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Program to find century for a year

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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)



Last Updated : 17 Feb, 2023
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