UNIT 1.pptx Programming for Problem Solving

Syllabus,
PROGRAMMING FOR
PROBLEM SOLVING
(PPS)
B.Tech I Sem

Unit
1
 Introduction to Programming :
 Compilers, compiling and executing a program.
 Representation of Algorithm –
 Algorithms for finding roots of quadratic equations,
 finding minimum and maximum numbers of a given set,
 finding if a number is a prime number
 Flowchart/Pseudocode with examples,
 Program design and structured programming.

Unit
1
 Introduction to C Programming Language:
 variables (with data types and space requirements),
 Syntax and Logical Errors in compilation,
 object and executable code,
 Operators,
 expressions and precedence,
 Expression evaluation,
 Storage classes (auto, extern, static, and register),
 type conversion,

Unit
1
 The main method and command line arguments Bitwise
operations:
 Bitwise AND, OR, XOR, and NOT operators.
 Conditional Branching and Loops:
 Writing and evaluation of conditionals and consequent branching
with if, if-else, switch-case,
 ternary operator,
 goto, Iteration with for, while, do-while loops.
 I/O: Simple input and output with scanf and printf, formatted I/O,
Introduction to stdin, stdout, and stderr. Command line arguments.

Textbooks &
References
1. Jeri R. Hanly and Elliot B. Koffman, “Problem Solving and
Program Design in C,” 7th Edition, Pearson.
2. A.B. Forouzan and F.S. Gilberg, “C Programming and Data
Structures,” Cengage Learning, 3rd Edition.

C is a programming
language developed
at Bell Laboratories
of USA in 1972 by
Dennis Ritchie

 C is a programming language developed
at Bell Laboratories of USA in 1972 by
Dennis Ritchie

UNIT 1.pptx Programming for Problem Solving

Decimal to
Binary
156 (10)
10011100 (2)

Binary representation (Decimal to
Binary
Decimal no.= 123
Binary no. =
1111011
Decimal no. = 255
Binary no. = 1111
1111

Binary to
Decimal
Binary no. =
110011 Decimal
no. = 51
Binary no. = 1111
0001 Decimal no. =
241

Algorith
m
Algorithm
 Algorithm is a finite set of instructions that if
followed accomplishes a particular task.
Problem
Inpu
t
Outpu
t
Comput
er

Problem: Shortest
distance Input
n nodes
connecting
nodes
dista
nce between
nodes
O
utput

Characteristics of
Algorithm
 Input
 Output
 Definitenes
s
 Effectivenes
s
 Finiteness
OD
euftipnui
tte−n
eAsns
a−lSghoorui
tlhdmbseh
oc
luel
da
rhaanv
de 1 or
lead to only one
meaning.
IEnfpf
euctt
i−v
eAnneas
sl
g−o
rAi
tnh
ma
lsgh
oorui
tl
dh
mh
as
hvoeu0l
dohra
v
e
ms
t
eopr
e-
bwye-
sl
lt-
edpe
fdi
ni
reedc
t
iion
pn
su,t
sw. hich should
be
independent of any programming
code
Finiteness − An algorithm must terminate
after a finite number of steps

Characteristics of
Algorithm
 Input − Should have 0 or more well-defined inputs.
 Output - Should have 1 or more well-defined outputs, and
should match the desired output.
 Definiteness −Should be clear and unambiguous.
 Effectiveness − Should have step-by-step directions, which
should be independent of any programming code.
 Finiteness − Must terminate after a finite number of steps.

Program to display “Welcome
to C”
#include<stdio.
h> int main()
{
printf(“Welcome to
C”); return 0;
}

Problem 1: Addition of two
numbers
 Input :
🞑 two numbers (num1
= 4, num2 = 6)
 Output:
🞑 Sum = 10
Algorithm
 Step 1: Start
 Step 2: Read the first
number(num1)
 Step 3: Read second
number(num2)
 Step 4: Sum  num1+num2
 Step 5: Print Sum

Sum = num1 +
num2

Program to accept two numbers from the user and
display the sum using C language
/*
Program to accept two numbers and display
the sum Programmer : --Your name--
Roll no: --your roll number--
Date of compilation:
27Jan2021 Class : B.Tech I
Sem „CST'
*/

#include<stdio.h>
/ / Main function begins
int main()
{
int num1,num2,sum;
printf("Enter first value:");
scanf("%d",&num1);
printf("Enter Second
value:");
scanf("%d",&num2);
sum=num1+num2;
printf(“Addition of %d and %d is:
%d",num1,num2,sum); return 0;
}

Problem 2: Algorithm to find the area of
rectangle
 Input :
🞑 two numbers.
(length = 8
breadth = 4)
 Output:
🞑 Area of rect = 32
Algorith
m
 Step 1: Start
 Step 2: Read first number(length)
 Step 3: Read second number
(breadth)
 Step 4: Area  length * breadth
 Step 5: Print “Area of rect”
, Area
 Step 6: Stop
Area = length *
breadth

Ex3: Problem: Find greater of two
no‟s
 Algorithm:
🞑 Step 1: Start
🞑 Step 2: Declare variables A, B
🞑 Step 3: Accept two values from user and store in A
and B respectively.
🞑 Step 4: Check whether A > B; True: 4.1; False: 4.2
1. Yes  Print A is greater
2. No  Print B is greater
🞑 Step5: Stop

Ex4: Find Largest of three
numbers
Step 1: Start
Step 2: Declare variables a,b
and c. Step 3: Read variables
a,b and c.
Step 4: If a>b ; True: 4.1; False:
4.2
4.1 If a>c True: 4.1.1;
False: 4.1.2
1. Display a is
the largest
number.
else
2. Display c is
the largest
number.

Ex5: Calculate Gross, Net
Salary
 Problem:
 Acceptthe employee details such as empid and
basic salary. Consider DA with 120% of basic,
HRA with 10% of basic. Calculate the gross
salary and net salary considering tax
deduction of Rs. 2% of gross.

Example: Gross,
Net
 Empid: 1001
 Basic: 10000
 DA: 120% of Basic=1.20*10000= 12000
 HRA: 10% of Basic=0.1*10000 = 1000
 Gross=Basic+DA+HRA =
10000+12000+1000=23000
 Tax: 2% of Gross = 0.02*23000=460

Algorithm: Gross,
Net
Step 1: Start
Step 2: Declare variables
empid,basic,da,hra,tax,gross,net Step 3: Print "Enter
Employee ID"
Step 4: Accept empid
Step 5: Print "Enter basic
salary" Step 6: Accept basic
Step 7: Compute
da=(120/100)*basic Step 8:
Compute hra=(10/100)*basic Step
9: Compute gross=(basic+da+hra)
Step 10: Print "Gross Salary"
Step 11: Compute
tax=(2/100)*gross Step 12:

#include<stdio.h>
int main()
{
int empid;
float
basic,da,hra,tax,gross,net;
printf("Enter Employee id:");
scanf("%d",&empid);
printf("Enter Basic Salary:");
scanf("%f",&basic);
da=1.2*basic;
hra=0.1*basic;
gross=(basic+da+hra)
printf("Gross Salary=
%f",gross); tax=0.02*gross;
net=gross-tax;
printf("Net Salary=%f",net);
return
0;
}

Ex 6: Calculate Simple
Interest
 Si = (P * T * R) /
100

Flowchar
t
 A Flowchart is a type of diagram (graphical or symbolic)
that represents an algorithm or process.
 Each step in the process is represented by a different
symbol and contains a short description of the process
step.
 The flow chart symbols are linked together with arrows
showing the process flow direction. A flowchart typically
shows the flow of data in a process, detailing the
operations/steps in a pictorial format which is easier to

Flowchar
t
 Flowchart is the pictorial representation of an
algorithm

49
Identif
y
 What do each of the following symbols
represent?
Decision
Terminal
Input/Output
Operation
Process
Connector
Module

Example 1: Algorithm &
Flowchart
 Problem: Display the Sum, Average,
Product of three given numbers
Algorithm
 Step1: Start
 Step 2: Read X, Y, Z
 Step 3: Compute Sum (S)  X + Y + Z
 Step 4: Compute Average (A)  S / 3
 Step 5: Compute Product (P) as X x Y x Z
 Step 6: Print S, A, P
 Step 7: Stop

Example 2: Algorithm &
Flowchart
 Problem: Display the largest of two given
numbers
Algorithm
 Step1: Start
 Step 2: Read A, B
 Step 3: If A is less than B
🞑 True: Assign A to BIG and B to SMALL
🞑 False: Assign B to BIG and A to SMALL
 Step 6: Print S, A, P
 Step 7: Stop

Control
Structures
 Sequence: A series of steps or statements that are
executed in the order they are written in an algorithm.
 Selection: Defines two courses of action depending
on the outcome of a condition. A condition is an
expression that is, when computed, evaluated to
either true or false. (ex: if, if else, nested if)
 Repetition/Loop: Specifies a block of one or more
statements that are repeatedly executed until a
condition is satisfied. (Ex: for, while, do while)

Ex: Sequence Control
Structure
 A series of steps or statements that are executed
in the order they are written in an algorithm.
 The beginning and end of a block of statements
can be optionally marked with the keywords
begin and end.
begin
statement 1
statement 2
statement n
.....

Sequence Ex: Calculate Person‟s
age

Sequence
Exampleint main()
{
int first, second, temp;
printf("Enter first number: ");
scanf("%d", &first);
printf("Enter second number: ");
scanf("%d", &second);
temp = first;
first = second;
second = temp;
printf("After swapping, firstno. = %dn", first);
printf("After swapping, secondno. = %d", second);
}
Swap two
numbers
using
temp var

Sequence
Example
int main()
{
int a, b;
printf("Enter a and b: ");
scanf("%d %d", &a, &b);
a = a - b;
b = a + b;
a = b - a;
printf("After swapping, a = %dn", a);
printf("After swapping, b = %d", b);
return 0;
}
Swap two
numbers
without
using
temp var

Selection (Decision) Control
structure
is
conditio
n
Y N
Print
stmt
1
Print
stmt
2

Syntax: if
if
(condition)
stmt1
else
stmt2
C programming
if (a>b)
printf(“a is greater than b”);
else
printf(“b is greater than a”);
Selection (Decision) Control
structure

Selection Ex: Program to print the given
number is negative or positive using if
else
int main()
{
int num;
printf("Enter a number to check.n");
scanf("%d",&num);
if (num<0)
printf(“Given Number = %d is negativen",num);
else
printf(“Given Number = %d is positiven",num);
return 0;
}

Selection
Ex:
 Problem: Write an algorithm to determine a students
‟ final
grade and display pass or fail. The final grade is calculated
as the average of four subject marks.
Algorithm
 Step1: Start
 Step 2: Input M1,M2,M3,M4
 Step 3: GRADE  (M1+M2+M3+M4)/4
 Step 4: if (GRADE > 60) then Print “Pass” else “Fail”
 Step 5: Stop

STA
RT
Input
M1,M2,M3,M
4
GRADE(M1+M2+M3+M4)/4
IS
GRADE>6
0
STO
P
Y
N
Print
“Fail
”
Print
“Pass
”
Algorithm
Step1: Start
Step 2: Input M1,M2,M3,M4
Step 3: GRADE  (M1+M2+M3+M4)/4
Step 4: if (GRADE > 60) then Print “Pass” else
“Fail”
Step 5: endif
Step 6: Stop
Selection
Ex:

Selection Ex:
Program to check for odd or
even int main()
{
int num;
printf("Enter an integer: ");
scanf("%d", &num);
if (num % 2 == 0)
printf("%d is even.",
num);
else
printf("%d is odd.", num);
return 0;

Selection Ex: Program to check
odd or
even using ternary operator
? :
int main()
{
int num;
printf("Enter an integer: ");
scanf("%d", &num);
(num % 2 == 0) ? printf("%d is even.", num) : printf("%d is odd.",
num);
return 0;
}

Selection Ex: Problem: Find greater of two
no‟s
 Algorithm:
🞑 Step 1: Start
🞑 Step 2: Declare variables A, B
🞑 Step 3: Accept two values from user and store in A
and B respectively.
🞑 Step 4: Check whether A > B; True: 4.1; False: 4.2
1. Yes  Print A is greater
2. No  Print B is greater
🞑 Step5: Stop

Selection Ex: Find Largest of three
numbers
Step 1: Start
Step 2: Declare variables a,b
and c. Step 3: Read variables
a,b and c.
Step 4: If a>b ; True: 4.1; False:
4.2
4.1 If a>c True: 4.1.1;
False: 4.1.2
1. Display a is
the largest
number.
else
2. Display c is
the largest
number.

int main()
{
double n1, n2, n3;
printf("Enter three different numbers: ");
scanf("%lf %lf %lf", &n1, &n2, &n3);
if (n1 >= n2 && n1 >= n3)
printf("%.2f is the largest number.", n1);
if (n2 >= n1 && n2 >= n3)
if (n3 >= n1 && n3 >= n2)
return 0;
}
Selection:
if

int main()
{
double n1, n2, n3; printf("Enter
three numbers: ");
scanf("%lf %lf %lf", &n1, &n2, &n3);
if (n1 >= n2 && n1 >= n3)
printf("%.2lf is the largest number.", n1);
else if (n2 >= n1 && n2 >= n3)
else
return 0;
}
Selection: if else
if

int main()
{
double n1, n2, n3; printf("Enter
three numbers: ");
scanf("%lf %lf %lf", &n1, &n2,
&n3);
if (n1 >= n2)
{ if (n1 >= n3)
else
}
else
{
if (n2 >= n3)
else
}
return 0;
}
Selection: nested
if

Loop:
for
for(initialization, condition,
incrementation)
{
code
statements;
}
int main()
{
int i;
for (i=0; i<10; i++)
{
printf("i=%dn",i);
}
return 0;
}

Loop Example: Multiplication
table
int main()
{
int n, i;
printf("Enter no. to print multiplication table: ");
scanf("%d",&n);
for(i=1;i<=10;++i)
{
printf("%d * %d = %dn", n, i, n*i);
}
}

for(i = 0; i < 5; i++)
{
printf("tttt");
for(j = 0; j < 5; j++)
printf("* ");
printf("n");
}

ANSI C:
Syntax
and
construc
ts

C basic
elements
 Valid character
set
 Identifiers
 Keywords
 Basic data types
 Constants
 Variables

C Valid character
set
 uppercase English alphabets A to Z,
 lowercase letters a to z,
 digits 0 to 9,
 certain special characters as building blocks to
form basic program elements viz. constants,
variables, operators, expressions and
statements.

C
Identifiers
 Identifiers are names given to various items in the program,
such as variables, functions and arrays.
 An identifier consists of letters and digits, in any order, except
that the first character must be a letter.
 Both upper and lowercase letters are permitted.
 C is a case sensitive, the upper case and lower case
considered different, for example code, Code, CODE etc.
are different identifiers.

The underscore character _ can also be included.
 Keywords like if, else, int, float, etc., have special meaning and
they cannot be used as identifier names.

Valid and invalid C
identifiers
 ANSI standard recognizes 31 characters
 valid identifiers: A, ab123, velocity,
stud_name, circumference, Average,
TOTAL.
 Invalid identifiers:
🞑 1st
🞑 "Jamshedpur"
🞑 stud-name

C Data types and
sizes
Data type Description Size Range
char single character 1 byte 0 - 255
int integer number 4 bytes
-2147483648 to
2147483647
float
single precision floating
point number (number
containing fraction & or
an exponent)
4 bytes 3.4E-38 to 3.4E+38
double
double precision floating
point number
8 bytes 1.7E-308 to 1.7E+308

C Data types and
sizesData type Size Range
short int 2 bytes -32768 to 32767
long int 4 bytes
-2147483648 to
2147483647
unsigned short int 2 bytes 0 to 65535
unsigned int 4 bytes 0 to 4294967295
unsigned long int 4 bytes 0 to 4294967295
long double (extended
precision)
8 bytes 1.7E-308 to1.7E+308

C
Constants
 C can be classified into four categories namely
integer constants, floating point constants,
character constants and string constants.
 A character constant is written as for example -
'A'
 A normal integer constant is written as 1234.
 A long int uses L (uppercase or lowercase) at the
end of the constant, e.g. 2748723L

Escape sequence
characters
Character Escape
Sequence
ASCII Value
Bell a 007
Backspace b 008
Null 0 000
Newline n 010
Carriage return r 013
Vertical tab v 011
Horizontal tab t 009
Form feed f 012

Symbolic
constants
 #define PI 3.141593
 #define TRUE 1
 #define PROMPT "Enter Your
Name :"

Accept name from the
userMethod 1
char name[20];
printf("Enter your
name:");
scanf("%s",name);
printf("Your name is:
%s",name);
Method 2
char name[20]
printf(“Enter your
name:”)
gets(name);
printf(“Your name
is:”);

Accept name from the
userMethod
3 #define
MAX_LIMIT 20 int
main()
{
char
name[MAX_LIMIT];
printf("Enter your
Method 4
char name[20];
printf("Enter your
name:");
scanf("%[^n]
%*c",name);
printf("Your name is:

Operators in
C
 Arithmetic
operators
 Relational
operators
 Logical operators
 Bitwise operators
 Assignment

int a = 10, b = 20, c = 25, d =
25; printf(“ %d" (a + b) );
printf(“ %d" (a - b) );
printf(“%d “ (a * b) );
printf(“ %d” (b / a) );
printf(“ %d” (b % a) );
printf(“ %d” (c % a) );
printf
(“%d“
printf(“%d
“
(a+
+) );
(a--) );
(d+
OUTPU
T
30
-10
20
0
2
0
5
10
11
25

Example: Bitwise
Operators
int a =
60; int b =
13;
int c =
0; c = a &
b; c = a |
b;
c = a
^ b; c =
~a;
c = a
printf(“%d" + c
);
printf(“%d" +
c );
printf(“%d" + c
);
printf(“%d" +
c );
c = a >> 2; printf(“%d" +
c );
OUTPU
T
a= 60 = 0011
1100
b= 13 = 0000
1101
c = a & b; 0000 1100
= 12
c = a | b; 0011 1101
= 61
c = a ^ b; 0011 0001
= 49
c = ~a; 1100 0011
= -61

Misc
Operators
sizeof(
) &
:
:
:
:
Returns the size of the
variable Returns the address
of a variable Pointer
variable
Conditional / Ternary
operator
*
?:

Operator Precedence int a = 20, b = 10, c = 15, d = 5;
e = (a + b) * c / d;
/ / Print value of e
e = ((a + b) * c) /
d;
e = (a + b) * (c /
d);
e = a + (b * c) / d;
int
e;
Value of (a + b) * c / d is :
90 Value of ((a + b) * c) / d
is :
90 Value of (a + b) * (c / d)
is :
90 Value of a + (b * c) / d is
: 50
( 30 * 15 ) /
5
(30 * 15 ) /
5
(30) *
(15/5)
20 +
(150/5)
associativity of operators determines the direction in
which an expression is evaluated. Example, b = a;
associativity of the = operator is from right to left (RL).

Operator Description Associativity
()
[ ]
.

Parentheses (grouping)
Brackets (array subscript)
Member selection
Member selection via pointer
LR
++ --
+ -
!
~
(type
)
Unary
preincrement/predecrement
Unary plus/minus
Unary logical
negation/bitwise
complement
Unary cast (change type)
Dereferen
RL

Arithmetic Operators
Multiplication operator, Divide
by, Modulus
*, /, % LR
Add, Subtract +, – LR
Relational Operators
Less Than <
Greater than >
Less than equal to <=
Greater than equal to >= LR
Equal to ==
Not equal !=
Logical Operators
AND && LR
OR || LR
NOT ! RL
1 == 2 != 3
operators ==
and
!= have the
same
precedence, LR
Hence, 1 == 2
is executed
first

Hungarian
Notation
 Hungarian is a naming convention for identifiers. Each
identifier would have two parts to it, a type and a
qualifier.
 Each address stores one element of the memory
array. Each element is typically one byte.
 For example, suppose you have a 32-bit quantity
written as 12345678, which is hexadecimal.
 Since each hex digit is four bits, eight hex digits are needed
to represent the 32-bit value. The four bytes are: 12, 34, 56,

Endiannes
s
 The endianness of a particular computer system is
generally described by whatever convention or
set of conventions is followed by a particular
processor or combination of
processor/architecture and possibly operating
system or transmission medium for the
addressing of constants and the
representations of memory addresses.


Big Endian
storage
 Big-endian: Stores most significant byte in smallest
address.
 The following shows how 12345678 is stored in big
endian Big Endian Storage
Address Value
1000 12
1001 34
1002 56
1003 78

Little Endian
storage
 Little-endian: Stores least significant byte in
smallest address.

The following shows how 12345678 is stored
in big endian Little Endian
Storage
Address Value
1000 78
1001 56
1002 34
1003 12

 For example 4A3B2C1D at address 100, they
store the bytes within the address range 100
through 103 in the following order:m

Type
Casting
 Type casting is a way to convert a variable
from one data type to another data type.
🞑 Implicit Conversions
🞑 Explicit Conversions

Implicit
Conversions
int main()
{
int i = 17;
char c = 'c'; /* ascii value is 99
*/ int sum;
sum = i + c;
printf("Value of sum : %dn",
sum );

Explicit
Conversions
General format /
Syntax: (type_name)
expression
Example:
main()
{
int sum = 17, count =
5; double mean;
mean = (double) sum /
count;

Examples: Type
Conversions
float a =
5.25;
int b = (int)a;
char c =
‟A ;
‟ int x =
(int)c;
int x=7,
y=5 ; float
z; z=x/y;
int x=7,
y=5; float
z;
z =
(float)x/(flo

break
Statement
The break statement in C programming language has the
following two usages:
When the break statement is encountered inside a loop, the
loop is immediately terminated and program control
resumes at the next statement following the loop.
It can be used to terminate a case in the switch statement
If you are using nested loops, the break statement will stop
the execution of the innermost loop and start executing the
next line of code after the block.
Syntax:
break;

int a =
10;
while( a < 20 )
{
printf("value of a: %dn",
a);
a++;
if( a > 15)
{
break;
}
}
OUTPUT
value of a:
10 value of
a: 11 value
of a: 12
value of a:
13 value of
a: 14 value
of a: 15

int num =0;
while(num<=10
0)
{
printf("value of variable num is: %d n",
num); if (num==2)
{ break
;
}
num++;
}
printf("Out of while-
loop");
Outpu
t:
value of variable num
is: 0 value of variable
num is: 1 value of
variable num is: 2 Out
of while-loop

int var;
for (var =100; var>=10; var
--)
{
printf("var: %dn",
var); if (var==99)
{
break;
}
}
printf("Out of for-loop");
Outpu
t:
var: 100
var: 99
Out of for-
loop

int a = 4;
while( a <
10 )
{
printf("valu
e of a:
%dn", a);
++a;
if( a > 8)
{
break;
printf("B
reak
n");

continu
e The continue statement in C
programming language works
somewhat like the break statement.
 Instead of forcing termination, continue
statement forces the next iteration of the
loop to take place, skipping any code in
between.

int a =
10;
do
{
if( a == 15)
{
a = a + 1;
continue;
}
printf("value of a: %dn",
a);
a++;
} while( a < 20 );
value of a:
10 value of
a: 11 value
of a: 12
value of a:
13 value of
a: 14 value
of a: 16
value of a:
17 value of
a: 18 value

for (int j=0; j<=8; j+
+)
{
if (j==4)
{
continu
e;
}
printf("%d ",
j);
}
Outpu
t:
0 1 2 3 5 6 7
8

int counter=10;
while (counter >=0)
{
if (counter==7)
{
counter-
-;
continue
;
}
printf("%d ",
counter); counter--;
}
Outpu
t:
10 9 8 6 5 4 3 2 1
0

goto
statement
 The goto statement is used to alter the normal sequence of
program execution by transferring control to some other
part of the program unconditionally.
 Syntax: goto label;

Control may be transferred to anywhere within
the current function.
 The target statement must be labeled, and a colon must
follow the label.
 label : statement;

int w;
printf("Enter the input 1 or 0:
n"); scanf("%d", &w);
if (w == 1)
goto CST;
if (w == 0) printf("Value entered
is 0 n"); return 0;
CST : printf("You belong to CST
Section n"); goto end;
end : printf("Have a nice Dayn");
return 0;
}

Example: Nested
Loop
int i, j;
for(i=2; i<10; i++)
{
for(j=2; j <= (i/j); j++)
if(!(i%j)) break;
if(j > (i/j)) printf("%d is
primen", i);
}

Program to check alphabet, digit or special character
Alphabet: a to z (or)
A to Z Digit : 0 to 9
else it is special character
if((ch >= 'a' && ch <= 'z') | | (ch >= 'A' && ch <= 'Z'))
printf("'%c' is alphabet.", ch);
else if(ch >= '0' && ch <= '9')
printf("'%c' is digit.", ch);
else
printf("'%c' is special

Program to check vowel of
consonant
Vowel : a (or) e (or) i (or) o (or)
u Consonant : a to z or A to Z
else: not an alphabet
if(ch=='a' | | ch=='e' | | ch=='i' | | ch=='o' | |
ch=='u' | | ch=='A' | | ch=='E' | | ch=='I' | |
ch=='O' | | ch=='U')
{
printf("'%c' is Vowel.", ch);
}
else if((ch >= 'a' && ch <= 'z') | | (ch >= 'A' && ch <=
'Z'))

Sum of digits of given number using
while
sum = 0;
/ / Accept number and store
in n num = n;
while( n > 0 )
{
rem = n % 10;
sum +=
rem; n /=
10;
}
printf("Sum of digits of %d is %d", num,
OUTPUT
Enter a number: 456
Sum of digits of 456 is
15

Print all ODD numbers from 1 to N using while
loop.
number=1;
while(number<=
n)
{
if(number%2 != 0)
printf("%d
",number);

Print its multiplication
table
i=1;
while(i<=10)
{
printf("%dn",
(num*i)); i++;
}

count total digits in a given integer using
loop
do
{
count++;
num /=
10;
} while(num
!= 0);
printf("Total

Program to print numbers between 1 and
100 which are multiple of 3 using the do while
loop
int i =
1; do
{
if(i %
3 ==
0)
{
prin

find sum of odd numbers from 1
to n
for(i=1; i<=n; i+=2)
{
sum += i;
}
printf("Sum of odd numbers = %d",
sum);

Exponential
series
int i, n;
float x, sum=1, t=1;
// Accept x
// Accept n
for(i=1;i<=n;i++)
{
t=t*x/i;
sum=sum+t;
}
printf(“Exponentia
l Value of %f =

Program to print its multiplication
table
i=1;
while(i<=10)
{
printf("%d n",
(num*i)); i++;
}
i=1;
do
{
printf("%dn",
(num*i)); i++;
}while(i<=10);
for(i=1;i<=10;i++)
{
printf("%d n",
(num*i));

k = 1;
for(i=1; i<=rows; i++)
{
for( j=1; j<=cols; j++, k++)
{
printf("%-3d", k);
}
printf("n");
}
Print number pattern as
shown

int main()
{
int i;
for (i=0; i<10; i+
+)
{
printf("i=%dn",i);
}
return 0;

Factorial using for
loop
fact=1;
for(i=num; i>=1; i--)
fact=fact*i;
printf("Factorial of %d is = %ld",num,fact);

S
u
m of n natural no‟s
using for
sum = 0;
for(i = 1; i <= n; i+
+) sum += i;
printf("Sum is: %d
n", sum);

Program to print its multiplication
table
i=1;
while(i<=10)
{
i++;
}
i=1;
do
{
printf("%dn",(num*i)); printf("%dn",(num*i));
i++;
}while(i<=10);
for(i=1;i<=10;i++)
{
printf("%dn",(num*i));
}

Ex: 2: Nested for
loop
for (int i=0; i<2; i++)
{
for (int j=0; j<4; j++)
{
printf("%d, %d
n",i ,j);
}

Ex: 3: Nested for
loop
k=1;
for (i=0; i<2; i++)
{
for (j=0; j<4; j++,k++)
{
printf("%d, %d, %dn",i ,j, k);
}
}

Print number pattern as
shown Output
:
1
12
123
1234
for(i=1; i<=rows; i++)
{
for( j=1; j<=i; j++)
printf("%d", j);
printf("n");
}

Structured vs Unstructured Programming
Structured Programming is a
programming paradigm which
divides the code into modules or
function.
Unstructured Programming
is the paradigm in which
the code is considered as
one single block.
Readabilit
y
Structured Programming
based
programs are easy to
read.
Unstructured Programming
based
programs are hard to read.
Purpos
e
Structured Programming is to
make the code more efficient
and easier to understand.
Unstructured programming is
just to program to solve the
problem. It does not create a
logical structure.
Complexit
y
Structured Programming is
easier because of modules.
Unstructured programming is
harder when comparing with
the structured programming

Application
Structured programming can be
used for small and medium scale
projects.
Unstructured programming is not
applicable for medium and complex
projects.
Modification
It is easy to do changes in
Structured Programming.
It is hard to do modifications in
Unstructured Programming.
Data Types
Structured programming uses
many data types.
Unstructured programming has a
limited number of data types.
Code Duplication
Structured programming
avoids code duplication.
Unstructured programming can
have code duplication.
Testing and Debug
It is easy to do testing and It is hard to do testing and

Palindrome
number
rev=0
origin=n;
while (n != 0)
{
rem = n
% 10;
rev = rev * 10 + rem;
n /= 10;
}
if (orig == rev) printf("%d is a palindrome.“,orig);

int i, j;
for(i = 2; i<100; i++)
{
for(j = 2; j <= (i/j); j++)
if(!(i%j)) break; // if factor found, not prime
if(j > (i/j)) printf("%d is primen", i);
}
Print Prime numbers upto
100

Fibonacci
series
for(i=3;i<=n;i++)
{
trm=prv+pre;
printf("% 5d",trm);
prv=pre;
pre=trm;
}
printf("n");

Palindrome
number
rev=0
origin=n;
while (n != 0)
{
rem = n
% 10;
rev =
rev * 10
+ rem;
n /= 10;
}
if (origin == rev) printf("%d is a palindrome.“,orig);
else printf("%d is not a palindrome.", orig);

goto statement: Unstructured
programming
 The goto statement is used to alter the normal
sequence of program execution by transferring
control to some other part of the program
unconditionally.
 Syntax: goto label;
Control may be transferred to anywhere
within the current function.
 The target statement must be labeled, and a
colon must follow the label.

UNIT 1.pptx Programming for Problem Solving

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UNIT 1.pptx Programming for Problem Solving