Lecture 3: If-Else, While loop & Lots of
Patterns (Part-1)
In this 3rd lecture, we will learn the syntax of conditional statements in
CPP. We will make a simple program , where we will ask computer to
print A if A is bigger than B , else print B. Coding Ninjas has supported
us in providing this free DSA course. If you are given a number A , and
you are told that print +ve if A is positive, otherwise, print not +ve. If
a block is not executed, then 'else ' block will be executed. If A is 5 ,
then 'if A is -2 and if a is -5, then it will print anything. If I enter -5 ,
then it did n't print anything because if block was not executed this
time. But if I add a 'else' block here, then this block is executed. Cin.
get ( ) takes input in the form of characters. It took ' 1 ' as a character
but we are assigning this value to an integer so that 's why it printed
ASCII value of that character. Cin do n't read SPACE , TAB , ENTER and
TAB. But we have an another method to take input. Let 's code that
tells us which number is greater.
If ( a > 0) { print +ve } else { if ( a < 0 ) { print negative } , otherwise
print 0 } } Let 's run it again. Entered A as 15 It printed A is positive
This time I entered -1. A is negative. Correct output. We solved it using
loop using loop to understand loops clearly. You just have to enter a
condition and the statements which you want to execute. You can
check your answer by applying this formula Bhaiya. Here is your
homework question. Q1- Find sum of all even numbers from 1 to N. We
have to tell whether n is prime or not prime number is a number which
has only 1 and N as it 's factor We will use this % operator for this
question. % tells us remainder when we divide 2 numbers. If any of
these numbers gave modules as 0, then N is not prime. If no such
number exists, then it is a prime number.
We will run a loop from 1 to n , ( for all rows ) We declared a variable i
to keep the track of the row number and then we are simply printing
star in each column Then we wrote this cout < endl , to come to a new
line , and after that this code will get executed for every row. We will
do many patterns problems in the next video. We will code this
triangular figure also in next video. But first let 's understand this
code. We will run a loop from 1 to n , ( for all rows ) Then we declared
a variable J for columns. J will also run from 1- N and we simply have to
print the row number. Then we incremented J inside the column loop
and Incrementing i inside the row loop after printing ENTER.
Patterns (Part-1)
In this 3rd lecture, we will learn the syntax of conditional statements in
CPP. We will make a simple program , where we will ask computer to
print A if A is bigger than B , else print B. Coding Ninjas has supported
us in providing this free DSA course. If you are given a number A , and
you are told that print +ve if A is positive, otherwise, print not +ve. If
a block is not executed, then 'else ' block will be executed. If A is 5 ,
then 'if A is -2 and if a is -5, then it will print anything. If I enter -5 ,
then it did n't print anything because if block was not executed this
time. But if I add a 'else' block here, then this block is executed. Cin.
get ( ) takes input in the form of characters. It took ' 1 ' as a character
but we are assigning this value to an integer so that 's why it printed
ASCII value of that character. Cin do n't read SPACE , TAB , ENTER and
TAB. But we have an another method to take input. Let 's code that
tells us which number is greater.
If ( a > 0) { print +ve } else { if ( a < 0 ) { print negative } , otherwise
print 0 } } Let 's run it again. Entered A as 15 It printed A is positive
This time I entered -1. A is negative. Correct output. We solved it using
loop using loop to understand loops clearly. You just have to enter a
condition and the statements which you want to execute. You can
check your answer by applying this formula Bhaiya. Here is your
homework question. Q1- Find sum of all even numbers from 1 to N. We
have to tell whether n is prime or not prime number is a number which
has only 1 and N as it 's factor We will use this % operator for this
question. % tells us remainder when we divide 2 numbers. If any of
these numbers gave modules as 0, then N is not prime. If no such
number exists, then it is a prime number.
We will run a loop from 1 to n , ( for all rows ) We declared a variable i
to keep the track of the row number and then we are simply printing
star in each column Then we wrote this cout < endl , to come to a new
line , and after that this code will get executed for every row. We will
do many patterns problems in the next video. We will code this
triangular figure also in next video. But first let 's understand this
code. We will run a loop from 1 to n , ( for all rows ) Then we declared
a variable J for columns. J will also run from 1- N and we simply have to
print the row number. Then we incremented J inside the column loop
and Incrementing i inside the row loop after printing ENTER.
