Muhammad al – Xorazmiy nomidagi Toshkent axborot texnologiyalari universiteti
Laboratoriya ishi № 5
Bajardi: Abduxamidov Muxriddin
Tekshirdi: Aliqulov Akmal
Guruh: 021 – L1
Toshkent – 2018
5-Labaratoriya ishi
Massivlar. Bir o’lchovli va ko’p o’lchovli massivlarni tashkil etish.
Massivlar ustida amallar bajarish.
Ishning maqsadi: darsning nazariy qismini mustahkamlash va massivlar bilan ishlash ko’nikmalarini oshirish . Massivlar ishtirokida dasturlar tuzish ko’nikmalarini xosil qilish, o’rganilayotgan til va uning boshqaruvchi konstruksiyasi bilan tanishish.
Topshiriq
Hisobot quyidagilardan iborat bo’lishi kerak:
Labaratoriya ishining nomi;
Labaratoriya ishiga topshiriq;
15,16,17,18,19,22 jadvallrda keltirilgan masalalarga mos algoritm va dasturlar tuzish;
Nazorat savollariga tayyorgarlik;
1-topshiriq
Berilgan {aj},j=1,2,...,10 massivda bitta bo’lsa ham manfiy element bor. Birinchi manfiy elementgacha bo’lagan sonlarning ko’paytmsini hisoblang.
Boshlang’ich ma’lumotlar:3,8,17,-10,-16,-1,5,8,9,10.
package com.company;
import java.util.Random;
import java.util.Scanner;
import static java.lang.Math.*;
public class Main {
public static void main(String[] args) {
int A[]={3,8,17,-10,-16,-1,5,8,9,10};
double s=1;
Random a=new Random();
for (int i=0;i<3;i++)
s*=A[i];
System.out.println("s="+s);
2-topshiriq
X(N) massivning Xi>M,shartini qanoatlantiruvchi elementlar yig’indisini hisoblang. (bunda M=maxX-minX).
package com.company;
import java.util.*;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner ha=new Scanner(System.in);
int a[]={10,5,6,3,8,7,9,-7,-10};
System.out.println("X="+Arrays.toString(a));
int max; max=a[0];
for (int i=0;iif (max}
System.out.println();
System.out.println("max="+max);
int min;min=a[0];
for (int i=0;iif (min>a[i])min=a[i];
}System.out.println("min="+min);
int M,S=0;
M=max-min;
System.out.println("M="+M);
for (int i=0;iif (MS+=a[i];}
System.out.println("S="+S); }}
3-topshiriq
A=(a1,a2,….,an)massivning juft indekslari elementlarini B=(b1,b2,…bn)massivga ketma ket yozing . Massivning juft elementlarining soni k-ni toping. ni hisoblang .
Begin
package javaapplication1;
P=1, a[]={1,2,3,4,5,6,7,8,9,10;
import static java.lang.Math.*;
import java.util.Scanner;
int i=1;ipublic class NewClass {
public static void main(String[] args) {
i%2==0
Scanner sc=new Scanner (System.in);
double P=1;
double a[]={1,2,3,4,5,6,7,8,9,10};
i
for(int i=1;i
P+=i
if(i%2==0)
System.out.print(" "+i+".");
P+=i;
P
}
System.out.println(" P="+P);
End
}
}
4-topshiriq
Berlgan A(n,m) boshlang’ich matritsaning barcha ustunlarida joylashganminimal elelmentlaridanB=(b1,b2,...,bm)vektorni tashkil eting
package com.company;
import java.lang.reflect.Array;
import java.util.*;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner ha=new Scanner(System.in);
int n,m;
System.out.println("Satirlar soni");
System.out.print("n=");n=ha.nextInt();
System.out.println("ustunlar soni");
System.out.print("m=");m=ha.nextInt();
int [][] Array=new int[n][m];
for (int i=0;ifor (int j=0;jSystem.out.print(Array[i][j]+" ");}
System.out.println();
}}}
5-topshiriq
BvaC vektorlar uzunlilklarini ko’paytmasini hisoblang.
package com.company;
import java.util.*;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner ha=new Scanner(System.in);
int b[]={10,5,6,3,8,7,9,-7,-10};
int c[]={1,8,9,3,4,6,5};
System.out.println("B="+Arrays.toString(b));
System.out.println("C="+Arrays.toString(c));
System.out.println("C ning uzunligi="+c.length);
System.out.println("B ning uzunligi="+b.length);
int S=(b.length)*(c.length);
System.out.println("ko'paytmasi");
System.out.println("S="+S);
}}
6-topshiriq
Berlgan {x}:-3.5,7.4,1.56,81,40,-0.15,20,-4.75,60,37.5. massivdan quyidagi formula asosida massiv hosil qiling va ustun ko’rinishida chop eting.
import static java.lang.Math.*;
public class JavaApplication{
public static void main(String[]args){
double a[]={-3.5,7.4,1.56,81,40,-0.15,20,-4.75,60,37.5};
double b[]=new double[10];
for(int i=0;i<10;i++){
b[i]=sqrt(abs(a[i]));
System.out.println(b[i]);
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