bir marta bajarilishini ta’minlaydi. do-while takrorlash operatori quyidagi sintaksisga ega:

do ;

while ();

Bunday takrorlash operatorining keng qo‘llaniladigan holatlari - bu takrorlashni boshlamasdan turib, takrorlash shartini tekshirishning iloji bo‘lmagan holatlar hisoblanadi.

Switch Tanlash operatori

Switch tanlash operatorida kiritilgan biror bir qiymat uning kalit so`zlari(case() qavs ichidagi ifoda) bilan solishtirib ko`riladi, agar ular ustma-ust tushsa unga mos amallar ketma-ketligi bajariladi, aks holda operator ichiga kirmaydi va undan keyingi operator bajarilib ketaveradi.