- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Here we will see how to check a number is divisible by 75 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 75, when the number is divisible by 3 and also divisible by 25. if the sum of digits is divisible by 3, then the number is divisible by 3, and if last two digits are divisible by 25, then the number is divisible by 25.Example Live Demo#include using namespace std; bool isDiv75(string num){ int n = num.length(); long sum = accumulate(begin(num), end(num), ... Read More
Here we will see how to check a number is divisible by 5 or not. In this case the number is very large number. So we put the number as string.To check whether a number is divisible by 5, So to check divisibility by 5, we have to see the last number is 0 or 5.Example Live Demo#include using namespace std; bool isDiv5(string num){ int n = num.length(); if(num[n - 1] != '5' && num[n - 1] != '0') return false; return true; } int main() { string num = "154484585745184258458158245285265"; if(isDiv5(num)){ cout
In this article, we will learn about the solution and approach to solve the given problem statement.Problem statementGiven a range, we need to print all the odd numbers in the given range.The brute-force approach is discussed below −Here we apply a range-based for loop which provides all the integers available in the input interval.After this, a check condition for odd numbers is applied to filter all the even numbers.This approach takes O(n) + constant time of comparison.Now let’s see the implementation below −Examplestart, end = 10, 29 # iteration for num in range(start, end + 1): # check ... Read More
The assert statement has the following syntax.assert , The line above is read as: If evaluates to False, an exception is raised and will be output.If we want to test some code block or an expression we put it after an assert keyword. If the test passes or the expression evaluates to true nothing happens. But if the test fails or the expression evaluates to false, an AssertionError is raised and the message is printed out or evaluated.Assert statement is used for catching/testing user-defined constraints. It is used for debugging code and is inserted at the start of ... Read More
Here we will see how to check a number is divisible by 3 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 3, if the sum of digits is divisible by 3.Example Live Demo#include using namespace std; bool isDiv3(string num){ int n = num.length(); long sum = accumulate(begin(num), end(num), 0) - '0' * n; if(sum % 3 == 0) return true; return false; } int main() { string num = "3635883959606670431112222"; if(isDiv3(num)){ cout
In this article, we will learn about the solution and approach to solve the given problem statement.Problem statementGiven a range, we need to print all the even numbers in the given range.The brute-force approach is discussed below −Here we apply a range-based for loop which provides all the integers available in the input interval.After this, a check condition for even numbers is applied to filter all the odd numbers.This approach takes O(n) + constant time of comparison.Now let’s see the implementation below −Examplestart, end = 10, 29 # iteration for num in range(start, end + 1): # check ... Read More
Let’s say we have a database “web” and we need to get all the tables having a specific column ’StudentFirstName’.For this, below is the query −mysql> select myColumnName.table_name from information_schema.columns myColumnName where myColumnName.column_name = 'StudentFirstName' and table_schema='web';This will produce the following output −+---------------+ | TABLE_NAME | +---------------+ | demotable109 | | demotable297 | | demotable335 | | demotable395 | | demotable418 | | demotable425 | | demotable436 | +---------------+ 7 rows in set (0.14 sec)Therefore, the above tables have one of the column names as “StudentFirstName”.Let us check the description ... Read More
For the above module, we need to prepare following setup.py script −from distutils.core import setup, Extension setup(name='helloworld', version='1.0', \ ext_modules=[Extension('helloworld', ['hello.c'])])Now, we use the following command,$ python setup.py installOnce we install the extension, we would be able to import and call that extension in our Python script test.py and catch the exception in it as follows −#test.py import helloworld try: print helloworld.helloworld() except Exception as e: print str(e)This would produce the following result −bad format char passed to Py_BuildValue
To convert MySQL timestamp to UNIX Timestamp, use the UNIX_TIMESTAMP(). Following is the syntax −select unix_timestamp(yourColumnName) from yourTableName;Let us first create a table −mysql> create table DemoTable( Duetimestamp timestamp ); Query OK, 0 rows affected (2.66 sec)Insert some records in the table using insert command −mysql> insert into DemoTable values(now()); Query OK, 1 row affected (1.53 sec) mysql> insert into DemoTable values('2016-01-21 12:34:00'); Query OK, 1 row affected (0.73 sec) mysql> insert into DemoTable values('2018-05-01 02:00:00'); Query OK, 1 row affected (0.42 sec) mysql> insert into DemoTable values('2017-03-02 01:10:20'); Query OK, 1 row affected (10.19 sec)Display all records from ... Read More
Here we will see how to check a number is divisible by 25 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 25, when the last two digits are 00, or they are divisible by 25.Example Live Demo#include using namespace std; bool isDiv25(string num){ int n = num.length(); int last_two_digit_val = (num[n-2] - '0') * 10 + ((num[n-1] - '0')); if(last_two_digit_val % 25 == 0) return true; return false; } int main() { string num = "451851549333150"; if(isDiv25(num)){ cout