Given a year the task is that we will find the century in the given year. The first century starts from 1 to 100 and the second-century start from 101 to 200 and so on.
Examples:
Input : year = 1970 Output : 20 century Input : year = 1800 Output : 18 century
Below is the Implementation:
CPP
// C++ code for find the century // in a given year #include <bits/stdc++.h> using namespace std;
void find_century( int year)
{ // No negative value is allow for year
if (year <= 0)
cout << "0 and negative is not allow"
<< "for a year" ;
// If year is between 1 to 100 it
// will come in 1st century
else if (year <= 100)
cout << "1st century\n" ;
else if (year % 100 == 0)
cout << year/ 100 << " century" ;
else
cout << year/ 100 + 1 << " century" ;
} // Driven code int main()
{ int year = 2001;
find_century(year);
return 0;
} |
Java
// Java code for find the century // in a given year class GFG {
static void find_century( int year) {
// No negative value is allow for year
if (year <= 0 )
System.out.print( "0 and negative is not allow"
+ "for a year" );
// If year is between 1 to 100 it
// will come in 1st century
else if (year <= 100 )
System.out.print( "1st century\n" );
else if (year % 100 == 0 )
System.out.print(year / 100 + " century" );
else
System.out.print(year / 100 + 1 + " century" );
} // Driver code public static void main(String[] args) {
int year = 2001 ;
find_century(year);
} } // This code is contributed by Anant Agarwal. |
Python3
# Python3 code for find the century # in a given year def find_century(year):
# No negative value is allow for year
if (year < = 0 ):
print ( "0 and negative is not allow for a year" )
# If year is between 1 to 100 it
# will come in 1st century
elif (year < = 100 ):
print ( "1st century" )
elif (year % 100 = = 0 ):
print (year / / 100 , "century" )
else :
print (year / / 100 + 1 , "century" )
# Driver code year = 2001
find_century(year) # This code is contributed by shubhamsingh10 |
C#
// C# code for find the century in a given year using System;
class GFG {
static void find_century( int year) {
// No negative value is allow for year
if (year <= 0)
Console.WriteLine( "0 and negative is not"
+ " allow for a year" );
// If year is between 1 to 100 it
// will come in 1st century
else if (year <= 100)
Console.WriteLine( "1st century\n" );
else if (year % 100 == 0)
Console.WriteLine(year / 100 + " century" );
else
Console.WriteLine(year / 100 + 1 +
" century" );
}
// Driver code
public static void Main() {
int year = 2001;
find_century(year);
}
} // This code is contributed by vt_m. |
PHP
<?php // PHP code for find the century // in a given year function find_century( $year )
{ // No negative value is
// allow for year
if ( $year <= 0)
echo "0 and negative is not allow"
, "for a year" ;
// If year is between 1 to 100 it
// will come in 1st century
else if ( $year <= 100)
echo "1st century\n" ;
else if ( $year % 100 == 0)
echo $year / 100 , " century" ;
else
echo floor ( $year / 100) + 1
, " century" ;
} // Driver Code
$year = 2001;
find_century( $year );
// This code is contributed by anuj_67. ?> |
Javascript
<script> // Java Script code for find the century // in a given year function find_century( year) {
// No negative value is allow for year
if (year <= 0)
document.write( "0 and negative is not allow"
+ "for a year" );
// If year is between 1 to 100 it
// will come in 1st century
else if (year <= 100)
document.write( "1st century\n" );
else if (year % 100 == 0)
document.write(parseInt(year / 100) + " century" );
else
document.write(parseInt(year / 100) + 1 + " century" );
} // Driver code let year = 2001;
find_century(year);
//contributed by sravan kumar </script> |
Output
21 century
Time Complexity: O(1)
Auxiliary Space: O(1)