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Here we will see how to check a number is divisible by 8 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 8, if the number formed by last three digits are divisible by 8.Example Live Demo#include using namespace std; bool isDiv8(string num){ int n = num.length(); int last_three_digit_val = (num[n-3] - '0') * 100 + (num[n-2] - '0') * 10 + ((num[n-1] - '0')); if(last_three_digit_val % 8 == 0) return true; return false; } int main() { string num = "1754586672360"; if(isDiv8(num)){ cout
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
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
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
To replace & with an ampersand, use MySQL REPLACE(). Let us first create a table −mysql> create table DemoTable( Id int NOT NULL AUTO_INCREMENT PRIMARY KEY, Value varchar(100) ); Query OK, 0 rows affected (1.06 sec)Insert some records in the table using insert command −mysql> insert into DemoTable(Value) values('@amp'); Query OK, 1 row affected (0.33 sec) mysql> insert into DemoTable(Value) values('&'); Query OK, 1 row affected (1.09 sec) mysql> insert into DemoTable(Value) values('#amp'); Query OK, 1 row affected (0.28 sec)Display all records from the table using select statement −mysql> select *from DemoTable;This will produce the following output −+----+-------+ ... Read More