Program to find century for a year Last Updated : 17 Feb, 2023 Comments Improve Suggest changes Like Article Like Report 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> Output21 century Time Complexity: O(1)Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Analysis of Algorithms S SHARIQ_JMI Follow Improve Article Tags : Misc Mathematical DSA Basic Coding Problems date-time-program +1 More Practice Tags : MathematicalMisc Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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