Sikllar hоsil qilish

for (int i = 1; i <= 1000; i++) {
for (int j = 1; j <= 1000; j++) {
if (i*i + j*j==c*c) {
a = i;
b = j;
found = true;
break;
}
}
}
Bu shaklda siklni sindiradigan bo’lsak faqat ichki sikl sinadi, tashqi sikl keyingi iteratsiyadan davom qiladi. Siklni butunlay sindirish uchun har bir tashqi sikl iteratsiyasidan oldin tekshiramiz, agar javob topilgan bo’lsa u holda tashqi siklni sindiramiz.
for (int i = 1; i <= 1000; i++) {
if (found)
break;
for (int j = 1; j <= 1000; j++) {
if (i*i + j*j==c*c) {
a = i;
b = j;
found = true;
break;
}
}
}
goto operatori.
Agar ichma-ich joylashgan sikllar soni ko’p bo’ladigan bo’lsa u holda siklni sindirish uchun har bir siklda break operatorini yozish lozim bo’ladi. Buning o’rniga sikldan keyingi amallar bajarilishiga o’tish uchun goto operatoridan foydalanish mumkin. goto operatoridan keyin nishon yoziladi. Agar goto operatori yozilsa, dastur nishon yozilgan nuqtadan bajarishni davom qildiradi. Nishon harflar, $(dollor belgisi), _(pastki chiziqcha), raqamlardan iborat bo’ladi va raqam bilan boshlanmaydi.
#include
using namespace std;
int main() {
int c;
cout<<"c=";
cin>>c;
bool found = false;
int a = 0, b = 0;
for (int i = 1; i <= 1000; i++) {
for (int j = 1; j <= 1000; j++) {
if (i*i + j*j==c*c) {
a = i;
b = j;
found = true;
goto label_1;
}
}
}
label_1:
if (found)
cout< else
cout<<"Bunday sonlar yo'q";
}

1. 
   c)
   
2. 
   
   

package sample.lab2_3_12.lab3_1_12;
import java.util.*;
public class cpp {
public static void main(String[] args)
{
Scanner cin = new Scanner(System.in);
double p = 1, s = 0, h = 0;
for(int i = 1; i <= 19; i++)
{
p*=(Math.pow(i,4) + i * i + 3)/Math.sqrt(Math.pow(i,i) + Math.exp(i));
}
for(int k = 1; k <= 20; k++)
{
s+=(k + 1)/(Math.pow(k,3) + 5 * k + 7);
}
for(int m = 1; m <= 25; m++)
{
double l =1;
for(int n = 1; n <= 30; n++)
{
l*=Math.sqrt((Math.pow(m,n) - Math.pow(n,m))/(Math.pow(m,n) + Math.pow(n,m)));
}


h+=l;
}


System.out.println("p="+p);
System.out.println("s="+s);
System.out.println("h="+h);


}}