edited computer project

7/30/2019 Edited Computer Project

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INDEX

S. no. TOPIC Page no.1 PROGRAM 1 2-82 PROGRAM 2 9-133 PROGRAM 3 14-184 PROGRAM 4 19-225 PROGRAM 5 23-316 PROGRAM 6 32-377 PROGRAM 7 38-428 PROGRAM 8 43-509 PROGRAM 9 51-59

10 PROGRAM 10 60-6711 PROGRAM 11 68-7012 PROGRAM 12 71-7513 PROGRAM 13 76-7814 PROGRAM 14 79-8215 PROGRAM 15 83-8616 PROGRAM 16 87-9217 PROGRAM 17 93-9718 PROGRAM 18 98-10119 PROGRAM 19 102-10420 PROGRAM 20 105-110


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Question 1- Define a classTWO_D with following data members:

N=size of the array

int arr[][]- a square matrix of N X N

Member functions:

TWO_D(int nn) : to initialize N=nn and to create matrix.void input( ) : to input matrix elements.

int has(int y) : returns 1 if y belongs to matrix arr[][] else it returns 0.

void check( ) : to remove all duplicate elements only single copy should

be there of each element in matrix and insert new

element at that position.

void print( ) : to print the matrix.

Also write main( ). In main both matrix the original one and after deletion.

ALGORITHM

START STEP 1- Input the elements into the square matrix of size 3 as arr[ ][ ].

STEP 2- Print the matrix.

STEP 3- Initialize first element of matrix into two variables-max and min which are to

store smallest and the biggest element of the matrix. STEP 4- Initialize a variable I and assign 0 to it.

STEP 5- Initialize another variable j and assign 0 to it.

STEP 6- If [i][j]th element of matrix arr is greater than max, assign the value of that

element to max.

STEP 7- If [i][j]th element of matrix arr is smaller than min ,assign the value of that

element to min.

STEP 7- Increment js value by 1 and repeat step 6 and 7 as long as it does not

become 2.

STEP 8- Increment is value by and repeat from step 6 to 8 as long as it does notbecome 2.

STEP 8- Initialize a variable m as 0.

STEP 9- Initialize a variable k as min.

STEP 10- Initaliaze a variable c as 0.

STEP 11- Initialize a variable i as 0.

STEP 12- Initialize a variable j as 0.

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STEP 13- If k is equals to [i][j]th element of the matrix, c is incremented.

STEP 14- Just when c gets greater than 1, user is asked to input a new number which

is again checked if its in already in matrix or no through linear search. If the new number

is already present in the matrix, user is again asked to input any number, not already

present in the matrix. But if the new number is not in the matrix, the new number isassigned to that [i][j]th position.

STEP 15- j is incremented by 1 and i4th step is reapeated as long as it does not

become equals to 2.

STEP 16- i is incremented by 1 and steps are reapeated from 12 to 15 as long as it

does not become equals to 2.

STEP 17- k is incremented by 1 and steps are reapeated from 10 to 16 as long as it

does not become equals to max.

STEP 18- m is incremented by 1 and steps are reapeated from 9 to 17 as long as it is

less than 9.

STEP 19- Again the matrix is printed in which all duplicate elements are replaced

with new numbers.

STOP

import java.io.*;

class TWO_D

{ //class startsint arr[][];

int N; //data members initialized

TWO_D(int nn)

{ //parametrizes constructor

N=nn;

arr=new int[N][N];

}

void input()throws IOException

{ //function to enter the elements into the matrix

BufferedReader in=new BufferedReader(new

InputStreamReader(System.in));

System.out.println("enter the elements in the matrix");

for (int i=0;i


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{

for (int j=0;j


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for (int m=0;m


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} //function over

public void print()

{ //function to print the matrix

for (int i=0;i


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OUTPUT

enter the elements in the matrix

11

22

33

11

22

33

1122

33

11 22 33

11 22 33

11 22 33

enter a new no.

22

enter a new no. not present in matrix121

enter a new no.

232

enter a new no.

343

enter a new no.

454

enter a new no.

33enter a new no. not present in matrix

565

enter a new no.

676

enter a new no.

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Question 2-Consider following inheritance path where Student is base class

Above classes are showing data members

Write contructors for passing various attributes for all the three classes and

Average( ) for the class CommerceStudent and ScienceStudent which returns

average marks. Also write display( ) for all the three classes to display the data

members and average too.

Student

Name

Address

rollno

CommerceStudentMarksAcc

MarksEco

ScienceStudentMarksphy

MarksMaths

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ALGORITHM

START STEP 1- Pass values to the data members of sciencestudent as

"Saumya","Maruti Puram",25,58,52.

STEP 2- Pas values to the data members of commercestudent as "Bittu","Janki

Puram",11,78,99.

STEP 2- In both the above steps, last two values are the marks of two subjects.

STEP 3- Average marks are calculated for both the objects by dividing the sum

of marks of both the subjects by 2.

STEP 4- Details are printed: name, address, roll no., marsk of both subjects and

their average. STOP

class mn

{ //main classvoid main()

{ //main function

Sciencestudent ob=new Sciencestudent("Saumya","Maruti Puram",25,58,52);

//object created for first derived class

Commercestudent obc=new Commercestudent("Bittu","Janki Puram"

,11,78,99); //object created for second derived class

ob.display(); //funtion calling

obc.display();}

} //main class over

class Student

{ //begins the base class

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String name;

String address;

int rollno; //data memebers intialized

Student(String u,String v,int w)

{ //parametrized constructorname=u;

address=v;

rollno=w;

}

void display()

{ // function to display the general details of the student

System.out.println("name of the student-"+name);

System.out.println("address-"+address);System.out.println("roll no.-"+rollno);

}

} //base class ends

class Commercestudent extends Student

{ //first derived class begins

int markseco;

int marksacc;Commercestudent(String a,String b,int c,int d,int e)

{ //parametrized constructor

super(a,b,c); //providing values to the base class data members

marksacc=d;

markseco=e;

}

double average()

{ //function to return the average of the 2 asked sunject marks

double ave=(markseco+marksacc)/2;return ave;

}

void display()

{ //function to display all the details

super.display(); //prints the general details

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double y=average(); //y holds the average marks

System.out.println("average marks-"+y);

} //function over

} //first derived class over

class Sciencestudent extends Student

{ //second derived class begins

int marksphy;

int marksmaths;

Sciencestudent(String f,String g,int h,int i,int j)


super(f,g,h); //providing values to the base class data members

marksphy=i;marksmaths=j;

}

double average()

{ //to calculate the average marks of the 2 asked marks

double ave=(marksphy+marksmaths)/2;

return ave;

}

void display(){ //to display all the details

super.display(); //display base class data members

double y=average();

System.out.println("average marks-"+y);

}

} //second derived class over

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OUTPUT

name of the student-Saumya

address-Maruti Puram

roll no.-25

average marks-55.0

name of the student-Bittuaddress-Janki Puram

roll no.-11

average marks-88.0

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Question 3- Define a classTWO_D with following data members :

in arr[][]- a square matrix of N X N

N- size of the array

Member functions:

TWO_D(int nn) : to initialize N=nn and to create matrix

void input( ) : to input matrix elements

void check( ) : to print the smallest in each row along with the

matrix.void sort( ) : to sort each row in descending order using any

sorting technique.


Also write main( ).

ALGORITHM

START

STEP 1- Create a square matrix of size 4.

STEP 2- Input elements into the matrix.

STEP 3- Initialize a variable k as 0.

STEP 4- Initialize [i][0]th element of the matrix as a variable min.

STEP 5- Initialize a variable m as 0.

STEP 6- If [k][m]th element is lesser than min, [k][m]th element is assigned to

the variable min.

STEP 7- Increment ms value by 1 and repeat step 6 as long as it is less then 4.

STEP 8- Increment ks value by 1 and repeat the steps from 4 to 7 as long as it

is less than 4.

STEP 9- At the end of each iteration of k loop, the smallest element of [k]th

row of the matrix is printed.

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STEP 11- If [i][j]th elements is greater than [i][j+1]th element of the matrix,

they are swapped. STEP 12- Increment js value by 1 and repeat step 11 as long as it is less than 3.

STEP 13- Increment Is value by 1 and repeat steps from 10 to 12 as long as it

is less than 4.

STEP 14- Increment ks value byy 1 and repeat steps from 9 to 13 as long as it

is less than 4.

STEP 15- Print the matrix.

STOP

import java.io.*;

class TWO_D1

{ //class starts

int arr[][];

int N; // data members initialized

TWO_D1(int nn)

{ //parametrized constructorN=nn;

arr=new int[N][N];

}


{ //function to input elements in the matrix



System.out.println("enter the elemets");for (int i=0;i


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}

void check()

{ //function to print the smallest element in each row

print(); //function calling to print the elements in the inputted order

for (int k=0;k


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{ //function to print the matrix

for (int i=0;i


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44 8 6 6

Question 4- Fractions are numbersand num is the numerator and den is thedenominator and denis not equal to 0. Some of the methods in class Division are

given below:

Class Name : Division

Data members : num and den of integer types.

Member functions-

Division(int, int) : constructor, first argument is numerator and

second argument is denominator.

int hcf(int, int) : find the hcf of two numbers.int lcm(int, int) : find the lcm(using hcf)

void reduce( ) : reduce current Division objects to its lowest

terms e.g. 1000/3000 is 1/3.

Division addto(Division f) : to add current Division object to Division f and

return the sum as a Division object. E.g. the sum

of 2/9 and 5/18 is .

Also write main( ).

ALGORITHM

START STEP 1- From the two numbers, c and d, find the smaller one and store it in

min. STEP 2- Initialize a variable i as 1 and find the highest number which when

divides both the number, c and d, leaves no remainder and store it in h.

STEP 3- Increment is value by 1 and repeat step 2 as long as it is less than

equals to min.

STEP 4- Find the lcm by using the formula: lcm=product of two numbers/their

hcf.

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STOP

import java.io.*;class main

{ //main class begin

void diplay()throws IOException

{

graduatestudent ob= new graduatestudent(); //base class object is created

System.out.println();

System.out.println();

ob.input();

ob.print();

}

} //main class over

class person

{ //base class begin

String name;

int age,rollno;

double per; //data members initialized

void input()throws IOException{ //function to input general details



System.out.println("enter general details");

name=in.readLine();

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age=Integer.parseInt(in.readLine());

rollno=Integer.parseInt(in.readLine());

per=Double.parseDouble(in.readLine());

}

} //base class overclass graduatestudent extends person

{ //derived class begin

double div; //% of first divisioners

String subject;

int s,t; //s is to store the index of top scorer

int year;

int employed;

//various arrays are initialized to store details of 10 detailsString nm[]=new String[10];

int a[]=new int[10];

int r[]=new int[10];

double marks[]=new double[10];

String sub[]=new String[10];


{ //function to input details

t=0; //t holds the no. of working graduatesBufferedReader in=new BufferedReader(new


System.out.println("enter % of first divisioners");

div=Double.parseDouble(in.readLine());

System.out.println("enter the year");

year=Integer.parseInt(in.readLine());

for(int i=0;i


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enter general details

"Payal"

10

12

99enter employed(1) or not(0)

0

enter subject

"Drawing"


"Nitika"

20

4498

enter employed(1) or not(0)

1

enter subject

"Accounts"


"Shruti"

1840

88.5


0

enter subject

"History"


"Aparna"

1022

66.7


0

enter subject

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"English"


"Kavya"

18

27100


1

enter subject

"Physics"


"Preeti"

1334

99


0

enter subject

"GK"


"Shreya"11

6

94.5


0

enter subject

"Biology"


"Shivani"17

33

98.5


0

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OUTPUT

enter the date

23

11

1995

enter the date

2310

2012

no. of days between 23/11/1995 and 23/10/2012 is 6179

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Question 7- A principal of a school needs to find the highest and the lowest marks

scored by the student and also the merit list on marks from highest to lowest in a

class of 10 students. The classes student and meritlist are declared for above

activities with following details:-

Class name : student

Data members/instance variables:

Tot[10] : (double) an array to store total marks of 10

students out 500 each.

Stud[10] : string array to store name of 10 students of a

Class.

Rollno[10] : string array to store roll no. of 10 students of

a class.

Member function/method:-

void readdetails( ) : to input names and marks of 10 students.

Marks are given out of 500 each. And to

assign rollnos. From 200101 to 200110.

void printresult( ) : to print students scoring highest and lowest

marks in the class along with names and

rollno. of students.

Class Name : meritlist

Data members/instance variables : NONE

Member function/methods:-

void generatemerit( ) : to generate merit list of 10 students on the

basis of total marks along with of students.

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void printmerit( ) : to print merit list of 10 students in three

columns in the following format:

ROLL NO. NAME MARKS SCOREDXXXX XXXXXX XXXXXX

XXXX XXXXXX XXXXXX

XXXX XXXXXX XXXXXX

ALGORITHM

STEP 1- Three arrays are created: tot, name and rollno of size 10.

STEP 2- Initialize a variable c as 200101.

STEP 3- Initialize a variable i as 0. STEP 4- In each iteration, a name and total marks are inpuuted as [i]th element

inot arrays tot and name.

STEP 5 - is value is incremented and step 4 is repeated With each iteration, c is

assigned as [i]th element into array rollno and then c is incremented by 1. This step is

performed as long as I is less than 10.



STEP 8- It is checked if [j]th element is lesser than [j+1]th element of array tot

or not. If it is, then [j+1]th element of arrays tot, name and rollno is assigned as [j]th

element of the arrays and former values of [j]th element of these arrays are assigned to[j+1]th elements of these arrays.

STEP 9- js value is incremented by and step 8 is repeated as long as it is less

than 9.

STEP 10- is value is incremented by 1 and steps are repeated from 7 to 9 as

long as it is less than 10.

STEP 8- All three arrays are printed in proper format.

STEP 9- [0]th and [9]th element of these arrays are printed as highest scorer

and lowest scorer respectively.

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import java.io.*;

class main

{ //class starts

void mn()throws IOException

{ //main functionname ob=new name(); //object created

ob.readlist();

ob.generatemerit(); //required functions are called

}

} //main class over

class student

{ //base class startsint tot[]=new int[10];

String name[]=new String[10];

int rollno[]=new int[10]; //data members initialized

void readlist()throws IOException

{ //function to input the details begins

BufferedReader in=new BufferedReader(new InputStreamReader(System.in));

System.out.println("Enter name,total out of 500");

int c=200101; //first roll no.for(int i=0;i


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if (tot[j]


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{

System.out.println(rollno[i]+ "\t\t" +name[i]+ "\t\t" +tot[i]);

}

} //function over

} //derived class over

OUTPUT

Enter name, total out of 500

111

Saumya222

Charul

333

Deeksha

444

Gauri

121

Kopal232

Himani

343

Shruti

12

Neha

345

Aish

454Aditi

Highest marks 454 are scored by Aditi having rollno.-200110

Lowest marks 12 are scored by Neha having rollno.-200108

ROLL NO. NAME MARKS OBTAINED

200110 Aditi 454

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200104 Gauri 444

200109 Aish 345

200107 Shruti 343

200103 Deeksha 333

200106 Himani 232200102 Charul 222

200105 Kopal 121

200101 Saumya 111

200108 Neha 12

Question 8- Define a class Binary with following data members:

String n1, n2

And following member functions:

Binary( ) :constructor

int check(String n) : returns 1 if n is a binary number else it returns 0.

void input( ) : to input 2 binary numbers in n1, n2 with decimals.

String Dec_to_bin(String h) : to return binary equivalent of h. Eg. If h=5.625

binary eq=101.101

String Bin_to_dec(String h) : to retuen decimal equivalent of h. Eg if h=101.101

decimal eq=5.625void sum( ) : to print sum of n1 and n2 in binary number system

and decimal number system.

Eg.

n1=111.101

n2=110.01

Sum in binary number system is 1101.111

Sum in decimal number system is 13.875

ALGORITHM

START STEP 1- Input two strings, n1 and n2 which consists of only numbers.

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STEP 2- Two strings, op1 and op2, are initialized.

STEP 3- A number is binary only and only if it has 1s, 0s or a decimal(.).

STEP 4- If a string is a binary number, 1 is returned else 0 is returned.

STEP 5- Step 4 is performed for n1 and n2 and returned values are stored in x

and y respectively. STEP 6- If x is 1, n1 is changed into a decimal number and stored in string form

in op1. Else if x is 0, then n1 is simply assigned in op1.

STEP 7- If y is 1, n2 is changed into a decimal number and stored in string form

in op2. Else if y is 0, then n2 is simply assigned in op2.

STEP 8- op1 and op2 are changed into double type and stored in a and b

respectively.

STEP 9- a nd b are added and stored in a variable of double type c.

STEP 10- Initialize a string to_bin.

STEP 11- c is changed into string type and stored in to_bin. STEP 12- Initialize a string type variable in_bin. Change the string to_bin into a

string which has binary equivalent of c and store it in in_bin.

STEP 13- Print c and in_bin.

STOP

import java.io.*;class Binary

{ //class begins

String n1;

String n2; //data members initialized

Binary()

{ //constructor

n1="";

n2="";}


{ //function to provide 2 strings

BufferedReader in=new BufferedReader(new InputStreamReader(System.in));

System.out.println("enter the two strings");

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n1=in.readLine();

n2=in.readLine();

}

int check(String n)

{ //function to check if the string is binary number or notint c=0;

int len=n.length();

for(int i=0;i


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System.out.println("given number is not binary");

}

else

{

int x=0;int len=h.length();

for(int i=0;i


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if(ch==49)

{

temp=temp+(1*Math.pow(2,y));

}

else if(ch==48)

{

temp=temp+(0*Math.pow(2,y));

}

y--;

}

sum=sum+temp;

}

else

{ //if decimal is not there

int a=Integer.parseInt(h);

int y=0;

while(a!=0)

{ sum+=a%10*Math.pow(2,y);

y++;

a=a/10;

}

}

}

String dec="";

dec=dec.valueOf(sum);

return dec;} //function over

String Dec_to_bin(String h)

{ //function to convert decimal no. to binary no.

String real="";

int y=check(h);

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if(y==1)

{

System.out.println("given string is binary already");

return "";

}else

{

int x=0;

int len=h.length();

for(int i=0;i


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Question 9- Define a class trr with following data members:

trr ar[4][4]

And following functions:

void input( ) : to input a 4 X 4 matrix

void transpose(trr M) : to print transpose of matrix M

void sum( ) : to print sum of each row and each column and

primary diagonal and print in following format:

1 1 1 2 5

2 2 2 4 7

1 2 1 2 6

2 2 3 3 10

Sum=6 7 7 11 7

void sort( ) : to sort only border elements in ascending order infollowing format of arr[][].

If arr[][] is

1 2 3 4 1 2 3 3

5 6 7 8 9 6 7 4

9 1 2 3 8 1 2 4

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4 5 6 7 7 6 5 5

Also write main( ).

ALGORITHM

START STEP 1- Create a square matrix with size of 4.

STEP 2- Input elements into the matrix


STEP 5- Initialize a variable sum as 0.


STEP 7- [i][j]th element of the matrix is printed and sum is incremented with [i]

[j]th element.

STEP 8- j is incremented by 1 and step 7 is repeated as long as it is less than 4.

STEP 9- i is incremented by 1 and steps 5 to 8 are repeated as long as it is less

than 4.

STEP 8- At one repeatition of steps from 5 to 8, sum has the sum of all the

elements of [i]th row of the matrix which is then printed..


STEP 10- Initialize a variable c_sum as 0. STEP 11- Initialize a variable j as 0.

STEP 13- c_sum is incremented with [i][j]th element.

STEP 14- Increment js value by 1 and repeat step 13 as long as it is less than 4.

STEP 15- Increment is value by 1 and repeat steps 10 to 14 as long as it is less

than 4.

STEP 16- At one repeatition of steps from 10 to 12, c_sum has the sum of all

the elements of [i]th row of the matrix which is then printed.

STEP 17- Initialize a variable p_sum as 0.

STEP 18- Initialize a variable as 0.

STEP 19- p_sum is then incremented with [i][i]th element of the matrix.

STEP 20- is value is incremented by 1 and step 19 is repeated as long as it is

less than 4.

STEP 21An array of 12 elements is created in which all the border elements of

the matrix are stored.


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{

for(int j=0;j


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System.out.print(c_sum+"\t");

}

int p_sum=0;

for(int i=0;i


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}

//border elements done

for(int i=1;i


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11

22

33

44

5566

77

88

99

99

88

77

6655

44

33

1 2 3 4 10

5 6 7 8 26

9 9 8 7 336 5 4 3 18

21 22 22 22 18

1 2 3 3

9 6 7 4

8 9 8 4

7 6 5 5

11 22 33 44

55 66 77 88

99 99 88 77

66 55 44 33

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11 55 99 66

22 66 99 55

33 77 88 44

44 88 77 33

Question 10Design a program to accept a day number (between 1 and 366), year

(4 digits) from the user to generate and display the corresponding date. Also accept

N (1


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STEP 5- Initialize an integer variable m as 2 which would be used later for

calculation of the month of the date

STEP 6- If s would less than or equals to 31 then mns value would be 1 and

dds value would be s itself.

STEP 7- If s is greater than 31 and less than or equals to 59 and the year is nota leap year or if s is greater than 31 and less than or equals to 60 and the year is a leap

year then dd would be 31 less day_no and mns value is 2

STEP 8- If s is greater than 59 and its not a leap year then s would be 59 less s

and then a loop is created which is iterated as long as s is greater than or equal to 31.

In each iteration, ss vaule is decremented by 31 (as maximum number of days in a

month is 31) and ms value is incremented by 1.

STEP 9- If s is greater than 60 and its a leap year then s would be 60 less s and

then a loop is created which is iterated as long as s is greater than or equal to 31. In

each iteration, ss vaule is decremented by 31 (as maximum number of days in a

month is 31) and ms value is incremented by 1.

STEP 10- If m is greater than 3 and less than oe equals to 5 then xs value is 1

STEP 11- If m is greater than 5 and less than or equals to 8 then xs value is 2.

STEP 12- If m is greater than 8 and less than equals to 10 then xs value is 3.

STEP 13- If m is greater than 10 the n xs value is 4.

STEP 14- Final value of dd would be sum of ss value and xs value and mns

value is one more than ms value.

STEP 15- Create a String array arr whose size is 13.

STEP 16- Keep the first value of arr empty and then sequentially assign all

twelve months names between elements os index no. 1 and 12.

STEP 17- Initialize a String variable suf which would store the suffix to be

written after after the day.

STEP 18- If dd lies between 11 and 19 then sufs value is th, else if dds last

digit is 1 then suf value is st, else if dds last digit is 2 then sufs value is nd, else

if dds last digit is 3 then sufs value is rd and in all other conditions sufs value is

th.

STEP 19- Print dd followed by suf, then a space followed by the mnth

element of the array arr and again a space followed by year.

STEP 20- A new integer variable nday_no is initialized which is the sum of

day_no and N.

STEP 21- If year is a leap year and nday_no is greater than 366 then nday_no

is 366 less than nday_no but if is not a leap year and nday_no is greater than 365 then

nday_no is 365 less than nday_no. In both the cases, year is incremented by 1.

STEP 22- Now all the steps from step no.2 till 19 are repeated.

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STOP

import java.io.*;

class calc

{//class started

int N;

int day_no;

int year; //data members initialized

calc()

{ //default constructor

N=0;

day_no=0;

year=0;} //constructor over


{ //function to input values to data members



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int s=days; //initialising day no. as 's'

if (yr%4==0)

{ //when its a leap year:

if(s31)&&(s60)

{ //when day no. is greater than 60:

s=s-60; //subtracting first 60 days of the year

while(s>=31)

{ //31 as maximum no. of days in a month can be 31 only

s=s-31;

m++;

}//calculating extra days as per the months

if((m>3)&&(m5)&&(m8)&&(m10)

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{

x=4;

}

dd=s+x; //day of the date would also include extra days

mn=m+1; //month of the date would be one more than the calculated as'm'

}

} //all probable conditions for the date in the leap year checked

else if(yr%4!=0)

{ //same conditions would be checked to calculate the date in an year which

is not a leap year

if(s31)&&(s59)

{ //59 as in a normal year total no. of days in first two months is 59

s=s-59;

while(s>=31)

{

s=s-31;

m++;

}

if((m>3)&&(m5)&&(m


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x=2;

}

else if((m>8)&&(m10)

{

x=4;

}

dd=s+x;

mn=m+1;

}}

String arr[]={"", "January", "February", "March", "April", "May", "June",

"July","August", "September", "October", "November",

"December"}; //array of all the months such a way that the

index number is the month number of the year

String suf=""; // 'suf' is the variable for the suffix after the day

if((dd>=11)&&(dd


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OUTPUT

enter limit such that 1


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START STEP 1- Input a value to data member limit.

STEP 2- If n is equal to c, then return 1. Else if n is equals to 1 or on being

divided by c leaves no remainder then return 0. If above two values are not satisfied

then this step is again repeated with variables n remaining the same and (c+1) actingas new c.

STEP 3- If m is equals to 0 then s is returned else remainder of m on being

divided by 10 is stored in a new variable r, s is multiplied by 10 and then to it is added

r and this value is again stored in s, and then this step is repeated with (m/10) acting as

new m.


STEP 5- Assign 0 as value to data member s.

STEP 6- Step 1 is repeated with variables i and 2 and returned value is stored in

a new variable x.

STEP 7- Step 2 is repeated with i acting as m and returned value is stored in anew variable y.

STEP 8- If y is equals to i and x is 1, then print i.

STEP 9- Increment is value by 1 and repeat steps from 5 to 8 as long as it is

less than or equals to limit.

STOP

import java.io.*;class Number

{ //class began

int limit,s;

Number(int l)

{ //constructor

limit=l;

s=0;

}int prime(int n,int c)

{ //recursive function to see if the no. is prime or not

if(n= =c)

{

return 1;

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}

else if((n%c= =0)||(n= =1))

{

return 0;

}else

{

return prime(n,c+1); //repeating the same function

}

}

int reverse(int m)

{ //function to return the reverse of the no.

if(m= =0){

return s;

}

else

{

int r=m%10;

s=s*10+r;

return reverse(m/10);}

}

void print()

{ //to print prime palindrome no. bet 1 and limit

System.out.println("prime palindrome numbers between 1 to "+limit);

for(int i=0;i


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}

}

void main()

{ //function to call the print function

print();}

} //class over

OUTPUT

prime palindrome numbers between 1 to 100

2

3

5

7

11

Question 12- NIC institutes resource manager has decided to network thecomputer resources like printer, storage media etc, so that minimum resources

and maximum sharing could be availed. Accordingly printers are linked to a

centralized system and the printing jobs are done on a first cum first served

basis only. This is like the first persons job will get done first and the next

persons job will be done as the next job in the list and so on. In order to avoid

collision, the restriction is that no more than 20 printing jobs can be added.Define the class PrintJob with the following detail:

Class name : PrintJobData members/instance variable:job[] : array of integers to hold the printing jobs.

newjob : to add a new printing job into the array.

capacity : the maximum capacity of the integer array.

front : to point to the index of the frony.

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rear : to point to the index of the last.

Member functions:

PrintJob( ) : constructor to initialize the data member

capacity=20, front=rear= -1 and call thefunction creatjob( ).

void createjob( ) : to create an array to hold the printing jobs.

void addjob( ) : adds the new printing job to the end of the last

printing job, if possible, otherwise displays

the message Printjob is full, cannot add any

more.

void removejob( ) : removes the printing job from the front if the

printing job is not empty, otherwise displaysthe message Printjob is empty.

Also write main( ) for above class.

ALGORITHM

START

STEP 1- Create an array of size 20.

STEP 3- If rears value is 19 a message is printed Printjob is full,cannot add

more" else rear is incremented by 1 and newjob is assigned as the value of [rear]th

element of the array and if fronts value is -1 then its incremented.

STEP 4- If fronts value is -1, a message is printed Printjob is empty. elsefronts value is incremented.

STEP 5- Now user is asked to input either 1 or 2 or 3.

STEP 6- If its 1 then step 3 is performed, else if its 2 step 4 is performed. Else if

its 3, array is printed. If user gives any other input than these 3, a message is

displayed- wrong input.

STEP 7- Step 5 and 6 are continued as long as user does not input 3.

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STOP

import java.io.*;

class printjob{ //class begin

int job[];

int newjob;

int capacity;

int front;

int rear; //data members initialized

printjob()throws IOException

{ //constructor

capacity=20;

front=-1;

rear=-1;

job=new int[capacity];

createjob();

}

void createjob()throws IOException

{ //function to input values into the array

job=new int[20];} //function ends

void addjob()throws IOException

{

if(rear= =19)

{

System.out.println("Printjob is full,cannot add more");

}

else{

BufferedReader in= new BufferedReader(new


System.out.println("enter the printjob");

int x=Integer.parseInt(in.readLine());

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2

enter choice

enter 1 to input

enter 2 to delete

enter 3 to display and exit3

wrong input

Question 13- Define a class Recursion with following data members

String str

Int v, w

And following functions:

Recursion( ) : constructor.

void accept( ) : to input any string.

void count(int ) : to count number of words starting with a

vowel in str and store in data member v andnumber of words in w using recursion.

void display( ) : to print number of words starting with vowels

and words in str.

Also write main( ).

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ALGORITHM

START STEP 1- Initialize three data members- string type str and two integer types v

and w.

STEP 2- Calculate the length of str and store it in a variable l.

STEP 3- Begin this step with i as 0. As long as i is less than l, we extract out the

[i]th character of str and store it in ch. If ch is any vowel then v is incremented by 1

else if ch is a space then w is incremented by 1 and again this step is repeated with

(i+1) acting as new i.

STEP 4- (w+1) is the no. of words in str which are printed and v is number of

words starting with vowel which is again printed.STOP

import java.io.*;

class recursion

{ //class begin

String str;

int v,w; //data members initialized

recursion()

{ //constructorstr="";

v=0;

w=0;

}

void accept()throws IOException

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} //class over

OUTPUT

enter the string

"Saumya is a good boy"

no. of words 5no. of words starting with vowels 2

Question 14- Define a class unique with following data members:

int arr[10]

And following methods:

void input( ) : to input 10 integers in arr.

int check(int n) : returns 1 if n is a unique no. else it returns 0.

A unique no. is a no. whose all the digits are

different (eg. 1273 is unique but 121 is notunique.)

void print( ) : to print the original array and then print all the

unique numbers in the array with proper

messages.

Also write main( ).

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{ //function to input all ten elements in an array



System.out.println("enter all the 10 elements of the array");

for(int i=0;i


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ALGORITHM

START

STEP 1- Create three objects ob1 with argument 5, ob2 with argument 10 andob3 with argument 0.

STEP 2- Input elements into the array of object ob1 and then print it.

STEP 3- Input elements into the array of object ob2 and then print it.

STEP 4-Initialize 0 as value of a new variable x.

STEP 5- Create a loop from 0 to 4.

STEP 6- Assign [i]th element of the array of object ob1 as value of [x]th

element of the array of object ob3.

STEP 7- x is incremented by 1.

STEP 8- Create a loop from 0 to 9. STEP 9- Assign [i]th element of the array of object ob2 as value of [x]th

element of the array of object ob3.

STEP 10- x is incremented by 1.

STEP 11- Print the array of object ob3.

STOP

import java.io.*;

class array

{ //class begin

int a[];

int size; //data members initialized

array(int n){ //parametrized constructor

size=n;

a=new int[size];

}


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for(int m=0;m


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} //function ends

} //class over

OUTPUT

enter the elements

12

3

4

5

enter the elements

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6

7

8

9

1011

2 4 6 8 10

enter the elements

1

2

2

2

31 2 3

Question 17- Define a class pattern with following data members and functions:

int size, int arr[ ][ ]

void input( ) : to input size(>=3) and to create array.

void print( ) : to generate following matrix.

If n is odd If n is even

If n=5 o/p if n=4 o/p

4 2 1 3 5 1234

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9 7 6 8 10

14 12 11 13 15 4123

19 17 16 18 20

24 22 21 23 25 3412

2341

ALGORITHM

START

STEP 1- Input a number as size

STEP 2- Create a matrix of that size.

STEP 3- If value of size is an odd then then initialize 1 as value of a new

variable y.

STEP 4- Find the middle of the size as x.

STEP 5- Start assigning values from middle of each row.

STEP 6- But if size is even values are assigned from the [i][i]th position where I

varies from 0 to the size.STOP

import java.io.*;

class pattern

{ //class started

int size; //data member for the size of the array

int arr[][]; //data member for the array

pattern()

{ //constructor

size=0;

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}


{ //function to input the value of size


InputStreamReader(System.in));System.out.println("enter any number greater than 3");

size=Integer.parseInt(in.readLine());

arr=new int[size][size]; //creates the array of size 'size'

} //function ends

void print()

{ //function to assign values in the matrix

if(size%2!=0)

{ //if the size of the array is an odd numberint y=1; //first element in the matrix

int x=size/2; //middle point array

for(int i=0;i


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}

}

else if(size%2= =0)

{ //if the size of the array is an even number

for(int i=0;i


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34 32 30 29 31 33 35

41 39 37 36 38 40 42

48 46 44 43 45 47 49

enter any number greater than 3

41234

4123

3412

2341

Question 18- A class arithmetic is defined to perform arithmetic operation on

the expression given. The expression given contains only one operator(+, -, *

or /). The expression should be input as a string and the output should be a

number. Assume that only integer constants and an operator is given as theexpression. E.g. INPUT 10*7

OUTPUT 70

The description of the class is given below:

Class name : arithmetic

Data members : String expr, double result

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STEP 14- If chs Unicode value is 42, then product of a and b is stored in result.

STEP 15- If chs Unicode value is 47, then a is divided by b and its quotient is

stored in result.

STEP 16- Result is then printed.

STOP

import java.io.*;

class arithmetic

{ //class starts

String expr;

double result; //data membersvoid accept()throws IOException

{ //function to input values to data members



System.out.println("enter a string containing digits and any of the arithmetic

operators only");

expr=in.readLine();

}void calculate( )

{ //function to calculate the result from mathematical expression

char ch=' '; //to store the arithmetic operator

int x=0; //to store the index of the arithmetic operator in the string

expression

int len=expr.length(); //to store the length of string expression

for(int i=0;i


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break; //stops the iteration once operator is found

}

}

String op1=expr.substring(0,x); //stores first half of the expression as op1

String op2=expr.substring(x+1); //stores the second half of the expression asop2

double a=Double.parseDouble(op1); //op1 is converted into double type as a

double b=Double.parseDouble(op2); //op2 is converted into double type as

b

int c=0;

c=(int) ch; //stores the unicode value of the operator

if (c==43)

{ //unicode value of + is 43result=a+b;

}

else if(c==45)

{ //unicode value of - is 45

result=a-b;

}

else if(c==42)

{ //unicode value of * is 42result=a*b;

}

else if(c==47)

{ //unicode value of / is 47

result=a/b;

}

} //function ends with the final value of result

void print()

{ //function to display the string expression and the resultcalculate(); //function is called

System.out.println("the expression given by the user "+expr); //string

expression is displayed

System.out.println(result); //result is displayed

} //function ends

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void main()throws IOException

{ //main function

accept(); //input function is called

print(); //print function is called

} //main function ends} //class ends

OUTPUT

enter a string containing digits and any of the arithmetic operators only

122/4

the expression given by the user 122/4

30.5

Question 19- Define a class Number with following specifications:

Data members:

int limit, (you can assume other data members also)

Member functions:

Number( ) : constructor to initialize limit=0.

Number(int L) : constructor to initialize limit=L.

int length(int n) : returns length of number n. Using recursion.

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limit=L;

}

int length(int n)

{ //function to return the number of digits to nif(n/10==0)

{

return 1;

}

else

{ //returns(1+calls again to the function)

return(1+length(n/10));

}} //function ends

int power(int h)

{ //function to return 10 to power h

if(h==1)

{ //when 10 to power is 1,the number is 10 itself

return 10;

}

else{ //else the number is 10*(calls the function again(h-1))

return(10*power(h-1));

}

} //function ends

void automorphic()

{ //function to check if the number is automorphic or not

for(int i=1;i


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}

}

} //function ends

void main()

{ //main functionNumber ob=new Number(100); //object is created with an arguement

ob.automorphic(); //automorphic function is called

} //function ends

} //class ends

OUTPUT

1

5

6

2576

Question 20- Define a class Matrix with following specifications:

Data members:

int arr[ ][ ] a matrix of 3 X 3.

Member functions:

void input( ) : to input a 3X3 matrix.

Matrix multiplication(Matrix M1) : to return multiplication of matrix M1

and current matrix object.

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Matrix sort(Matrix M1) : to sort each row of matrix M1 in

ascending order using selection sort and

returns it.


Write main( ) for above class.

ALGORITHM

START

STEP 1- Create three matrices. STEP 2- For first two matrices, the size is 3.

STEP 3- Values are inputted into the matrices and then printed.



STEP 5- 0 is initialized as a value of new integer variable sum.


STEP 7- sum is incremented with product of [k][j]th element of first matrix and

[j][i]th element of second matix.

STEP 8- Increment js value by 1 and repeat step 7 as long as it is less than 3.

STEP 9- sums value is assigned to the [k][i]th element of third matrix.

STEP 10- Increment is value by 1 and repeat steps from 5 to 9 as long as it is

less than 3.

STEP 11- Increment ks value by 1 and repeat steps from 4 to 10 as long as it is

less than 3.

STEP 12- Third matrix is printed.


STEP 14- Selection sort of [i]th row of first matrix is performed

STEP 15- Increment is value by 1 and repeat step 14 as long as it is less than 3.

STEP 16- Print this sorted matrix.STOP

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import java.io.*;

class matrix

{ //class begins

int arr[][]=new int[3][3]; //matrix is created

void input()throws IOException{ //values are inputted into the matrix



System.out.println("enter elements into the matrix");

for(int i=0;i


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int sm=m1.arr[i][j];

int pos=j;

for(int k=j+1;km1.arr[i][k]){

sm=m1.arr[i][k];

pos=k;

}

}

int t=m1.arr[i][j];

m1.arr[i][j]=m1.arr[i][pos];

m1.arr[i][pos]=t; //swapping done} //selection sort done

}

return m1; //object is returned

} //function over

matrix multiplication(matrix m1)

{ //function to multiply curent matrix matrix and M1 and then return it

matrix obj=new matrix();

for(int k=0;k


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{ //main function

matrix ob1=new matrix(); //first object

matrix ob2=new matrix(); //second object

matrix ob3=new matrix(); //third object

matrix ob4=new matrix(); //fourth objectob1.input();

ob1.print();

ob2.input();

ob2.print();

ob3=ob1.multiplication(ob2);

System.out.println("multiplication of two matrices:");

ob3.print();

ob4=sort(ob1);System.out.println("sorted matrix");

ob4.print();

} //function over

} //class ends

OUTPUT

enter elements into the matrix1

5

0

11

33

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55

67

99

3

1 5 0

11 33 55

67 99 3

enter elements into the matrix

11

55

88

4477

1

5

8

6

11 55 88

44 77 15 8 6

multiplication of two matrices:

231 440 93

1848 3586 1331

5108 11332 6013

sorted matrix

0 1 511 33 55

3 67 99

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edited computer project

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