Week 1: Basic syntax, variables and data types
Today's topics:
- Input Output
- Data Types
- If Else
- If Else Ladder
- Nested If Else
Problem 1:
You need to print the lines given below in the same format as it is - “Recently I heard that you’ve achieved 95% marks in your exam. This is brilliant! I wish you’ll shine in your life! Good luck with all the barriers(/\) in your life.”
Solution:
#include< stdio.h>
int main()
{
printf("Recently I heard that you’ve achieved 95%% marks
in your exam.\nThis is brilliant!\nI wish you’ll shine in
your life!\n Good luck with all the barriers(/\\) in your life.");
return 0;
}
Probelm: 2
You need to take two integer values as input and show the summation, subtraction, multiplication and division in the given format below.
For example:
Inputs are 5 and 2
Then you’ll give output as:
5 + 2 = 7
5 - 2 = 3
5 * 2 = 10
5 / 2 = 2.50
Sample Input Sample Output
5 2 5 + 2 = 7
5 - 2 = 3
5 * 2 = 10
5 / 2 = 2.50
10 3 10 + 3 = 13
10 - 3 = 7
10 * 3 = 30
10 / 3 = 3.33
Solution:
#include< stdio.h>
int main()
{
int a,b,sum,sub,mul;
scanf("%d",&a);
scanf("%d",&b);
sum = a+b;
sub = a-b;
mul = a*b;
float div = a*1.0/b;
printf("%d %d %d %.2f", sum,sub,mul,div);
return 0;
}
Probelm: 3
You need to take one non-negative integer (0 or greater than 0) value as input and tell if it is even or odd. See the sample input and output for more clarification.
Sample Input Sample Output 10 even 3 odd
Solution:
#include< stdio.h>
int main()
{
int a;
scanf("%d",&a);
if(a>=0)
{
if(a%2==0)
{
printf("even");
}
else
{
{
printf("odd");
}
}
}
else
{
printf("Please enter a non-negative number!\nThank you.");
}
return 0;
}
Probelm: 4
You need to take one integer value as input and tell if the value is positive or negative or zero. See the sample input and output for more clarification.
Sample Input Sample Output 10 positive -50 negative 0 zero
Solution:
#include< stdio.h>
int main()
{
int a;
scanf("%d",&a);
if(a>0)
{
printf("positive");
}
else if(a<0)
{
printf("negative");
}
else
{
printf("zero");
}
return 0;
}
Probelm: 5
Suppose your friend has told you that she will buy a Gucci Bag if she has 10 thousand taka or more. Otherwise if she has 5 thousand taka or more, she will buy a Levis Bag. Otherwise she will buy Something from New Market. She also told you that, if she could buy the Gucci bag and she has more than 20 thousand taka she will also buy a Gucci Belt. Now, if you know the amount of money she has, can you tell which item/items she will buy? See the sample input and output for more clarification.
Input Output
20000 Gucci Bag
6500 Levis Bag
200 Something
25400 Gucci Bag
Gucci Belt
Solution:
#include< stdio.h>
int main()
{
int a;
scanf("%d",&a);
if(a>0)
{
printf("positive");
}
else if(a<0)
{
printf("negative");
}
else
{
printf("zero");
}
return 0;
}
Github link here
Click here to go the practice contest.
Problem: 1
Problem Statement I know that you're loving practice days! So this task is for you. You need to print "I Love Practice" without the quotation marks. I know that you can do it! Input Format There is no input in this problem Output Format Output "I Love Practice" Sample Output 0 I Love Practice
Solution:
#include< stdio.h>
int main()
{
printf("I Love Practice");
return 0;
}
Problem: 2
Problem Statement Take two integers A and B as input and output their summation. Input Format You will be given A and B separated by a space. Constraints -10^9 <= A,B <= 10^9 Output Format Output their summation Sample Input 0 2 3 Sample Output 0 5 Sample Input 1 -10 5 Sample Output 1 -5
Solution:
#include< stdio.h>
int main()
{
int a,b;
scanf("%d %d",&a,&b);
int ans;
ans = a+b;
printf("%d", ans);
return 0;
}
Problem: 3
Problem Statement I know and you also know that you love practice day so much. So this task is for you. You will be given a positive integer N, you need to print "I Love Practice" N times. Here positive integer means those integers that are greater than 0. Input Format You will be given a positive integer N. Constraints 1 <= N <= 1000 Output Format Output "I Love Practice" N times. Don't forget to put a new line after every line. Sample Input 0 5 Sample Output 0 I Love Practice I Love Practice I Love Practice I Love Practice I Love Practice Sample Input 1 2 Sample Output 1 I Love Practice I Love Practice
Solution:
#include< stdio.h>
int main()
{
int n;
scanf("%d", &n);
for (int i = 0; i
Problem: 4
Problem Statement You've learned about variables, right? Now its time to practice them. You need to take an integer A, a very big integer B, a floating value C and a character D as input and output them serially. Input Format First line will contain A Second line will contain B Third line will contain C Fourth line will contain D Constraints -10^9 <= A <= 10^9 -10^18 <= B <= 10^18 -10^9 <= C <= 10^9 Output Format Output them serially and put a new line after each value. Output the floating value 2 points after decimal. Sample Input 0 100 1234567891234567 23.5675 A Sample Output 0 100 1234567891234567 23.57 A
Solution:
#include< stdio.h>
int main()
{
int n;
long long b;
float c;
char d;
scanf("%d %lld %f %c", &n, &b, &c, &d);
printf("%d\n", n);
printf("%lld\n", b);
printf("%.2f\n", c);
printf("%c\n", d);
return 0;
}
Problem: 5
Problem Statement You will be given a positive integer N, you need to print from 1 to N and besides the value, print Yes or No. Print Yes if the value is divisible by 5 and print No otherwise. Input Format Input will contain a positive integer N. Constraints 1 <= N <= 1000 Output Format Output as mentioned in the question. See the sample input output for more clarifications. Put a new line after every line. Sample Input 0 10 Sample Output 0 1 No 2 No 3 No 4 No 5 Yes 6 No 7 No 8 No 9 No 10 Yes Sample Input 1 5 Sample Output 1 1 No 2 No 3 No 4 No 5 Yes
Solution:
#include< stdio.h>
int main()
{
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
if(i%5==0)
{
printf("%d Yes\n", i);
}
else
{
printf("%d No\n",i);
}
}
return 0;
}
Github link here
Click to the below link and complete the assignment. Don't see the solution before!
Click here to go the assignment
Problem: 1
Problem Statement Welcome to the "Panta Vat" assignment. In this task you just need to print the following lines as it is. Hello, world! I am learning C programming language. ^_^ Programming is fun and challenging. /\/\/\ I want to give my 100% dedication to learn! I will succeed one day. Note: Here you will see 4 spaces in the last line which is a tab, you need to print a tab there. Input Format There is no input Output Format Output the lines. Sample Output 0 Hello, world! I am learning C programming language. ^_^ Programming is fun and challenging. /\/\/\ I want to give my 100% dedication to learn! I will succeed one day.
Solution:
#include< stdio.h>
int main()
{
printf("Hello, world! I am learning C programming
language. ^_^\nProgramming is fun and
challenging. /\\/\\/\\\nI want to give
my 100%% dedication to learn!\tI will
succeed one day.");
return 0;
}
Problem: 2
Problem Statement You will be given two integers A and B. You need to give output their multiplication. Input Format Input will contain A and B Constraints -10^9 <= A,B <= 10^9 Output Format Output the multiplication Sample Input 0 10 50 Sample Output 0 500 Sample Input 1 10000000 10000000 Sample Output 1 100000000000000 Sample Input 2 -100 62 Sample Output 2 -6200
Solution:
#include< stdio.h>
int main()
{
long long a,b;
scanf("%lld %lld",&a,&b);
long long sum = a*b;
printf("%lld", sum);
return 0;
}
Problem: 3
Problem Statement You will be given a non-negative integer N, you need to tell if this number is divisible by 3 or not. If it is divisible by 3 output "YES" otherwise output "NO" without the quotation mark. Input Format Input will contain N. Constraints 0 <= N <= 10^9 Output Format Output "YES" or "NO" without the quotation mark according to the question. Sample Input 0 33 Sample Output 0 YES Sample Input 1 29 Sample Output 1 NO Sample Input 2 0 Sample Output 2 YES
Solution:
#include< stdio.h>
int main()
{
int a;
scanf("%d",&a);
if(a%3==0)
{
printf("YES");
}
else
{
printf("NO");
}
return 0;
}
Problem: 4
Problem Statement
You will be given a non-negative integer
N, you need to print all numbers from 1
to N that are divisible by both 3 and 7.
Input Format
Input will contain N.
Constraints
21 <= N <= 10000
Output Format
Output all numbers from 1 to N that are
divisible by both 3 and 7. Don't forget
to print a new line after every number.
Sample Input 0
30
Sample Output 0
21
Sample Input 1
50
Sample Output 1
21
42
Sample Input 2
100
Sample Output 2
21
42
63
84
Solution:
#include< stdio.h>
int main()
{
int a;
scanf("%d",&a);
for (int i = 1; i <= a; i++)
{
if(i%3==0 && i%7==0)
{
printf("%d\n",i);
}
}
return 0;
}
Problem: 5
Problem Statement Alisa and you have gone out for shopping, and Alisa wants to buy a new pair of shoes for Eid. She has enough money to buy anything. However, Alisa will only buy shoes if you also buy a pair. And you will buy a pair of shoes if you can buy a Punjabi. That means, everything is depending on the Punjabi. You have decided that you will buy a Punjabi only if you have more than 1000 Taka. After purchasing the Punjabi the amount of your money will be reduced by 1000. Suppose you have 1600 taka with you, after buying the Punjabi you will have 600 taka left with you. Then you will only buy shoes if you have 500 Taka or more left with you. That means, if you can't buy your Punjabi you can't buy shoes. Now if I inform you the amount N Taka that your mother will give you, can you tell me what will happen next? If you buy a punjabi print "I will buy Punjabi". If you buy a pair of shoes print "I will buy new shoes" If Alisa buy a pair of shoes print "Alisa will buy new shoes" If no one can buy anything print "Bad luck!" Note: Don't forget to print new line after every line you print. Input Format Input will contain a non-negative integer N. Constraints 1 <= N <= 2^31 Output Format Output the events that will happen as asked in the question. Sample Input 0 1000 Sample Output 0 Bad luck! Sample Input 1 1450 Sample Output 1 I will buy Punjabi Sample Input 2 1500 Sample Output 2 I will buy Punjabi I will buy new shoes Alisa will buy new shoes
Solution:
#include< stdio.h>
int main()
{
int a;
scanf("%d",&a);
if(a<1001)
{
printf("Bad luck!");
}
else if(a>=1500)
{
printf("I will buy Punjabi\nI will buy new shoes\nAlisa will buy new shoes\n");
}
else
{
printf("I will buy Punjabi");
}
return 0;
}