javapravticalfile.doc

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Algorithm-
Step1 Start.
Step2 Declare suitable variables like n,i,jand str variable of string type.
Step3 Start a loop from i=65 to i<=90.
Step4 Start a loop from j=0 to j<=length of string.
Step5 Check if int(ch)==i and if true then increment p by 1 and then print
the value of p.
Step6 Repeat step3 and step4 with i=97 and i<= 122 and then repeat
step5.

49 | 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
PREM PRAKASH SAINI ISC COMPUTER SCIENCE PROJECT
2
Step7 End.
Variable Discription :
* TYPE NAME PURPOSE
* int n Store length of str1
* String str1 to add character.
* Int i looping variable
* Int j looping variable
* Int p counter variable
* char ch to store present character
* char ch2 to compare with ch2
PROGRAM-
import java.io.*;
class frequencyofeachletter // defining class
{
public static void main(String args[])throws Exception //defining main function
{
BufferedReader num=new BufferedReader(new InputStreamReader(System.in));
String str,str1=new String();//declaring string
str1=" ";
int n,i,j,p=0;
System.out.println("Enter your string");
str=num.readLine(); //accepting data from the user
n=str.length();
for(i=65;i<=90;i++) //generating loop
{
for(j=0;j<n;j++) //generating loop
{
char ch=str.charAt(j); // extracting a character
if((int)ch==i)
{
p++;
}
}
char ch2=(char)i;
System.out.print("frequency of "+ch2+"= "+" "+" "); //printing
System.out.print(p+" ");
System.out.println();
p=0;
}
for(i=97;i<=122;i++)// generating loop
{
for(j=0;j<n;j++)
{
char ch=str.charAt(j);

c t
3
if((int)ch==i)
{
p++;
}
}
char ch2=(char)i;
System.out.print("frequency of "+ch2+"= "+" "+" ");//printing
System.out.print(p+" ");
p=0;
}
}
}
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O
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T—
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f
fr
re
eq
qu
ue
en
nc
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y o
of
f H
H=
= 2
2
f
fr
re
eq
qu
ue
en
nc
cy
y o
of
f I
I=
= 5
5
f
fr
re
eq
qu
ue
en
nc
cy
y o
of
f J
J=
= 0
0
f
fr
re
eq
qu
ue
en
nc
cy
y o
of
f K
K=
= 1
1
f
fr
re
eq
qu
ue
en
nc
cy
y o
of
f L
L=
= 3
3
f
fr
re
eq
qu
ue
en
nc
cy
y o
of
f M
M=
= 1
1
f
fr
re
eq
qu
ue
en
nc
cy
y o
of
f N
N=
= 8
8
f
fr
re
eq
qu
ue
en
nc
cy
y o
of
f O
O=
= 6
6
f
fr
re
eq
qu
ue
en
nc
cy
y o
of
f P
P=
= 1
1
f
fr
re
eq
qu
ue
en
nc
cy
y o
of
f Q
Q=
= 0
0
f
fr
re
eq
qu
ue
en
nc
cy
y o
of
f R
R=
= 7
7
f
fr
re
eq
qu
ue
en
nc
cy
y o
of
f S
S=
= 9
9

c t
4
f
fr
re
eq
qu
ue
en
nc
cy
y o
of
f T
T=
= 3
3
f
fr
re
eq
qu
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en
nc
cy
y o
of
f U
U=
= 1
1
f
fr
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eq
qu
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nc
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y o
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f V
V=
= 1
1
f
fr
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eq
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en
nc
cy
y o
of
f W
W=
= 1
1
f
fr
re
eq
qu
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nc
cy
y o
of
f X
X=
= 0
0
f
fr
re
eq
qu
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en
nc
cy
y o
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f Y
Y=
= 1
1
f
fr
re
eq
qu
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en
nc
cy
y o
of
f Z
Z=
= 0
0
(
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gr
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to
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np
pu
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t t
th
hr
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ee
e b
bi
in
na
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ry
y n
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um
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be
er
rs
s a
an
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d d
di
is
sp
pl
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ay
y t
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e s
su
um
m o
of
f t
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em
m.
.
F
Fo
or
r e
eg
g-
-
1
11
10
01
1+
+1
10
00
01
1+
+1
11
10
00
0=
=1
10
00
00
01
10
0.
.
Algorithm-
Step1. Initialize array arr [] with 100.
Step 2. Declare a,b,c,i,s,s1,s2,d,d1,d2,c,c0,c1,c2,sum,k.
Step 3. Initilize s,s1,s2,d,d1,d2,sum,k with 0 and c0,c1,c2 with 1;
Step 4. Input three binary number from user.

c t
5
Step 5. Begin loop for i from a to1 and repeat step 6 to 8.
Step 6. Assign value of i%10 to s.
Step 7. d=d+s*c0.
Step 8. Assign value of c0*2 to c0.
Step 9. Begin loop for i from b to1 and repeat step 10 to 12.
Step 10. Assign value of i%10 to s1.
Step 11. d1=d1+s1*c1.
Step 13. Begin loop for i from c to1and repeat step 14 to 16.
Step 14. Assign value of i%10 to s2.
Step 15. d2=d2+s2*c2.
Step 17. Add d,d1,d2 and store the value to sum.
Step 18 Begin while loop for sum upto 0 and repeat step 19 to 21.
Step 19. Assign value of sum%2 to sum.
Step 20. Assign value of sum/2 to sum.
Step 21. k++.
TYPE NAME PURPOSE
int c To get the remainder
Int i looping variable
Int n1,n2,n3 To store 3 nos. by user
Int sum1,sum2,sum3 To get the sum of binary nos.
int p1,p2,p3 to store powers
int k array variable
int sum4 sum of all 3 nos.
int a[] array
PROGRAM-
import java.io.*;
class Sum_Binary // class begins
{
public static void main()throws Exception
{
int c,i,n1,n2,n3,sum1=0,sum2=0,sum3=0,p1=0,p2=0,p3=0,p,k=0,sum4;
// initializing variables.
int a[]=new int[10];
System.out.println("Enter three numbers"); // get input by user
n1=Integer.parseInt(num.readLine());

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6
for(i=n1;i>0;i/=10) // for loop for first number
{
c=i%10;
sum1=sum1+(int)Math.pow(2,p1)*c;
p1++;
}
for(i=n2;i>0;i/=10) // for loop foe second number
{
c=i%10;
sum2=sum2+(int)(Math.pow(2,p2)*c);
p2++;
}
for(i=n3;i>0;i/=10) // for loop for third number
{
c=i%10;
sum3=sum3+(int)(Math.pow(2,p3)*c);
p3++;
}
sum4=sum1+sum2+sum3; // sum of all three numbers
System.out.print(sum4);
while(sum4>0)
{
p=sum4%2;
a[k]=p;
k++;
sum4=sum4/2;
}
System.out.println("The added binary number is = "); // printing comment
for(i=k-1;i>=0;i--)
{
System.out.print(a[i]);
}
}
} // class ends
output-
I
In
np
pu
ut
t—
—
E
En
nt
te
er
r t
th
hr
re
ee
e n
nu
um
mb
be
er
rs
s
1
10
00
01
1
1
11
10
01
1
1
11
10
00
0
O
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tp
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t-
--
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t
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d b
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na
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10
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01
10
0.
.

c t
7
(
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3)
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g p
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at
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er
rn
n.
.
N
N=
=4
4
1
1 8
8 1
13
3 1
16
6
5
5 2
2 9
9 1
14
4
1
11
1 6
6 3
3 1
10
0
1
15
5 1
12
2 7
7 4
4
Algorithm-
1).Declare a matrix a[][] of size 10.
2).Declare variable r,c,n and initialize n with 4.
3).Begin loop for r from 0 to 3 and repeat step 4.
4).Assign r+1 to a[r][r].
6).Assign n+1 to a[r+1][r],n+4 to a[r][r+1] and n+1 to n.
8).Assign n+4 to a[r+2][r],n+6 to a[r][r+2] and n+1 to n.
9). Begin loop for r from 3 to less than 1 and repeat step 10 and 11.
10).Begin loop for c from 0 to less than 1 and repeat step 11.
11).Assign n+6 to a[r][c] and n+7 to a[c][r].
12).Print a[][] for 4 rows and 4 columns.
TYPE NAME PURPOSE
int a[][] Store 2-D array.
int c1,c2 counter variable
Int n input by user
Int r row of 2-D array
int c column of 2-D array
PROGRAM-

c t
8
import java.io.*;
class matrix // class begins
{
public static void main()throws Exception // defining main function
{
int a[][]=new int[20][20]; // declaring matrix
int p=1,c1=0,c2=0,k1=1,k=1,r,c,n,i;
System.out.println("enter your limit");
n=Integer.parseInt(num.readLine()); // accepting values by the user
for(r=0;r<n;r++) // loop generated
{
for(c=0;c<n;c++) // inner loop
{
if(r==c) // conditional statement
{
a[r][c]=p++;
}
}
}
for(i=1;i<n;i++) // loop generated
{
for(r=k;r<n;r++) // inner loop
{
a[r][c1]=p++;
c1++;
}
k++;
c1=0;
for(r=k1;r<n;r++) // loop generated
{
a[c2][r]=p++;
c2++;
}
k1++;
c2=0;
}
for(r=0;r<n;r++) // loop generated
{
for(c=0;c<n;c++) // inner loop
{
System.out.print(a[r][c]+" "); // printing of matrix
}
}
}
} // class ends

c t
9
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--
-
1
1 1
10
0 1
17
7 2
22
2 2
25
5
6
6 2
2 1
11
1 1
18
8 2
23
3
1
14
4 7
7 3
3 1
12
2 1
19
9
2
20
0 1
15
5 8
8 4
4 1
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3
2
24
4 2
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1 1
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6 9
9 5
5
4
4)
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a m
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ag
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l m
ma
at
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ix
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.A
A m
ma
ag
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ic
ca
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l m
ma
at
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ix
x i
is
s s
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na
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is
s a
al
lw
wa
ay
ys
s s
sa
am
me
e.
.
Eg.
8+1+6
3+5+7
4+9+2
Algorithm-
Step1 Start.
Step2 Declare suitable variables like n,r=0,I,r1,c1,c and declare an 2-d array of limit 10.
Step3 Take an odd limit in n and perform c=n/2.
Step4 Start a loop from i=1 to i<=n*n.
Step5 Strore a[r][c] in I and then put r1=r,c1=c.
Step6 Perform r-- and c++.
Step7 Check if r==-1 then put r=n-1.
Step8 Check if r>n-1 then put r=0.
Step9 Check if c>n-1 then put c=0.
Step10 Check if a[r][c]!=0,if so then put r=r1,c=c1,r++.
Step11 Now print the elements of 2-d array.
Step12 End.

c t
10
* TYPE NAME PURPOSE
* int a[][] Store 2-D array.
* int i looping variable
* int n input by user
* int r row of 2-D array
* int c column of 2-D array
PROGRAM-
import java.io.*;
class magic // defining class
{
public static void main(String args[])throws Exception
{ // defining main function
BufferedReader num=new BufferedReader
(new InputStreamReader(System.in));
int a[][]=new int[10][10]; // declaring array
int n,r=0,i,r1,c1,c;
System.out.println("Enter your odd limit");
n=Integer.parseInt(num.readLine()); // taking input from the user
c=n/2;
for(i=1;i<=n*n;i++) // for loop
{
a[r][c]=i;
r1=r;
c1=c;
r--;
c++;
if(r==-1) // conditional statement
{
r=n-1;
}
if(r>n-1) // conditional statement
{
r=0;
}
if(c>n-1) // conditional statement
{
c=0;

c t
11
}
if(a[r][c]!=0) // conditional statement
{
r=r1;
c=c1;
r++;
}
}
for(r=0;r<n;r++) // for loop
{
for(c=0;c<n;c++)// inner for loop
{
System.out.print(a[r][c]); // printing of array
}
}
} // main function ends
} // class ends
o
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En
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3
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-
8
8 1
1 6
6
3
3 5
5 7
7
4
4 9
9 2
2

c t
12
(
(5
5)
) A
A s
sq
qu
ua
ar
re
e o
of
f s
sq
qu
ua
ar
re
e m
ma
at
tr
ri
ix
x c
ca
an
n b
be
e g
ge
en
ne
er
ra
at
te
ed
d b
by
y t
ta
ak
ki
in
ng
g a
an
ny
y s
sq
qu
ua
ar
re
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ma
at
tr
ri
ix
x o
of
f e
ev
ve
en
n d
di
im
me
en
ns
si
io
on
ns
s (
(>
>=
=2
2)
)
a
an
nd
d f
fi
il
ll
l t
th
he
e i
in
nn
ne
er
rm
mo
os
st
t n
nu
um
mb
be
er
r n
n i
is
s c
ci
ir
rc
cu
ul
la
ar
r o
or
rd
de
er
r.
. T
Th
he
en
n f
fi
il
ll
l i
it
ts
s s
su
ur
rr
ro
ou
un
nd
di
in
ng
g s
sq
qu
ua
ar
re
e o
on
ne
e b
by
y o
on
ne
e.
.
E
Eg
g N
N=
=4
4 s
si
iz
ze
e=
=6
6
2
20
0 2
21
1 2
22
2 2
23
3 2
24
4 2
25
5
3
39
9 8
8 9
9 1
10
0 1
11
1 2
26
6
3
38
8 1
19
9 4
4 5
5 1
12
2 2
27
7
3
37
7 1
18
8 7
7 6
6 1
13
3 2
28
8
3
36
6 1
17
7 1
16
6 1
15
5 1
14
4 2
29
9
3
35
5 3
34
4 3
33
3 3
32
2 3
31
1 3
30
0
Algorithm-
Step1 Start.
Step2 Declare suitable variables like p,c,x,n,r1,c1,c2,c3,k,n,r,i,j and array of limit
declare a 2-d 10.
Step3 Take an limit in N and perform p=N and c=n/2-1.
Step4 Start a loop from r=n/2-1 to r<=n/2 and perform a[c][r]=p++ and after
executing the loop do c++;
Step5 Start a loop from r=n/2 to r>=n/2-1,r—and perform a[c][r]=p++.
Step6 Perform c++.
Step7 Start the loops with j,c1,c2,c3 and i to store the numbers in circular order.
Step8 Now print the elements of 2-d array.
Step9 End.
* TYPE NAME PURPOSE
* int a[][] 2-D array
* int c,p counter variable
* int c1,c2,c3 columns of array
* int i,j looping variable
* int N input from user
PROGRAM-
import java.io.*;
class circular // class definition starts
{
public static void main(String args[])throws Exception//main function starts
{
int a[][]=new int[10][10];

c t
13
int p,c,x,n,r1,c1,c2,c3,k,N,r,i,j; // initializing variables
System.out.println("Enter the value of N");
N=Integer.parseInt(num.readLine());// input from the user
p=N;
System.out.println("enter the dimensions");
n=Integer.parseInt(num.readLine()); // input from the user
c=n/2-1;
for(r=n/2-1;r<=n/2;r++) // generating for loop
{
a[c][r]=p++;
}
c++; // counter variable
for(r=n/2;r>=n/2-1;r--) // generating for loop
{
a[c][r]=p++; // incrementing value
}
r=n/2-2;
r1=n/2+1;
x=n/2-2;
for(j=1;j<=n/2-1;j++) // generating for loop
{
for(c1=r;c1<=r1;c1++) // generating for loop
{
a[x][c1]=p++;
}
for(c2=x+1;c2<=r1;c2++) // generating for loop
{
a[c2][r1]=p++;
}
for(c3=r1-1;c3>=r;c3--) // generating for loop
{
a[r1][c3]=p++;
}
k=r1-1;
for(i=r1;i>r+1;i--) // generating for loop
{
a[k][c3+1]=p++;
k--;
}
r--;
r1=r1+1;
x--;
}
for(r=0;r<n;r++) // generating for loop
{
for(c=0;c<n;c++) // generating for loop
{
System.out.print(a[r][c]+" "+" "); // printing array
}

c t
14
}
}
}
OUTPUT-
I
In
np
pu
ut
t-
--
-
E
En
nt
te
er
r t
th
he
e v
va
al
lu
ue
e o
of
f N
N
5
5
e
en
nt
te
er
r t
th
he
e d
di
im
me
en
ns
si
io
on
ns
s
6
6
O
Ou
ut
tp
pu
ut
t-
--
-
2
21
1 2
22
2 2
23
3 2
24
4 2
25
5 2
26
6
4
40
0 9
9 1
10
0 1
11
1 1
12
2 2
27
7
3
39
9 2
20
0 5
5 6
6 1
13
3 2
28
8
3
38
8 1
19
9 8
8 7
7 1
14
4 2
29
9
3
37
7 1
18
8 1
17
7 1
16
6 1
15
5 3
30
0
3
36
6 3
35
5 3
34
4 3
33
3 3
32
2 3
31
1
(
(6
6)
) W
Wr
ri
it
te
e a
a p
pr
ro
og
gr
ra
am
m t
to
o s
so
or
rt
t t
th
he
e e
el
le
em
me
en
nt
ts
s i
in
n a
a t
tw
wo
o d
di
im
me
en
ns
si
io
on
na
al
l a
ar
rr
ra
ay
y.
.
F
Fo
or
r e
eg
g—
—

c t
15
Input--
2 31 54 87
1 2 3 47
64 58 27 30
34 21 2 28
Output—
87 64 58 54
47 34 31 30
28 27 21 3
2 2 2 1
Algorithm-
Step 1 Declare and initialize variables like r1,c1,temp,r,c and a double dimensional array
a[][].
Step 2 Store the numbers from the user in 2-d array.
Step 3 Begin a loop from r1=0 till r<4.
Step 4 Begin a loop from c1=0 till c1<4.
Step 5 Begin a loop from r=0 till r<4.
Step 6 Begin a loop from c=0 till c<4.
Step 7 Check if a[r1][c1]>a[r][c].
Step 8 If step7 is true then perform the swapping of elements.
Step 9 Print the final 2-d array at the end.
Step 10 End of process.
* TYPE NAME PURPOSE
* int a[][] Store 2-D array.

c t
16
* int temp for swapping aray elements
* int r,r1 row of 2-D array
* int c,c1 column of 2-D array
PROGRAM-
import java.io.*;
class twodsorting // class definition starts
{
{
int a[][]=new int[6][5];
int r,c,temp,r1,c1;
for(r=0;r<4;r++) // generating for loop
{
for(c=0;c<4;c++)
{
System.out.println("Enter your number");
a[r][c]=Integer.parseInt(num.readLine()); // input by user
}
}
for(r1=0;r1<4;r1++) // for loop
{
for(c1=0;c1<4;c1++)// for loop
{
for(r=0;r<4;r++)// for loop
{
for(c=0;c<4;c++)// for loop
{
if(a[r1][c1]>a[r][c]) // condition
{
/* swapping process begins */
temp=a[r1][c1];
a[r1][c1]=a[r][c];
a[r][c]=temp;
/* swapping process ends */
}
}
}
}
}
System.out.println("The sorted array is");
for(r=0;r<4;r++)
{
for(c=0;c<4;c++)
{
System.out.print(a[r][c]+" ");
}

c t
17
}
}
}
o
ou
ut
tp
pu
ut
t-
-
I
In
np
pu
ut
t—
—
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
2
2
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
3
31
1
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
5
54
4
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
8
87
7
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
1
1
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
2
2
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
3
3
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
4
47
7
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
6
64
4
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
5
58
8
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
2
27
7
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
3
30
0
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
3
34
4
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
2
21
1
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
2
2
E
En
nt
te
er
r y
yo
ou
ur
r n
nu
um
mb
be
er
r
2
28
8
O
Ou
ut
tp
pu
ut
t—
—
T
Th
he
e s
so
or
rt
te
ed
d a
ar
rr
ra
ay
y i
is
s
8
87
7 6
64
4 5
58
8 5
54
4
4
47
7 3
34
4 3
31
1 3
30
0
2
28
8 2
27
7 2
21
1 3
3
2
2 2
2 2
2 1
1

c t
18
(
(7
7)
) W
Wr
ri
it
te
e a
a p
pr
ro
og
gr
ra
am
m i
in
n i
in
np
pu
ut
t t
tw
wo
o m
ma
at
tr
ri
ix
x a
an
nd
d p
pr
ri
in
nt
t t
th
he
e p
pr
ro
od
du
uc
ct
t o
of
f t
th
he
em
m.
Algorithm-
Step-1 Declare array A[][],B[][],C[][] and m,n,p,q for size.
Step-2 Input size of matrix A[][] in m,n.
Step-3 Input size of matrix B[][] in p,q.
Step-4 Input elements of matrix A[][] of m rows and n columns.
Step-5 Input elements of matrix B[][] of p rows and q columns.
Step-6 Check if n=p then goto step 7 otherwise goto step 15.
Step-7 Begins a loop for i from 0 to m-1 and repeat step 8 to 11.
Step-8 Begins a loop for j from 0 to q-1 and repeat step 9 to 11.
Step-9 Assign 0 to C[i][j].
Step-10 Begins a loop for k from 0 to n-1 and repeat step 11.
Step-11 Multiply matrix A[][] and B[][] as
C[i][j]=C[i][j] +(A[i][k]*B[k][j]).
Step-12 Print matrix A[][] of m rows and n columns using nested loops.
Step-13 Print matrix B[][] of p rows and q columns using nested loops.
Step-14 Print matrix C[][] of m rows and q columns using nested loops.
Step-15 End of algorithm.
* TYPE NAME PURPOSE
* a[][]
* int b[][] Store 2-D array.
* cr[][]
* int m,n row &column of matrix A
* int p,q row &column of matrix B
* int i,j looping variable
* int r,c row &column of cr[][]
PROGRAM-
import java.io.*;
class Product_Matrix // class starts
{
public static void main(String args[])throws Exception // main function starts
{
int m,n,p,q,i,j,k,r,c; // declaring variables
int a[][]=new int[20][20];
int b[][]=new int[20][20];
int cr[][]=new int[20][20];
System.out.println("Input no. of rows and columns matrix A");
m=Integer.parseInt(num.readLine()); // input from user
n=Integer.parseInt(num.readLine()); // input from user
System.out.println("Input no. of rows and columns matrix B");

c t
19
p=Integer.parseInt(num.readLine()); // input from user
q=Integer.parseInt(num.readLine()); // input from user
System.out.println("Input elements of A");
for(r=0;r<m;r++)
for(c=0;c<n;c++)
{
System.out.println("Enter numbers for matrix A");
a[r][c]=Integer.parseInt(num.readLine()); // input from user
}
System.out.println("Input elements of B");
for(r=0;r<p;r++)
for(c=0;c<q;c++)
{
System.out.println("Enter numbers for matrix B");
b[r][c]=Integer.parseInt(num.readLine()); // input from user
}
if(n==p) // condition
{
for(i=0;i<m;i++) // for loop
{
for(j=0;j<q;j++) // for loop
{
for(k=0;k<q;k++) // for loop
{
cr[i][j]+=a[i][k]*b[k][j]; // updating new 2-D array
}
}
}
}
System.out.println("Matrix A");
for(r=0;r<m;r++) // for loop
{
for(c=0;c<n;c++) // for loop
System.out.print(a[r][c]+"t"); // printing array
}
System.out.println("Matrix B");
for(r=0;r<p;r++) // for loop
{
for(c=0;c<q;c++) // for loop
System.out.print(b[r][c]+"t"); // printing array
}
System.out.println("Matrix C");
for(r=0;r<m;r++) // for loop
{
for(c=0;c<q;c++) // for loop
System.out.print(cr[r][c]+"t"); // printing array
}
}

c t
20
}

c t
21
I
In
np
pu
ut
t—
—
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x A
A
2
2
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x A
A

3
3
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x A
A
4
4
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x A
A
5
5
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x A
A
6
6
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x A
A
7
7
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x A
A
8
8
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x A
A
9
9
I
In
np
pu
ut
t e
el
le
em
me
en
nt
ts
s o
of
f B
B
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x B
B
1
1
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x B
B
2
2
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x B
B
3
3
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x B
B
4
4
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x B
B
5
5
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x B
B
6
6
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x B
B
7
7
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x B
B
8
8
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r m
ma
at
tr
ri
ix
x B
B
9
9
O
Ou
ut
tp
pu
ut
t-
--
-
M
Ma
at
tr
ri
ix
x A
A
1
1 2
2 3
3
4
4 5
5 6
6
7
7 8
8 9
9

c t
22
(
(8
8)
) W
Wr
ri
it
te
e a
a p
pr
ro
og
gr
ra
am
m t
to
o i
in
np
pu
ut
t a
a f
fo
ou
ur
r l
le
et
tt
te
er
r w
wo
or
rd
d a
an
nd
d p
pr
ri
in
nt
t a
al
ll
l t
th
he
e p
po
os
ss
si
ib
bl
le
e c
co
om
mb
bi
in
na
at
ti
io
on
ns
s o
of
f t
th
ha
at
t w
wo
or
rd
d.
.
M
Ma
at
tr
ri
ix
x B
B
1
1 2
2 3
3
4
4 5
5 6
6
7
7 8
8 9
9
M
Ma
at
tr
ri
ix
x C
C
3
30
0 3
36
6 4
42
2
6
66
6 8
81
1 9
96
6
1
10
02
2 1
12
26
6 1
15
50
0

c t
23
Algorithm-
Step 1 Input a string of not more than 4 characters.
Step 2 Declare variables like i,j,k,l and in n store the length of string.
Step 3 Begin a loop for i till n.
Step 4 Begin a loop for j till n.
Step 5 Begin a loop for k till n.
Step 6 Begin a loop for l till n.
Step 7 Check if (i!=j&&i!=k&&i!=l&&j!=k&&j!=l&&k!=l) then repeat steps 8,9,10,11.
Step 8 Print the character at i position.
Step 9 Print the character at j position.
Step 10 Print the character at k position.
Step 11 Print the character at l position.
Step 12 End of algorithm.
TYPE NAME PURPOSE
int n Store length of str1
String str1 to add character.
Int j looping variable
Int p counter variable
char ch to store present character
char ch2 to compare with ch2
PROGRAM-
import java.io.*;
class permutation
{
{
String str;
int n,j,k,l,i;
System.out.println("Enter your four letter word");
str=num.readLine();
n=str.length();
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
for(k=0;k<n;k++)

c t
24
{
for(l=0;l<n;l++)
{
if(i!=j&&i!=k&&i!=l&&j!=k&&j!=l&&k!=l)
{
System.out.print(str.charAt(i));
System.out.print(str.charAt(j));
System.out.print(str.charAt(k));
System.out.print(str.charAt(l));
}
}
}
}
}
}
}
o
ou
ut
tp
pu
ut
t-
-
I
In
np
pu
ut
t-
-
E
En
nt
te
er
r y
yo
ou
ur
r f
fo
ou
ur
r l
le
et
tt
te
er
r w
wo
or
rd
d
R
Ra
aj
ju
u
O
Ou
ut
tp
pu
ut
t-
--
-
r
ra
aj
ju
u
r
ra
au
uj
j
r
rj
ja
au
u
r
rj
ju
ua
a
r
ru
ua
aj
j
r
ru
uj
ja
a

c t
25
a
ar
rj
ju
u
a
ar
ru
uj
j
a
aj
jr
ru
u
a
aj
ju
ur
r
a
au
ur
rj
j
a
au
uj
jr
r
j
jr
ra
au
u
j
jr
ru
ua
a
j
ja
ar
ru
u
j
ja
au
ur
r
j
ju
ur
ra
a
j
ju
ua
ar
r
u
ur
ra
aj
j
u
ur
rj
ja
a
9
9)
) W
Wr
ri
it
te
e a
a p
pr
ro
og
gr
ra
am
m t
to
o i
in
np
pu
ut
t a
a m
ma
at
tr
ri
ix
x a
an
nd
d r
re
em
mo
ov
ve
e t
th
he
e c
ce
en
nt
tr
ra
al
l n
nu
um
mb
be
er
rs
s a
an
nd
d t
th
he
en
n p
pr
ri
in
nt
t t
th
he
e s
su
um
m o
of
f r
re
em
ma
ai
in
ni
in
ng
g
n
nu
um
mb
be
er
rs
s i
in
n t
th
he
e m
ma
at
tr
ri
ix
x.
.
I
If
f t
th
he
e i
in
np
pu
ut
t i
is
s-
-
O
Or
ri
ig
gi
in
na
al
l M
Ma
at
tr
ri
ix
x

c t
26
1
1 2
2 3
3
4
4 5
5 6
6
7
7 8
8 9
9
T
Th
he
en
n t
th
he
e o
ou
ut
tp
pu
ut
t i
is
s-
-
S
Sq
qu
ua
ar
re
e M
Ma
at
tr
ri
ix
x
1
1 2
2 3
3
4
4 6
6
7
7 8
8 9
9
S
Su
um
m i
is
s :
: 4
40
0
Algorithm-
Step 1 Start.
Step 2 Declare variables like r,c,m=0,s=0,l=n and a 2d array of limit n which is to be
entered from the user.
Step 3 Start a while loop such that n>1.
Step 4 Start a loop in which r=m,c=m tll c<l-m and then perform s=s+a[r][c] and then
print it.
Step 5 Change the line..
Step 6 Start for loop from r=m+1 till r<l-m-1.
Step7 Start another loop from c=m till c<l-m.
Step8 Check if (c==m||c==l||c==l-m-1).
Step9 Start another loop from r=l-m-1 till c=m and repeat second half of step4.
Step10 Initialize n=n-2,m++ and s=0.
Step11 End.
TYPE NAME PURPOSE
int a[][] Store 2-D array.
int n input by user
int r row of 2-D array
int c column of 2-D array
int s Sum of elements
PROGRAM-
import java.io.*;

c t
27
class Square_sum
{
{
System.out.println("Enter order of matrix(odd)");
int n=Integer.parseInt(num.readLine());
int a[][]=new int[n][n];
int r,c,m=0,s=0,l=n;
for(r=0;r<n;r++)
for(c=0;c<n;c++)
{
System.out.println("Enter numbers for array");
a[r][c]=Integer.parseInt(num.readLine());
}
System.out.println("Original Matrix");
for(r=0;r<n;r++)
{
for(c=0;c<n;c++)
System.out.print(a[r][c]+"t");
}
System.out.println("Square Matrix");
while(n>1)
{
for(r=m,c=m;c<l-m;c++)
{
s=s+a[r][c];
}
for(r=m+1;r<l-m-1;r++)
{
for(c=m;c<l-m;c++)
{
if(c==m||c==l-m-1)
{
s=s+a[r][c];
}
else
System.out.print("t");
}
}
for(r=l-m-1,c=m;c<l-m;c++)
{
s=s+a[r][c];
}
System.out.print("Sum is : "+s);

c t
28
System.out.print("n");
n-=2;
m++;
s=0;
}
}
}
o
ou
ut
tp
pu
ut
t-
-
(
(1
10
0)
) W
Wr
ri
it
te
e a
a p
pr
ro
og
gr
ra
am
m t
to
o i
in
np
pu
ut
t a
a l
li
im
mi
it
t a
an
nd
d p
pr
ri
in
nt
t a
al
ll
l t
th
he
e l
lu
uc
ck
ky
y n
no
os
s.
.
I
In
np
pu
ut
t-
--
-
E
En
nt
te
er
r o
or
rd
de
er
r o
of
f m
ma
at
tr
ri
ix
x(
(o
od
dd
d)
)
3
3
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r a
ar
rr
ra
ay
y
1
1
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r a
ar
rr
ra
ay
y
2
2
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r a
ar
rr
ra
ay
y
3
3
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r a
ar
rr
ra
ay
y
4
4
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r a
ar
rr
ra
ay
y
5
5
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r a
ar
rr
ra
ay
y
6
6
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r a
ar
rr
ra
ay
y
7
7
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r a
ar
rr
ra
ay
y
8
8
E
En
nt
te
er
r n
nu
um
mb
be
er
rs
s f
fo
or
r a
ar
rr
ra
ay
y
9
9
O
Or
ri
ig
gi
in
na
al
l M
Ma
at
tr
ri
ix
x
1
1 2
2 3
3
4
4 5
5 6
6
7
7 8
8 9
9
S
Sq
qu
ua
ar
re
e M
Ma
at
tr
ri
ix
x
1
1 2
2 3
3
4
4 6
6
7
7 8
8 9
9 S
Su
um
m i
is
s :
: 4
40
0

c t
29
F
Fo
or
r e
eg
g-
-i
if
f t
th
he
e i
in
np
pu
ut
t i
is
s 1
10
0.
.
T
Th
he
en
n t
th
he
e O
Ou
ut
tp
pu
ut
t i
is
s=
=1
1,
,3
3,
,7
7.
.
Algorithm-
Step 1 Declare variables like i,a[100],n, k,j,ctr=0,step=1.
Step 2 Input the limit in n.
Step 3 Begin a for loop for i from 0 to n and store natural numbers till limit in the array
a[].
Step 4 Begin a for loop for i from 0 to n-ctr
Step 5 Begin a for loop for j from i+1 to n-ctr and j=j+step.
Step 6 Begin a for loop for k from j to n and store numbers as a[k]=a[k+1].
Step 7 Add 1 to ctr at the end of k loop.
Step 8 Add 1 to step at the end of j loop.
Step 9 Print lucky numbers by the loop i from 0 to n-ctr.
Step 10 End of algorithm.
TYPE NAME PURPOSE
int n input from user
int i,j looping variable
int ctr,step counter variable
PROGRAM-
import java.io.*;
class lucky
{
{
int n,i,j,k,ctr=0,step=1;
System.out.println("Enter the limit");
n=Integer.parseInt(num.readLine());
for(i=0;i<n;i++)
a[i]=i+1;
for(i=0;i<n-ctr;i++)
{
for(j=i+1;j<(n-ctr);j+=step)
{
for(k=j;k<n;k++)
{
a[k]=a[k+1];
}
ctr++;
}

c t
30
step++;
}
System.out.println("Lucky number are=");
for(i=0;i<(n-ctr);i++)
System.out.println(a[i]+" ");
}
}
output-
I
In
np
pu
ut
t-
-
E
En
nt
te
er
r t
th
he
e l
li
im
mi
it
t
1
10
0
O
Ou
ut
tp
pu
ut
t
L
Lu
uc
ck
ky
y n
nu
um
mb
be
er
r a
ar
re
e=
=
1
1
3
3
7
7
(
(1
11
1)
) W
Wr
ri
it
te
e a
a p
pr
ro
og
gr
ra
am
m t
to
o i
in
np
pu
ut
t n
n n
nu
um
mb
be
er
rs
s i
in
n a
a a
ar
rr
ra
ay
y a
an
nd
d a
ar
rr
ra
an
ng
ge
e i
in
n g
gi
iv
ve
en
n m
ma
an
nn
ne
er
r w
wi
it
th
ho
ou
ut
t u
us
si
in
ng
g a
an
no
ot
th
he
er
r
a
ar
rr
ra
ay
y.
.
S
Sh
hi
if
ft
t t
th
he
e l
la
ar
rg
ge
es
st
t n
nu
um
mb
be
er
r a
at
t t
th
he
e m
mi
id
dd
dl
le
e p
po
os
si
it
ti
io
on
n (
(i
i.
.e
e N
N/
/2
2 i
in
nd
de
ex
x n
nu
um
mb
be
er
r)
) s
se
ec
co
on
nd
d l
la
ar
rg
ge
es
st
t b
be
ef
fo
or
re
e t
th
he
e m
mi
id
dd
dl
le
e
a
an
nd
d s
so
o o
on
n.
.
Algorithm-

c t
31
Step1. Declare variables n,i,j,k,k1,t.
Step 2. Input size of array in n from user.
Step 3. Declare array a[] of size (n*2).
Step 4. Assign (n+n)-4 to k and (n+n)-2 to k1.
Step 5. Input n numbers from user.
Step 6. Begin loop for i from 0 to n-1 and repeat step 7 to 12.
Step 7. Begin loop for j from i+1 to n-1 and repeat step 8 to12.
Step 8. if(a[i]<a[j])
Step 9. Assign value of a[i] to t.
Step 10. Assign value of a[j] to a[i].
Step 11. Assign value of t to a[j].
Step 12. Assign 0 to t.
Step 13. Assign a[0] to a[(n+n)-3].
Step 14. Begin loop for i from 1 to n-1 and repeat step 15 to 20.
Step 15. if(i%2!=0)
Step 16. Assign a[i] to a[k].
Step 17. Assign k-1 to k.
Step 18. else if(i%2= =0)
Step 19. Assign a[i] to a[k1].
Step 20. Assign k1+1 to k1.
Step 21. Print elements of array a[] from n to (n+n)-1.
STEP 22. END
TYPE NAME PURPOSE
int a[] array
int temp for swapping values
int x,y,z looping variables
PROGRAM-
import java.io.*;
class arraypat
{
{
int n,x,y,z,p,temp;
System.out.println("enter the number of elements");
for(x=0;x<n;x++)
{
System.out.println("enter the element");
a[x]=Integer.parseInt(num.readLine());
}
p=(n/2);

c t
32
n--;
for(x=0;x<p;x++)
{
for(y=x;y<=(n-x);y++)
{
for(z=y;z<=(n-x);z++)
{
if(a[y]>a[z])
{
temp=a[y];
a[y]=a[z];
a[z]=temp;
}
}
}
for(y=x+1;y<=(n-x);y++)
{
for(z=y;z<=(n-x);z++)
{
if(a[y]<a[z])
{
temp=a[y];
a[y]=a[z];
a[z]=temp;
}
}
}
}
if(n%2==0)
{
for(x=0;x<p;x++)
{
temp=a[x];
a[x]=a[n-x];
a[n-x]=temp;
}
}
System.out.println("the final array is");
for(x=0;x<=n;x++)
{
System.out.println(a[x]);
}
}
}
output-
Input—
enter the number of elements
6

c t
33
enter the element
45
enter the element
69
enter the element
5
enter the element
88
enter the element
4
enter the element
1
Output--
the final array is
1
5
69
88
45
4
(
(1
12
2)
) W
Wr
ri
it
te
e a
a p
pr
ro
og
gr
ra
am
m t
to
o i
in
np
pu
ut
t a
a p
po
os
si
it
ti
iv
ve
e i
in
nt
te
eg
ge
er
r a
an
nd
d p
pr
ri
in
nt
t a
al
ll
l t
th
he
e c
co
om
mb
bi
in
na
at
ti
io
on
ns
s o
of
f c
co
on
ns
se
ec
cu
ut
ti
iv
ve
e n
na
at
tu
ur
ra
al
l
n
nu
um
mb
be
er
r.
.
E
Eg
g-
-2
27
7;
;
2
2+
+3
3+
+4
4+
+5
5+
+6
6+
+7
7=
=2
27
7.
.
8
8+
+9
9+
+1
10
0=
=2
27
7.
.
1
13
3+
+1
14
4=
=2
27
7.
.
Algorithm-
Step1.Declare array a[][] and variable i,j,i1,s,x,k.
Step 2.Initialize s and k with value 0.
Step 3.Input value of x from user.

c t
34
Step 4.Begin loop for i1 from 1 to x-1 and repeat step 5 to 15.
Step 5.Begin loop for i from i1 to x-1 and repeat step 6 to14.
Step 6.if(s!=x)
Step 7.Assign value of s+i to s.
Step 8.Assign value of I to a[k].
Step 9.Assign value of k+1 to k.
Step 10.else
Step 11.Begin loop for j from 0 to k-1 and repeat step 12.
Step 12. Print the value of a[j].
Step 13.Print value of a[k-1].
Step 14.Assign 0 to k and s.
Step 15. Assign 0 to k and s.
step 16. end
* TYPE NAME PURPOSE
* int n input from user
* int sum to find sum
* int c (for loop)
* to print consecutive pairs
PROGRAM-
import java.io.*;
class consnatural
{
{
int c,i,j,sum=0,n;
System.out.println("Enter a number");
System.out.println("Consecutive pairs are");
for(i=1;i<n;i++)
{
for(j=i;j<n;j++)
{
sum=sum+j;
if(sum==n)
{
for(c=i;c<=j;c++)
{
System.out.print(c+" ");
}

c t
35
}
}
sum=0;
}
}
}
output-
Input—
Enter a number
27
Output--
Consecutive pairs are
2 3 4 5 6 7
8 9 10
13 14
(
(1
13
3)
) W
Wr
ri
it
te
e a
a p
pr
ro
og
gr
ra
am
m t
to
o p
pr
ri
in
nt
t t
th
he
e d
di
ia
am
mo
on
nd
d p
pa
at
tt
te
er
rn
n a
as
s s
sh
ho
ow
wn
n.
.
*
***
*****
*******
********
*******
*****
***
*
Algorithm-

c t
36
Step1 Start.
Step2 Declare suitable variables I,sp,j,n=4,i1,sp1,j1,n1.
Step3 Start a loop from i=1 till i<=9
Step4 Start a loop from sp=1 till sp<=n.
Step5 Start a loop from j=1 till j<=i .
Step6 Start a loop from i1=7 till i1>=1.
Step7 Start a loop from sp=1 till n1 to give the required spacing.
Step9 Start another loop from j1=1 till i1 and then print “*” and then perform n1 and
then change line.
Step 10 End.
TYPE NAME PURPOSE
int i,j looping variables for upper triangle
int i1,j1 looping variables for lower triangle
int sp to give limited spaces
PROGRAM-
import java.io.*;
class diamondpat // class definition starts
{
public static void main()throws Exception // main function starts
{
int i,sp,j,n=4,i1,sp1,j1,n1=1; // initializing variables
for(i=1;i<=9;i=i+2) // generating loop
{
for(sp=1;sp<=n;sp++) // loop to give spaces
{
System.out.print(" ");
}
for(j=1;j<=i;j++) // inner loop to print *
{
System.out.print("*");
}
System.out.println(); // chaning line

c t
37
n--;
}
for(i1=7;i1>=1;i1=i1-2) // generating loop
{
for(sp1=1;sp1<=n1;sp1++)
{
System.out.print(" "); // loop to give spaces
}
for(j1=1;j1<=i1;j1++)
{
System.out.print("*"); // inner loop to print *
}
System.out.println(); // chaning line
n1++;
}
} // main function ends
} // class ends
Output—
*
***
*****
*******
********
*******
*****
***
*

c t
38
PROGRAM 14
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 5 - IF i=0 THEN GOTO STEP 6
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

c t
39
s o l u t i o n
import java.io.*;
class Decode
{public void compute()throws IOException //compute() function
{BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
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;
}}}}

40
c t
var i abl e description
No. Name Type Method Description
1 br BufferedReader compute() BufferedReader object
2 name String compute() input string
3 n int compute() decode number
4 j int compute() loop variable
5 i int compute() loop variable
6 l int compute() length of string
7 c int compute() ASCII of c1
8 y int compute()
9 n1 int compute()
10 c1 char compute() character at index i
11 e NumberFormatException compute() NumberFOrmatException object
output

c t
41
PROGRAM 15
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

c t
42
s o l u t i o n
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
{System.out.println();
for(int i=0;i<l;i++)
{System.out.print(c[i]);
}}
public static void main(String args[]) throws IOException //main function
{Alpha obj = new Alpha();
obj.readword();
obj.arrange();
obj.display();
}}

c t
43
1 br BufferedReader readword() BufferedReader object
2 str String - input string
3 l int - length
4 c char[] - character array
5 i int readword() loop variable
6 j int readword() loop variable
7 temp char readword() temporary storage
8 obj Alpha main() Alpha object
output

c t
44
PROGRAM 16
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
s o l u t i o n
import java.io.*;
class Vowels
{public static void main(String args[])throws IOException //main function
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);
}}

c t
45
1 br BufferedReader main() BufferedReader object
2 a String main() input string
3 z Int main() length of string
4 y Int main() loop variable
5 b Int main() no. of consonants
6 x Int main() no. of vowels
output

46
c t
PROGRAM 17
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
s o l u t i o n
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
}}

c t
47
2 z int main() length of string
3 a String main() input string
4 x int main() space counter
5 y int main() loop variable
output

c t
48
PROGRAM 18
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
s o l u t i o n
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
}}

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49
2 a StringBuffer main() StringBuffer object of input string
3 z Int main() loop variable
output

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50
PROGRAM 19
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
s o l u t i o n
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

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51
a[x][y]=z;
}}
System.out.println("Array -");
for(x=0;x<5;x++)
{for(y=0;y<5;y++)
//loop for printing array
{System.out.print(a[x][y]+" ");
}
}
for(x=0;x<5;x++)
{for(y=0;y<5;y++)
{Sum=Sum+a[x][y];
}}
//loop for printing sum of array elements
System.out.println("Sum of Array elements="+Sum);
}}
//displaying sum
1 aa BufferedReader main() BufferedReader object
2 a int[][] main() input array
3 x Int main() loop variable
5 z Int main() input element
6 Sum main() main() Sum of all elements
output

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PROGRAM 20
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
s o l u t i o n
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;
}}

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53
System.out.println("Array -"); //printing the array in matrix form
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];
}
System.out.println("Sum of column "+(y+1)+" is "+Sum);
Sum=0;
}}}
//printing sum of coloumn
2 a int[][] main() input array
3 x int main() loop variable
4 y int main() loop variable
5 z int main() input element
6 Sum int main() Sum of each couloumn
output

54
c t
PROGRAM 21
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
s o l u t i o n
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 -");

55
o j e c t
for(x=0;x<4;x++)
{for(y=0;y<4;y++)
}
}
y=0;
for(x=0;x<4;x++)
{Sum=Sum+a[x][y];
y=y+1;
}
//displaying array
//loop for finding sum of diagonal
System.out.println("Sum of diagonal is "+Sum); //displaying the sum of diagonal
Sum=0;
}}
2 a int[][] main() input matrix
3 x Int main() loop variable
5 Sum Int main() Sum of diagonals
6 z Int main() input element
output

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PROGRAM 22
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
s o l u t i o n
import java.io.*;
class SalesComission
{double sales,comm;
System.out.println(“Enter sales”);
sales=Double.parseDouble(aa.readLine()); //reading sales from the keyboard

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/*calculating commission*/
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*sales;
System.out.println("Commission of the employee="+comm); //displaying commission
}}
1 Aa BufferedReader main() BufferedReader object
2 Sales double main() sales
3 comm. double main() commision
output

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PROGRAM 23
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
s o l u t i o n
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]);
}}

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1 in DataInputStream main() DataInputStream object
2 num Int main() input number
3 hund String[] main() array storing 100th
position
4 ten String[] main() array storing 10th
position
5 unit String[] main() array storing units position
output

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PROGRAM 24
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
s o l u t i o n
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;
}}

c t
2 ob C2F main() C2F object
3 temp double main() calculated Fahrenheit temperature
4 celcius double convert() input temperature in Celsius
5 far double convert() Calculated Fahrenheit temperature
output

javapravticalfile.doc

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