81818088 isc-class-xii-computer-science-project-java-programs

COMPUTER
SCIENCE
PROJECT
For ISC
Programming
in BLUEJ
Tirthanu Ghosh
12 A
Roll no. 11

Computer
Science
project
Tirthanu Ghosh
12 A
Roll No. 11

contents
No. Program Page No.
1 Pascal’s Triangle 1
2 Number in Words 3
3 AP Series 5
4 Calendar of Any Month 8
5 Factorial (Using Recursion) 11
6 Fibonacci Series (Using Recursion) 13
7 GCD (Using Recursion) 15
8 Spiral Matrix 17
9 Magic Square 20
10 Linear Search 23
11 Binary Search 26
12 Selection Sort 29
13 Bubble Sort 32
14 Decimal to Binary Number 35
15 Date Program 37
16 Star Pattern Using Input String 40
17 Palindrome Check 42
18 Frequency of Each String Character 44
19 Word Search in String 47
20 Decoding of String 49
21 String in Alphabetical Order 52
22 Number of Vowels and Consonants 55
23 Word Count 57
24 Replacing Vowels with * 59
25 Sum of All Matrix Elements 61
26 Sum of Matrix Column Elements 63
27 Sum of Matrix Diagonal Elements 65
28 Sales Commission 67
29 Decimal to Roman Numerical 69
30 Celsius to Fahrenheit (Using Inheritance) 71

ACKNOWLEDGEMENTS
First of all, I’d like to thank my parents for helping me out
with the project.
Secondly, I’d like to thank our Computer teachers Pinaki Sir
and Archan Sir for helping us with the programs.
Lastly, I’m really grateful to Tabish Haider Rizvi whose help
was imperative for myself making this project.

1 | I S C C o m p u t e r S c i e n c e P r o j e c t
PROGRAM 1
To Create Pascal’s Triangle
ALGORITHM
STEP 1 - START
STEP 2 - pas[0] = 1
STEP 3 - IF i=0 THEN GOTO STEP 4
STEP 4 - IF j=0 THEN GOTO STEP 5
STEP 5 - PRINT pas[j]+" "
STEP 6 - i++& IF i<n GOTO STEP 4
STEP 7 - j=0 & IF j<=i GOTO STEP 5
STEP 8 - IF j=i+1 THEN GOTO STEP 7
STEP 9 - pas[j]=pas[j]+pas[j-1]
STEP 10 - j--& IF j>0 GOTO STEP 9
STEP 11 – END
solution
import java.io.*;
class Pascal
{public void pascalw()throws IOException //pascalw() function
{BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.println(“Enter a no.”);
int n=Integer.parseInt(br.readLine()); //accepting value
int [ ] pas = new int [n+1];
pas[0] = 1;
for (int i=0; i<n; i++) //loop evaluating the elements
{for (int j=0; j<=i; ++j)
System.out.print(pas[j]+" "); //printing the Pascal Triangle elements
System.out.println( );
for (int j=i+1; j>0; j--)
pas[j]=pas[j]+pas[j-1];
}}}

varia
No. Name Type
1 br Buffere
2 n int
3 pas int[]
4 i int
5 j int
outpu
2 | I S C C o m p u t
able descr
Method Description
edReader pascalw() BufferedReader object
pascalw() Input value
pascalw() Matrix storing pascal num
pascalw() Loop variable
pascalw() Loop variable
ut
e r S c i e n c e P r o j e c t
ription
mbers

PROGRAM 2
To Display Entered Number
in Words
ALGORITHM
STEP 1 - START
STEP 2 - INPUT amt
STEP 3 - z=amt%10 , g=amt/10
STEP 4 - IF g!=1 THEN GOTO STEP 5 OTHERWISE GOTO STEP 6
STEP 5 - PRINT x2[g-1]+" "+x1[z]
STEP 6 - PRINT x[amt-9]
STEP 7 – END
solution
import java.io.*;
class Num2Words
{public static void main(String args[])throws IOException //main function
System.out.println("Enter any Number(less than 99)");
int amt=Integer.parseInt(br.readLine()); //accepting number
int z,g;
String x[]={“”,"Ten","Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"};
String x1[]={"","One","Two","Three","Four","Five","Six","Seven","Eight","Nine"};
String x2[]={"","Twenty","Thirty","Fourty","Fifty","Sixty","Seventy","Eighty","Ninety"};
z=amt%10; //finding the number in words
g=amt/10;
if(g!=1)
System.out.println(x2[g-1]+" "+x1[z]);
else System.out.println(x[amt-9]);
}}

varia
No. Name Type
1 br Buffere
2 z int
3 g int
4 x String[]
5 x1 String[]
6 x2 String[]
7 amt int
outpu
able descr
Method Description
edReader main() BufferedReader object
main() amt%10
main() amt/10
] main() String array storing no.in w
main() input number
ut
ription
words
words
words

PROGRAM 3
To Display A.P. Series and
Its Sum
ALGORITHM
STEP 1 - START
STEP 2 - a = d = 0
STEP 3 - IMPORT a, d
STEP 4 - this.a = a & this.d = d
STEP 5 - IMPORT n
STEP 6 - RETURN (a+(n-1)*d)
STEP 7 - IMPORT n
STEP 8 - RETURN (n*(a+nTHTerm(n))/2)
STEP 9 - IMPORT n
STEP 10 - PRINT ntSeriesnt"
STEP 11 - IF i=1;i<=n;i++ GOTO STEP 12
STEP 12 - PRINT nTHTerm(i)+" "
STEP 13 - i++ & IF i<=n GOTO STEP 12
STEP 14 - PRINT ntSum : "+Sum(n)
STEP 15 – END

solution
class APSeries
{private double a,d;
APSeries() //default constructor
{a = d = 0;
}
APSeries(double a,double d) //parameterized constructor
{this.a = a;
this.d = d;
}
double nTHTerm(int n) //final AP term
{return (a+(n-1)*d);
}
double Sum(int n) //function calculating sum
{return (n*(a+nTHTerm(n))/2);
}
void showSeries(int n) //displaying AP Series
{System.out.print("ntSeriesnt");
for(int i=1;i<=n;i++)
{System.out.print(nTHTerm(i)+" ");
}
System.out.print("ntSum :"+Sum(n));
}
}
void main()throws IOException //main function
{BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter 1st term");
a=Integer.parseInt(br.readLine()); //accepting 1st
term
System.out.println("Enter Common difference");
d=Integer.parseInt(br.readLine()); //accepting common difference
System.out.println("Enter no.of terms");
int n=Integer.parseInt(br.readLine()); //accepting no. of terms
nTHTerm(n);
Sum(n);
showSeries(n);
}

varia
No. Name Type
1 a int
2 d int
3 n int
4 i int
outpu
able descr
Method D
- 1s
- co
Sum(), showSeries(), nTHTerm() to
showSeries() lo
ut
ription
escription
st
term
ommon difference
otal terms
oop variable

PROGRAM 4
To Display Calendar of Any
Month of Any Year
ALGORITHM
STEP 1 - START
STEP 2 - INPUT int month,int year
STEP 3 - int i,count=0,b,c,d=1 & String w="SMTWTFS"
STEP 4 - IF (year%100==0 && year%400==0) || (year%100!=0 && year%4==0)
STEP 5 - days[1]=29
STEP 6 - PRINT "================The Calendar of"+month1[month-1]+" "+year+"is==================")
STEP 8 - PRINT (i)+"t" & " "
STEP 9 - IF i=1 GOTO STEP 10
STEP 10 - IF (year%100==0 && year%400==0) || (year%100!=0 && year%4==0)THEN GOTO STEP 11OTHERWISE GOTO STEP 12
STEP 11 - count+=2
STEP 12 - count+=1
STEP 14 - count+=days[i] , count+=1, count%=7 & b=7-count
STEP 15 - IF b!=1 || b!=7 GOTO STEP 16
STEP 16 - IF count>0 GOTO STEP 17,18
STEP 17 - PRINT ' '+"t")
STEP 18 - count--
STEP 20 - IF b>0 && IF d<=days[month-1] GOTO STEP 21,22
STEP 21 - PRINT d+"t"
STEP 22 - d++ & b--
STEP 23 - b=7
STEP 24 - i++ & IF i<MONTH GOTO STEP14
STEP 25 - PRINT " "
STEP 26 – END

solutionimport java.io.*;
class Calendar
{public void dee()throws IOException //dee() function
{int i,count=0,b,d=1;
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.println(“Enter month”); //accepting month and year
int month=Integer.parseInt(br.readLine());
System.out.println(“Enter Year”);
int year=Integer.parseInt(br.readLine());
/* Computing and displaying calendar*/
String w="SMTWTFS";
int days[]={31,28,31,30,31,30,31,31,30,31,30,31};
String month1[]={"January","February","March","April","May","June","July","August","September","October","November","December"};
if((year%100==0 && year%400==0) || (year%100!=0 && year%4==0))
days[1]=29;
System.out.println("================The Calendar of"+month1[month-1]+" "+year+"is==================");
for(i=0;i<w.length();i++)
System.out.print(w.charAt(i)+"t");
System.out.println(" ");
for(i=1;i<year;i++)
if((year%100==0 && year%400==0) || (year%100!=0 && year%4==0))
count+=2;
else count+=1;
for(i=0;i<month;i++)
count+=days[i];
count+=1;
count%=7;
b=7-count;
if(b!=1 || b!=7)
while(count>0)
{System.out.print(' '+"t");
count--;
}
for(i=1;i<7;i++)
{while(b>0 && d<=days[month-1])
{System.out.print(d+"t");
d++;
b--;
}
b=7;
}}}

varia
No. Name Type
1 br Buffe
2 i int
3 count int
4 b int
5 d int
6 month int
7 year int
8 w String
9 days String
10 month1 String
outpu
able descr
Method Description
eredReader dee() BufferedReader object
dee() loop variable
dee() counter
dee() week counter
dee() day counter
dee() input month
dee() input year
g dee() week days
g[] dee() array storing days
g[] dee() array storing months
ut
ription

PROGRAM 5
To Calculate Factorial Using
Recursion
ALGORITHM
STEP 1 - START
STEP 2 - INPUT n
STEP 3 - IF(n<2) THEN return 1 OTHERWISE return (n * fact(n-1))
STEP 4 – END
solution
import java.io.*;
class Factorial
{public static void main(String args[]) throws IOException //main function
{BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("enter no =");
int n = Integer.parseInt(br.readLine()); //accepting no.
Factorial obj = new Factorial();
long f = obj.fact(n);
System.out.println("Factorial ="+f); //displaying factorial
}
public long fact(int n) //recursive fact()
{if(n<2)
return 1;
else return (n*fact(n-1));
}}

varia
No. Name Type
1 br Buffere
2 n int
3 obj Factoria
4 f long
5 n int
outpu
able descr
Method Description
main() input number
al main() Factorial object
main() variable storing factorial
fact() parameter in recursive fun
ut
ription
nction fact()

PROGRAM 6
To Display Fibonacci Series
Using Recursion
ALGORITHM
STEP 1 - START
STEP 2 - INPUT n
STEP 3 - IF(n<=1) THEN return 1 OTHERWISE return (fib(n-1) +fib(n-2))
STEP 4 – END
solution
import java.io.*;
class Fibonacci
{Fibonacci obj = new Fibonacci();
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("enter no of term ="); //accepting no. of terms
int n = Integer.parseInt(br.readLine());
System.out.println();
for(int i=1;i<=n;i++) //Fibonacci element display loop
{int f = obj.fib(i);
System.out.print(f+" ");
}}
public int fib(int n) //Recursive function fib() for calculation of Fibonacci element
{if(n<=1)
return n;
else
return (fib(n-1) +fib(n-2));
}}

varia
No. Name Type
1 br Buffere
2 obj Fibonac
3 n int
4 i int
5 f int
6 n int
outpu
able descr
Method Description
cci main() Fibonacci object
main() input number
main() loop variable for Fibonacci
main() Fibonacci element
fib() recursive function fib() par
ut
ription
element
rameter

PROGRAM 7
To Calculate GCD Using
Recursion
ALGORITHM
STEP 1 - START
STEP 2 - INPUT p,q
STEP 3 - IF(q=0) THEN return p OTHERWISE return calc(q,p%q)
STEP 4 – END
solution
import java.io.*;
class GCD
System.out.println("enter the numbers =");
int p = Integer.parseInt(br.readLine()); //accepting nos.
int q = Integer.parseInt(br.readLine());
GCD obj = new GCD();
int g = obj.calc(p,q);
System.out.println("GCD ="+g);
}
public int calc(int p,int q) //recursive function calculating GCD
{if(q==0)
return p;
else return calc(q,p%q);
}}

varia
No. Name Type
1 br Buffere
2 p int
3 q int
4 obj GCD
5 g int
outpu
able descr
Method Description
edReader main() BufferedReader obje
main() ,calc() input number
main() ,calc() input number
main() GCD object
main() variable storing the G
ut
ription
ect
GCD

PROGRAM 8
To Display Spiral Matrix.
ALGORITHM
STEP 1 - START
STEP 2 - INPUT a[][]
STEP 3 - IF p!=(int)Math.pow(l,2) GOTO STEP 4
STEP 4 - IF co!=0 GOTO STEP 5
STEP 5 - re=1
STEP 6 - IF ri=1;ri<=k1-re;ri++ GOTO STEP 7
STEP 7 - p++,c++
STEP 8 - IF c==l GOTO STEP 9
STEP 9 - BREAK
STEP 10 - a[r][c]=p
STEP 11 - IF c==l GOTO STEP 12
STEP 12 - BREAK
STEP 13 - IF dw=1 GOTO STEP 14
STEP 14 - p++,r++,a[r][c]=p
STEP 15 - IF le=1 GOTO STEP 16
STEP 16 - p++,c--,a[r][c]=p
STEP 17 - IF up=1 GOTO STEP 18
STEP 18 - p++,r--,a[r][c]=p
STEP 19 - k1=k1+2, k2=k2+2 & co++
STEP 20 - up++ & IF up<=k2-1 GOTO STEP 18
STEP 21 - le++ & IF le<=k2-1 GOTO STEP 16
STEP 22 - dw++ & IF dw<=k1-1 GOTO STEP 14
STEP 23 - IF y=0 GOTO STEP 24
STEP 24 - IF yy=0 GOTO STEP 25
STEP 25 - PRINT "t"+a[y][yy]) & ()
STEP 26 - yy++ & IF yy<l GOTO STEP 25
STEP 27 - y++ & IF y<l GOTO STEP 24
STEP 28 – END

solution
import java.io.*;
class SpiralMatrix
{public static void main(String[] args) throws IOException //main function
{int a[][],r,c,k1=2,k2=3,p=0,co=0,re=0;
System.out.println("enter the dimension of matrix A x A =");
int l = Integer.parseInt(br.readLine()); //accepting dimension of square spiral matrix
a=new int[l][l];
r=l/2;c=r-1;
if(l%2==0)
{System.out.println("wrong entry for spiral path");
System.exit(0);
}
/*Calculating and displaying spiral matrix*/
while(p!=(int)Math.pow(l,2))
{if(co!=0)
re=1;
for(int ri=1;ri<=k1-re;ri++)
{p++;c++;if(c==l)break;a[r][c]=p;}
if(c==l)break;
for(int dw=1;dw<=k1-1;dw++)
{p++;r++;a[r][c]=p;}
for(int le=1;le<=k2-1;le++)
{p++;c--;a[r][c]=p;}
for(int up=1;up<=k2-1;up++)
{p++;r--;a[r][c]=p;}
k1=k1+2;
k2=k2+2;
co++;
}
for(int y=0;y<l;y++) //Displaying matrix
{for(int yy=0;yy<l;yy++)
System.out.print("t"+a[y][yy]);
}}}

varia
No. Name Type
1 br Buffere
2 a int[][]
3 r int
4 c int
5 k1 int
6 k2 int
7 p int
8 co int
9 re int
10 l int
11 ri int
12 le int
13 dw int
14 up int
15 y int
16 yy int
outpu
able descr
Method Description
main() spiral matrix
main() l/2
main() r-1
main() stores 2
main() stores 3
main() loop gate
main() coloumn index
main() row index
main() dimensions of thr matrix
main() right side matrix loop varia
main() left side matrix loop variab
main() down side matrix loop vari
main() up side matrix loop variabl
main() loop variable to print matr
main() loop variable to print matr
ut
ription
able
ble
iable
le
rix
rix

PROGRAM 9
To Display Magic Square
ALGORITHM
STEP 1 - START
STEP 2 - arr[][]=new int[n][n],c=n/2-1,r=1,num
STEP 3 - IF num=1;num<=n*n;num++ GOTO STEP 4
STEP 4 - r--,c++
STEP 5 - IF r==-1 GOTO STEP 6
STEP 6 - r=n-1
STEP 7 - IF c>n-1 GOTO STEP 8
STEP 8 - c=0
STEP 9 - IF arr[r][c]!=0 GOTO STEP 10
STEP 10 - r=r+2 & c--
STEP 11 - num++ & IF num<=n*n GOTO STEP 4
STEP 12 - arr[r][c]=num
STEP 13 - IF r==0&&c==0 GOTO STEP 14
STEP 14 - r=n-1, c=1 & arr[r][c]=++num
STEP 15 - IF c==n-1&&r==0 GOTO STEP 16
STEP 16 - arr[++r][c]=++num
STEP 17 - PRINT ()
STEP 18 - IFr=0 GOTO STEP 19
STEP 19 - IF c=0 GOT STEP 20
STEP 20 - PRINT arr[r][c]+" " & ()
STEP 21 - c++ & IF c<n GOTO STEP 20
STEP 21 - r++ & r<n GOTO STEP 19
STEP 22 – END

solution/*A Magic Square is a square whose sum of diagonal elements, row elements and coloumn
elements is the same*/
import java.io.*;
class MagicSquare
{public static void main(String args[])throws Exception //main function
System.out.println("enter the dimension of magical square=");
int n = Integer.parseInt(br.readLine()); //accepting dimensions
int arr[][]=new int[n][n],c=n/2-1,r=1,num;
for(num=1;num<=n*n;num++) //loop for finding magic square elements
{r--;
c++;
if(r==-1)
r=n-1;
if(c>n-1)
c=0;
if(arr[r][c]!=0)
{r=r+2;
c--;
}
arr[r][c]=num;
if(r==0&&c==0)
{r=n-1;
c=1;
arr[r][c]=++num;
}
if(c==n-1&&r==0)
arr[++r][c]=++num;
}
for(r=0;r<n;r++) //loop displaying magic square
{for(c=0;c<n;c++)
System.out.print(arr[r][c]+" ");
}}}

varia
No. Name Type
1 br Buffere
2 n int
3 arr int[][]
4 num int
5 r int
6 c int
outpu
able descr
Method Description
main() input dimensions
main() magic square matrix
main() loop variable for magic squ
main() row
main() coloumn
ut
ription
uare

PROGRAM 10
To Search an Array Using
Linear Search
ALGORITHM
STEP 1 - START
STEP 2 - INPUT a[]
STEP 3 - FROM i=0 to i<n REPEAT STEP 4
STEP 4 - PRINT a[i]+" "
STEP 5 - flag=-1
STEP 7 - IF (a[i] == v) THEN flag =i
STEP 8 - IF (flag=-1) THEN GOTO STEP 9 OTHERWISE GOTO STEP 10
STEP 9 - PRINT “ not found”
STEP 10 - PRINT v+" found at position - "+flag
STEP 11 – END

solution
import java.io.*;
class LinearSearch
{int n,i;
int a[] = new int[100];
static BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
public LinearSearch(int nn)
{n=nn;
}
public void input() throws IOException //function for obtaining values from user
{System.out.println("enter elements");
for(i=0;i<n;i++)
{a[i] = Integer.parseInt(br.readLine());
}}
public void display() //function displaying array values
{System.out.println();
for(i=0;i<n;i++)
{System.out.print(a[i]+" ");
}}
public void search(int v) //linear search function
{int flag=-1;
for(int i=0; i<n ; i++)
{if(a[i] == v)
flag =i;
}
if(flag== -1 )
System.out.println("not found");
else System.out.println(v+" found at position - "+flag);
}
public static void main(String args[]) throws IOException //main function
{LinearSearch obj = new LinearSearch(10);
obj.input();
obj.display();
System.out.println("enter no. to be searched -"); //accepting the values to be searched
int v = Integer.parseInt(br.readLine());
obj.search(v);
}}

varia
No. Name Type
1 br Buffere
2 n int
3 i int
4 a[] int[]
5 nn int
6 v int
7 flag int
8 obj LinearS
outpu
able descr
Method Description
edReader - BufferedReader o
- array length
- loop variable
- input array
LinearSearch() parameter in cons
search(), main() search element
search() flag
Search main() LinearSearch obje
ut
ription
object
structor
ect

PROGRAM 11
To Search an Array Using
Binary Search
ALGORITHM
STEP 1 - START
STEP 2 - INPUT a[]
STEP 5 - flag=-1 , l=0, u=n-1
STEP 6 - IF(l<=u && flag=-1) REPEAT STEP 7 AND Step 8
STEP 7 - m = (l+u)/2
STEP 8 - IF (a[m] == v) THEN flag =m OTHERWISE GOTO STEP 9
STEP 9 - IF (a[m] < v) THEN l = m+1 OTHERWISE u =m-1
STEP 10 - IF (flag=-1) THEN GOTO STEP 11 OTHERWISE GOTO STEP 12
STEP 11 - PRINT “ not found”
STEP 12 - PRINT v+" found at position - "+flag
STEP 13 - END

solution
import java.io.*;
class BinarySearch
{int n,i;
public BinarySearch(int nn) //default constructor
{n=nn;
}
public void input() throws IOException //function accepting array elements
{System.out.println("enter elements");
for(i=0;i<n;i++)
}}
public void display() //displaying array elements
for(i=0;i<n;i++)
}}
public void search(int v) //function to search array elements using binary search
technique
{int l=0;
int u = n-1;
int m;
int flag=-1;
while( l<=u && flag == -1)
{m = (l+u)/2;
if(a[m] == v)
flag = m;
else if(a[m] < v)
l = m+1;
else u = m-1;
}
if(flag== -1 )
System.out.println("not found");
else System.out.println(v+" found at position - "+flag);
}
{BinarySearch obj = new BinarySearch(10);
obj.input();
obj.display();
System.out.println("enter no. to be searched -");
int v = Integer.parseInt(br.readLine()); //accepting integer to be searched by binary search
obj.search(v);

}}
varia
No. Name Type
1 br Buffere
2 n int
3 i int
4 a[] int[]
5 nn int
6 v int
7 flag int
8 l int
9 u int
10 m int
11 obj BinaryS
outpu
able descr
Method Description
edReader - BufferedReader o
- array length
- loop variable
- input array
BinarySearch() parameter in con
search(), main() search element
search() flag
search() lower limit
search() upper limit
search() middle index
Search main() BinarySearch obj
ut
ription
object
nstructor
ect

PROGRAM 12
To Sort an Srray Using
Selection Sort
ALGORITHM
STEP 1 - START
STEP 2 - INPUT a[]
STEP 5 - flag=-1
STEP 6 - FROM i=0 to i<n-1 REPEAT STEP 7 to STEP 11
STEP 7 - min =i
STEP 8 - FROM j=i+1 to j<n REPEAT STEP 8
STEP 9 - IF(a[j]<a[min]) then min =j
STEP 10 - IF (min!=i) GOTO STEP 11
STEP 11 - temp = a[i], a[i] =a[min], a[min] = temp

solution
import java.io.*;
class SelectionSort
{int n,i;
public SelectionSort(int nn) //parameterized constructor
{n=nn;
}
{BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
System.out.println("enter elements");
for(i=0;i<n;i++)
}}
public void display() //function displaying array elements
for(i=0;i<n;i++)
}}
public void sort() //function sorting array elements using selection sort technique
{int j,temp,min;
for(i=0;i<n-1;i++)
{min =i;
for(j=i+1;j<n;j++)
{if(a[j]<a[min])
min =j;
}
if(min!=i)
{temp = a[i];
a[i] =a[min];
a[min] = temp;
}}}
{SelectionSort x = new SelectionSort(5);
x.input();
System.out.print("Before sorting - ");
x.display();
System.out.print("After sorting - ");
x.sort();
x.display();
}}

varia
No. Name Type
1 br Buffere
2 n int
3 i int
4 a[] int[]
5 nn int
6 j int
7 temp int
8 min int
9 x Selectio
outpu
able descr
Method Description
edReader input() BufferedReader obje
- array length
- loop variable
- input array
SelectionSort() parameter in constr
sort() sort index
sort() temporary storage
sort() minimum value
onSort main() SelectioSort object
ut
ription
ect
ructor

PROGRAM 13
To Sort an Array Using
Bubble Sort
ALGORITHM
STEP 1 - START
STEP 2 - INPUT a[]
STEP 5 - flag=-1
STEP 6 - FROM i=0 to i<n-1 REPEAT STEP 7 to STEP 9
STEP 7 - FROM j=i+1 to j<n REPEAT STEP 8
STEP 8 - IF(a[j] > a[j+1]) THEN GOTO STEP 9
STEP 9 - temp = a[i], a[i] =a[min], a[min] = temp
STEP 10 - END

solution
import java.io.*;
class BubbleSort
{int n,i;
public BubbleSort(int nn) //parameterized constructor
{n=nn;
}
System.out.println("enter elements");
for(i=0;i<n;i++)
}}
public void display() //function displaying array elements
for(i=0;i<n;i++)
}}
public void sort() //function sorting array elements using Bubble Sort technique
{int j,temp;
for(i=0 ; i<n-1 ; i++)
{for(j=0 ; j<n-1-i ; j++)
{if(a[j] > a[j+1])
{temp = a[j];
a[j] =a[j+1];
a[j+1] = temp;
}}}}
{BubbleSort x = new BubbleSort(5);
x.input();
System.out.print("Before sorting - ");
x.display();
System.out.print("After sorting - ");
x.sort();
x.display();}}

varia
No. Name Type
1 br Buffere
2 n int
3 i int
4 a[] int[]
5 nn int
6 j int
7 temp int
8 x Selectio
outpu
able descr
Method Description
edReader input BufferedReader objec
- array length
- loop variable
- input array
BubbleSort() parameter in construc
sort() sort index
sort() temporary storage
onSort main() SelectionSort object
ut
ription
t
ctor

PROGRAM 14
To Convert a Decimal no. Into
its Binary Equivalent
ALGORITHM
STEP 1 - START
STEP 2 - n = 30
STEP 3 - INPUT int no
STEP 4 - c =0 , temp = no
STEP 5 - IF (temp!=0) REPEAT STEP 6
STEP 6 - a[c++] = temp%2, temp = temp / 2
STEP 7 - FROM i=c-1 to i>0 REPEAT STEP 8
STEP 8 - PRINT a[i]
STEP 9 – END
solution
import java.io.*;
class Dec2Bin
{int n,i;
public Dec2Bin(int nn) //parameterized contructor
{n=nn;
}
public void dectobin(int no) //function converting decimalto binary number
{int c = 0;
int temp = no;
while(temp != 0)
{a[c++] = temp % 2;
temp = temp / 2;
}
System.out.println("Binary eq. of "+no+" = ");
for( i = c-1 ; i>=0 ; i--) //Displaying binary number
System.out.print( a[ i ] );

}
public static void mai
{Dec2Bin obj = new D
System.out.println("e
int no = Integer.parse
obj.dectobin(no);
}}
varia
No. Name Type
1 br Buffere
2 n int
3 i int
4 a[] int[]
5 nn int
6 no int
7 temp int
8 c int
9 obj Dec2Bi
outpu
in(String args[]) throws IOException //m
Dec2Bin(30);
enter decimal no -");
eInt(br.readLine());
able descr
Method Description
edReader BufferedReader
- array length
- loop variable
- array storing bin
Dec2Bin() parameter in co
main(), dectobin() input number
dectobin() temporary stora
dectobin() counter
n main() Dec2Bin object
ut
main function
ription
r object
nary no.
onstructor
age

PROGRAM 15
To Display Date From
Entered Day Number
ALGORITHM
STEP 1 - START
STEP 2 - INITIALISE a[ ] , m[ ]
STEP 3 - INPUT n , yr
STEP 4 - IF ( yr%4=0) THEN a[1] = 29
STEP 5 - t =0 , s = 0
STEP 6 - IF ( t<n) REPEAT STEP 7
STEP 7 - t =t + a[s++]
STEP 8 - d = n + a[--s] - t
STEP 9 - IF ( d ==1|| d == 21 || d == 31 ) then PRINT d + "st" + m[s] + "
, "+yr
STEP 10 - IF ( d ==2|| d == 22 ) then PRINT d + "nd" + m[s] + " , "+yr
STEP 11 - IF ( d ==3|| d == 23 ) then PRINT d + "rd" + m[s] + " , "+yr
OTHERWISE GOTO STEP 12
STEP 12 - PRINT d + "th" + m[s] + " , "+yr
STEP 13 – END

solution
import java.io.*;
class Day2Date
{static BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
public void calc(int n, int yr) //function to calculate date
{int a[ ] = { 31,28,31,30,31,30,31,31,30,31,30,31 } ;
String m[ ] = { "Jan", "Feb", "Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec" } ;
if ( yr % 4 == 0)
a[1] =29;
int t=0,s=0;
while( t < n) //loop calculating date
{t =t + a[s++];
}
int d = n + a[--s] - t;
if( d == 1|| d == 21 || d == 31 )
{System.out.println( d + "st" + m[s] + " , "+yr);
}
if( d == 2 || d == 22 )
{System.out.println( d + "nd" + m[s] + " , "+yr);
}
if( d == 3|| d == 23 )
{System.out.println( d + "rd" + m[s] + " , "+yr);
}
else {System.out.println( d + "th" + m[s] + " , "+yr);
}}
{Day2Date obj = new Day2Date();
System.out.println( "Enter day no = "); //accepting day no.
int n = Integer.parseInt(br.readLine());
System.out.println( "Enter year = "); //accepting year
int yr = Integer.parseInt(br.readLine());
obj.calc(n,yr);
}}

varia
No. Name Type
1 br Buffere
2 n int
3 yr int
4 a int[]
5 m int[]
6 t int
7 s int
8 d int
9 obj Day2Da
outpu
able descr
Method Description
edReader - BufferedReader obje
calc(), main() Day number
calc(), main() year
calc() array storing day
calc() array storing month
calc() array index
calc() array index
calc() n+a[--s]+t
ate main() Day2Date object
ut
ription
ect

PROGRAM 16
To Create a Star Pattern
From Entered String
solution
import java.io.*;
class Pattern
{public static void main (String args[]) throws IOException
{int i,sp,j,k,l;
System.out.println("enter the string ="); //accepting string
String s = br.readLine();
l=s.length();
/*printing the pattern*/
for(i=0;i<l;i++)
if(i==l/2)
System.out.println(s);
else {sp=Math.abs((l/2)-i);
for(j=sp;j<l/2;j++)
System.out.print(" ");
k=0;
while(k<3)
{System.out.print(s.charAt(i));
for(j=0;j<sp-1;j++)
System.out.print(" ");
k++;
}
}}}

varia
No. Name Type
1 br Buffere
2 s String
3 i int
4 sp int
5 j int
6 k int
7 l int
outpu
able descr
Method Description
main() input string
main() loop variable for printing t
main() Math.abs((l/2)-i)
main() length of string
ut
ription
he pattern
he pattern
he pattern

PROGRAM 17
To Check if Entered String is
Palindrome or Not
ALGORITHM
STEP 1 - START
STEP 2 - INPUT string s
STEP 3 - StringBuffer sb = s
STEP 4 - sb.reverse
STEP 5 - String rev = sb
STEP 6 - IF rev = s GOTO STEP 7 OTHERWISE GOTO STEP 8
STEP 7 - PRINT " Palindrome"
STEP 8 - PRINT " Not Palindrome"
STEP 9 – END
solution
import java.io.*;
class Palindrome
System.out.println("enter the string=");
String s = br.readLine(); //accepting the string
StringBuffer sb = new StringBuffer(s);
sb.reverse(); //reversing the string
String rev = new String(sb);
if(s.equalsIgnoreCase(rev)) //checking for palindrome
System.out.println("Palindrome " ); //displaying the result
else System.out.println("Not Palindrome " );
}}

varia
No. Name Type
1 br Buffere
2 s String
3 sb StringB
4 rev String
outpu
able descr
Method Description
main() input string
Buffer main() StringBuffer object of s
main() revese string
ut
ription

PROGRAM 18
To Display a Frequency of
Each Character in Entered
String
ALGORITHM
STEP 1 - START
STEP 2 - INPUT str
STEP 3 - l=str.length()
STEP 4 - PRINT str
STEP 5 - IF i=0 THEN GOTO STEP 4 OTHERWISE GOTO STEP 22
STEP 6 - char a=str.charAt(i)
STEP 7 - IF ii=0 THEN GOTO STEP 4 OTHERWISE GOTO STEP 22
STEP 8 - char b = str.charAt(ii)
STEP 9 - IF a==b GOTO STEP 10
STEP 10 - freq=freq+1
STEP 11 - ii++ & IF ii<1 GOTO STEP 8
STEP 12 - i++ & IF i<1 GOTO STEP 6
STEP 13 - DISPLAY a+" occurs "+freq+" times"
STEP 14 – END

solution
import java.io.*;
class Frequency
{private int i,a1,l,p,j,freq;
public Frequency() //default constructor
{p=0;
freq=0; // initialise instance variables
}
public void count(String str) //counting character frquency
{int ii;
l=str.length();
System.out.print(str);
for(i=0;i<l;i++)
{char a=str.charAt(i);
for(ii=0;ii<l;ii++)
{char b = str.charAt(ii);
if (a==b)
freq=freq+1;
}
System.out.println(a+" occurs "+freq+" times"); //displaying frequency
freq=0;
}}
System.out.println("enter string");
String str = br.readLine();
Frequency x = new Frequency();
x.count(str);
}}

varia
No. Name Type
1 br Buffere
2 i int
3 a1 int
4 l int
5 p int
6 freq int
7 ii int
8 a char
9 b char
10 str String
11 x Freque
outpu
able descr
Method Description
- loop variable
- instance variable
- length of string
- instance variable
- frequency of characters
count() loop variable
count() character at index i
count() character at index ii
main() input string
ncy main() Frequency object
ut
ription

PROGRAM 19
To Find a Word in Entered
String
ALGORITHM
STEP 1 - START
STEP 2 - INPUT string s
STEP 3 - StringTokenizer st = s
STEP 4 - l =str.length()
STEP 5 - INPUT look
STEP 6 - flag = -1
STEP 7 - IF (st.hasMoreElements()) REPEAT STEP 8
STEP 8 - IF (look.equals(st.nextElement())) THEN flag =1
STEP 9 - IF flag = - 1 GOTO STEP 10 OTHERWISE STEP 11
STEP 10 - PRINT "word not found"
STEP 11 - PRINT "word found"
STEP 12 – END
solutionimport java.util.StringTokenizer;
import java.io.*;
public class WordSearch
{public static void main(String[] args) throws IOException //main function
System.out.println("enter the string=");
String s = br.readLine(); //accepting string
StringTokenizer st = new StringTokenizer(s," "); //StringTokenizer initialization
System.out.println("enter the word to be searched =");
String look = br.readLine();
int flag = -1;
while(st.hasMoreElements()) //searching for word
{if(look.equals(st.nextElement()))
flag =1;

}
if(flag ==-1)
{System.out.println("
}
else {
System.out.println("t
}}}
varia
No. Name Type
1 br Buffere
2 s String
3 st StringT
4 look String
5 flag int
outpu
"the word not found"); //display
the word found");
able descr
Method Description
main() input string
okenizer main() StringTokenizer object
main() word to be searched
main() flag
ut
ying the result
ription

PROGRAM 20
To Decode the Entered String
ALGORITHM
STEP 1 - START
STEP 2 - INPUT name, n
STEP 3 - l=name.length()
STEP 4 - PRINT original string is "+name
STEP 6 - char c1=name.charAt(i)
STEP 7 - c=(int)c1
STEP 8 - IF n>0 THEN GOTO STEP 9 THERWISE GOTO STEP 12
STEP 9 - IF (c+n)<=90 THEN GOTO STEP 10 OTHERWISE GOTO STEP 11
STEP 10 - PRINT (char)(c+n)
STEP 11 - c=c+n;c=c%10,c=65+(c-1) & PRINT (char)(c)
STEP 12 - ELSE IF n<0 THEN GOTO STEP 13 OTHERWISE GOTO STEP 19
STEP 13 - n1=Math.abs(n)
STEP 14 - IF (c-n1) >=65 THEN GOTO STEP 15 OTHERWISE GOTO STEP 16
STEP 15 - DISPLAY (char) (c-n1)
STEP 16 - IF c>65 THEN GOTO STEP 17 OTHERWISE GOTO STEP 18
STEP 17 - c=c-65,
STEP 18 - c=n1 & PRINT (char)(90-(c-1))
STEP 19 - ELSE IF n==0
STEP 20 - DISPLAY "no change "+name
STEP 21 - END

solution
import java.io.*;
class Decode
{public void compute()throws IOException //compute() function
System.out.println(“Enter name:”);
String name=br.readLine();
System.out.println(“Enter number:”);
int n=Integer.parseInt(br.readLine());
int j,i,l,c=0,y,n1;
l=name.length();
System.out.println("original string is "+name);
for(i=0;i<l;i++)
{char c1=name.charAt(i);
try //trying for NumberFormatException
{c=(int)c1 ;
}
catch(NumberFormatException e)
{}
if(n>0)
{if((c+n)<=90)
/*Decoding String*/
System.out.print((char)(c+n));
else {c=c+n;
c=c%10;
c=65+(c-1);
System.out.print((char)(c));
}}
else if(n<0)
{n1=Math.abs(n);
if((c-n1) >=65)
System.out.print((char) (c-n1));
else {if(c>65)
c=c-65;
else c=n1;
System.out.print((char)(90-(c-1)));
}}
else if (n==0)
{System.out.println("no change "+name);
break;
}}}}

varia
No. Name Type
1 br Buffere
2 name String
3 n int
4 j int
5 i int
6 l int
7 c int
8 y int
9 n1 int
10 c1 char
11 e Numbe
outpu
able descr
Method Description
edReader compute() BufferedReade
compute() input string
compute() decode numbe
compute() loop variable
compute() loop variable
compute() length of string
compute() ASCII of c1
compute()
compute()
compute() character at ind
erFormatException compute() NumberFOrma
ut
ription
er object
er
g
dex i
atException object

PROGRAM 21
To Display the Entered String
in Alphabetical Order
ALGORITHM
STEP 1 - START
STEP 2 - str = "" , l = 0
STEP 3 - INPUT string str
STEP 4 - l =str.length()
STEP 5 - FROM i=0 to i<l REPEAT STEP 6
STEP 6 - c[i] = str.charAt(i)
STEP 7 - FROM i=0 to i<l-1 REPEAT STEP 8
STEP 8 - FROM j=0 to i<l-1 REPEAT STEP 9
STEP 9 - temp =c[j], c[j] = c[j+1] , c[j+1] = temp
STEP 10 - FROM i=0 to i<l REPEAT STEP 11
STEP 11 - PRINT c[i]
STEP 12 – END

solution
import java.io.*;
class Alpha
{String str;
int l;
char c[] = new char[100];
public Alpha() //Alpha() constructor
{str = "";
l =0;
}
public void readword() throws IOException //function to read input string
{System.out.println("enter word - ");
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
str = br.readLine();
l = str.length();
}
public void arrange() //function to arrange string in ascending order
{int i,j;
char temp;
for(i=0;i<l;i++)
{c[i]= str.charAt(i);
}
for(i=0;i<l-1;i++) //loops for swapping of characters
{for(j=0;j<l-1-i;j++)
{if(c[j] > c[j+1])
{temp = c[j];
c[j] = c[j+1];
c[j+1] = temp;
}}}}
public void display() //function to display the rearranged string
for(int i=0;i<l;i++)
{System.out.print(c[i]);
}}
{Alpha obj = new Alpha();
obj.readword();
obj.arrange();
obj.display();
}}

varia
No. Name Type
1 br Buffere
2 str String
3 l int
4 c char[]
5 i int
6 j int
7 temp char
8 obj Alpha
outpu
able descr
Method Description
dReader readword() BufferedReader object
- input string
- length
- character array
readword() loop variable
readword() loop variable
readword() temporary storage
main() Alpha object
ut
ription

PROGRAM 22
To Create a String and Count
Number of Vowels and
Consonants
ALGORITHM
STEP 1 - START
STEP 2 - a = "Computer Applications"
STEP 3 - z = a.length()
STEP 4 - x= 0 , b= 0
STEP 5 - FROM y =0 to y<z REPEAT STEP 6
STEP 6 - IF (a.charAt(y)=='a'||a.charAt(y)=='e'||a.charAt(y)=='i'||a.charAt(y)=='o'||a.charAt(y)=='u') THEN x =x +1 OTHERWISE b = b+1
STEP 7 - PRINT x
STEP 8 - PRINT b
STEP 9 – END
solutionimport java.io.*;
class Vowels
System.out.println("Enter a string");
String a= br.readLine(); //Accepting string
int z=a.length(),y,x=0,b=0;
for(y=0;y<z;y++) //loop for counting number of vowels
{if(a.charAt(y)=='a'||a.charAt(y)=='e'||a.charAt(y)=='i'||a.charAt(y)=='o'||a.charAt(y)=='u')
x++;
else b++;
}
System.out.println("Number of vowels in string ="+x); //displaying result
System.out.println("Number of consonants in string ="+b);
}}

varia
No. Name Type
1 br Buffere
2 a String
3 z int
4 y int
5 b int
6 x int
outpu
able descr
Method Description
main() input string
main() loop variable
main() no. of consonants
main() no. of vowels
ut
ription

PROGRAM 23
To Create a String and Count
Number of Words
ALGORITHM
STEP 1 - START
STEP 3 - z = a.length()
STEP 4 - x= 0
STEP 5 - FROM y =0 to y<z REPEAT STEP 6
STEP 6 - IF (a.charAt(y)==' ' ) then x =x+1
STEP 7 - PRINT "Number of words in string ="+(x+1)
STEP 8 – END
solution
import java.io.*;
class NoOfWords
{public static void main(String args[])throws IOException
System.out.println("Enter Sentence");
String a=br.readLine(); //accepting string
System.out.println("The string is -"+a);
int z=a.length(),y,x=0;
for(y=0;y<z;y++) //loop for counting number of spaces
{if(a.charAt(y)==' ')
x=x+1;
}System.out.println("Number of words in string ="+(x+1)); //displaying result
}}

varia
No. Name Type
1 br Buffere
2 z int
3 a String
4 x int
5 y int
outpu
able descr
Method Description
main() input string
main() space counter
ut
ription

PROGRAM 24
To create a string and replace
all vowels with *
ALGORITHM
STEP 1 - START
STEP 3 - x= 0
STEP 4 - FROM z =0 to z<a.length() REPEAT STEP 5
STEP 5 - if(a.charAt(z)=='a'||a.charAt(z)=='e'||a.charAt(z)=='i'||a.charAt(z)=='o'||a.charAt(z)=='u’) THEN a.setCharAt(z,'*')
STEP 6 - PRINT "New String -"+a
STEP 7 – END
solution
import java.io.*;
class VowelReplace
System.out.println(“Enter a String”);
StringBuffer a=new StringBuffer(br.readLine()); //accepting a string
System.out.println("Original String -"+a);
int z=0;
for(z=0;z<a.length();z++) //loop for replacing vowels with "*"
{if(a.charAt(z)=='a'||a.charAt(z)=='e'||a.charAt(z)=='i'||a.charAt(z)=='o'||a.charAt(z)=='u')
a.setCharAt(z,'*');
}System.out.println("New String -"+a); //displaying the result
}}

varia
No. Name Type
1 br Buffere
2 a StringB
3 z int
outpu
able descr
Method Description
Buffer main() StringBuffer object of inpu
ut
ription
ut string

PROGRAM 25
To Generate Sum of All
Elements of a Double
Dimensional Array of 5*5
Subscripts
ALGORITHM
STEP 1 - START
STEP 2 - INPUT a[]
STEP 3 - FROM x =0 to x<5 REPEAT STEP 4
STEP 4 - FROM y =0 to y<5 REPEAT STEP 5
STEP 5 - PRINT (a[x][y]+" "
STEP 8 - Sum=Sum+a[x][y]
STEP 9 - PRINT Sum
STEP10 – END
solution
import java.io.*;
class MatrixSum
{ int a[][]=new int[5][5];
BufferedReader aa=new BufferedReader(new InputStreamReader(System.in));
int x,y,z,Sum=0;
System.out.println("Enter the array");
for(x=0;x<5;x++) //loop for reading array
{for(y=0;y<5;y++)
{ z=Integer.parseInt(aa.readLine()); //accepting array element

a[x][y]=z;
}}
System.out.println("A
for(x=0;x<5;x++)
{for(y=0;y<5;y++)
{System.out.print(a[x
}
System.out.print("n
}
for(x=0;x<5;x++)
{for(y=0;y<5;y++)
{Sum=Sum+a[x][y];
}}
System.out.println("S
}}
varia
No. Name Type
1 aa Buffere
2 a int[][]
3 x int
4 y int
5 z int
6 Sum main()
outpu
Array -");
//loop for printing array
x][y]+" ");
");
//loop for printing sum of array ele
Sum of Array elements="+Sum); /
able descr
Method Description
main() input array
main() input element
main() Sum of all elements
ut
ements
//displaying sum
ription

PROGRAM 26
To Find Sum of Each Column
of a Double Dimensional
Array
ALGORITHM
STEP 1 - START
STEP 2 - INPUT a[]
STEP 5 - PRINT (a[x][y]+" "
STEP 6 - FROM x =0 to x<4 REPEAT STEP 7 , STEP 9 and STEP 10
STEP 8 - Sum=Sum+a[x][y] ,
STEP 9 - PRINT Sum
STEP 10 - Sum = 0
STEP11 – END
solution
import java.io.*;
class ColoumnSum
{int a[][]=new int[4][4];
int x,y,z,Sum=0;
System.out.println("Enter the array"); //reading array
for(x=0;x<4;x++)
{for(y=0;y<4;y++)
{z=Integer.parseInt(aa.readLine());
a[x][y]=z;
}}

System.out.println("A
for(x=0;x<4;x++)
{for(y=0;y<4;y++)
}System.out.print("n
}
for(y=0;y<4;y++)
{for(x=0;x<4;x++)
{Sum=Sum+a[x][y];
}
Sum=0;
}}}
varia
No. Name Type
1 aa Buffere
2 a int[][]
3 x int
4 y int
5 z int
6 Sum int
outpu
Array -"); //printing the arra
x][y]+" ");
n");
Sum of column "+(y+1)+" is "+Sum);
able descr
Method Description
main() input array
main() Sum of each couloumn
ut
ay in matrix form
//printing sum of coloumn
ription

PROGRAM 27
To Find Sum of Diagonal of
a Double Dimensional Array
of 4*4 Subscripts
ALGORITHM
STEP 1- START
STEP 2- INPUT a[]
STEP 3- FROM x =0 to x<4 REPEAT STEP 4
STEP 4- FROM y =0 to y<4 REPEAT STEP 5
STEP 5- PRINT (a[x][y]+" "
STEP 6- FROM x =0 to x<4 REPEAT STEP 7
STEP 7 - Sum=Sum+a[x][y] , y=y+1
STEP 9- PRINT Sum
STEP 10 - Sum = 0
STEP11- END
solution
import java.io.*;
class DiagonalSum
{int a[][]=new int[4][4];
int x,y,z,Sum=0;
System.out.println("Enter the array");
for(x=0;x<4;x++) //Reading array
{for(y=0;y<4;y++)
{z=Integer.parseInt(aa.readLine());
a[x][y]=z;
}}
System.out.println("Array -");

for(x=0;x<4;x++)
{for(y=0;y<4;y++)
}
System.out.print("n
}
y=0;
for(x=0;x<4;x++)
{Sum=Sum+a[x][y];
y=y+1;
}
Sum=0;
}}
varia
No. Name Type
1 aa Buffere
2 a int[][]
3 x int
4 y int
5 Sum int
6 z int
outpu
//displaying array
x][y]+" ");
");
//loop for finding sum of diagonal
Sum of diagonal is "+Sum); //displayin
able descr
Method Description
main() input matrix
main() Sum of diagonals
ut
g the sum of diagonal
ription

PROGRAM 28
To Calculate the Commission
of a Salesman as per the
Following Data
Sales Commission
>=100000 25% of sales
80000-99999 22.5% of sales
60000-79999 20% of sales
40000-59999 15% of sales
<40000 12.5% of sales
ALGORITHM
STEP 1 - START
STEP 2 - INPUT sales
STEP 3 - IF (sales>=100000) THEN comm=0.25 *sales OTHERWISE GOTO STEP 4
STEP 4 - IF (sales>=80000) THEN comm=0.225*sales OTHERWISE GOTO STEP 5
STEP 7 - comm=0.125*sales
STEP 8 - PRINT "Commission of the employee="+comm
STEP 9 – END
solution
import java.io.*;
class SalesComission
{double sales,comm;
System.out.println(“Enter sales”);
sales=Double.parseDouble(aa.readLine()); //reading sales from the keyboard

/*calculating commis
if(sales>=100000)
comm=0.25*sales;
else if(sales>=80000)
comm=0.225*sales;
comm=0.2*sales;
comm=0.15*sales;
else comm=0.125*sa
System.out.println("C
}}
varia
No. Name Type
1 aa Buffer
2 sales double
3 comm. double
outpu
ssion*/
les;
Commission of the employee="+comm); //d
able descr
Method Description
e main() sales
e main() commision
ut
displaying commission
ription

PROGRAM 29
To Convert a Decimal
Number to a Roman Numeral
ALGORITHM
STEP 1 – START
STEP 2 – Enter number num
STEP 3 -- hund[]={"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"}
STEP 4 -- ten[]={"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};
STEP 5 -- unit[]={"","I","II","III","IV","V","VI","VII","VIII","IX"};
STEP 6 – Display hund[num/100] and ten[(num/10)%10] and unit[num%10]
STEP 7 – END
solution
import java.io.*;
public class Dec2Roman
{public static void main() throws IOException //main function
{DataInputStream in=new DataInputStream(System.in);
System.out.print("Enter Number : ");
int num=Integer.parseInt(in.readLine()); //accepting decimal number
String hund[]={"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};
String ten[]={"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};
String unit[]={"","I","II","III","IV","V","VI","VII","VIII","IX"};
/*Displaying equivalent roman number*/
System.out.println("Roman Equivalent= "+hund[num/100]+ten[(num/10)%10]+unit[num%10]);
}}

varia
No. Name Type
1 in DataInp
2 num int
3 hund String[]
4 ten String[]
5 unit String[]
outpu
able descr
Method Description
putStream main() DataInputStream object
main() input number
] main() array storing 100th
positio
] main() array storing 10th
position
] main() array storing units positio
ut
ription
on
n
on

PROGRAM 30
To Convert Celsius into
Fahrenheit Using Inheritence
ALGORITHM
STEP 1 – START
STEP 2 -- Input temperature ‘celcius’ in celcius
STEP 3 – far=1.8*celcius + 32
STEP 4 – Display far
STEP 5 -- END
solution
import java.io.*;
class C2F
{ public static void main(String args[])throws IOException //main function
{Temperature ob= new Temperature();
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter temperature in Celsius"); //accepting temperature
double temp=ob.convert(Double.parseDouble(br.readLine()));
System.out.println("The temperature in fahrenheit is = "+temp);
}}
class Temperature extends C2F
{double convert(double celcius) //function to convert Celsius to fahrenheit
{double far=1.8*celcius+32.0;
return far;
}}

varia
No. Name Type
1 br Buffere
2 ob C2F
3 temp double
4 celcius double
5 far double
outpu
able descr
Method Description
main() C2F object
e main() calculated Fahrenheit tem
e convert() input temperature in Cel
e convert() Calculated Fahrenheit tem
ut
ription
mperature
sius
mperature

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