给定年份范围内的Le年计数

给定为L和R的两年，任务是找出(L，R)范围内(包括L和R)的of年总数。

例子：

Input: L = 1, R = 400
Output: 97

Input: L = 400, R = 2000
Output: 389

天真的方法：这个想法是通过L到R进行迭代，并使用该方法检查年份是否为a年。
时间复杂度： O(N)

高效方法：

1. 如果满足以下条件，则一年是a年：
   • 年是400的倍数。
   • Year是4的倍数，而不是100的倍数。
2. 因此计算范围(1，L)和(1，R)中满足上述条件的数字为

   Number of Leap years in (1, year) = (year / 4) – (year / 100) + (year / 400)

3. 范围(1，R)的Le年数与范围(1，L)的leap年数之差将是所需的输出。

下面是上述方法的实现。

C++

// C++ implementation to find the 
// count of leap years in given
// range of the year 
  
#include 
  
using namespace std;
  
// Function to calculate the number
// of leap years in range of (1, year)
int calNum(int year)
{
    return (year / 4) - (year / 100) +
                        (year / 400);
}
  
// Function to calculate the number
// of leap years in given range
void leapNum(int l, int r)
{
    l--;
    int num1 = calNum(r);
    int num2 = calNum(l);
    cout << num1 - num2 << endl;
}
  
// Driver Code
int main()
{
    int l1 = 1, r1 = 400;
    leapNum(l1, r1);
  
    int l2 = 400, r2 = 2000;
    leapNum(l2, r2);
  
    return 0;
}

Java

// Java implementation to find the 
// count of leap years in given
// range of the year 
class GFG
{
  
// Function to calculate the number
// of leap years in range of (1, year)
static int calNum(int year)
{
    return (year / 4) - (year / 100) +
                        (year / 400);
}
  
// Function to calculate the number
// of leap years in given range
static void leapNum(int l, int r)
{
    l--;
    int num1 = calNum(r);
    int num2 = calNum(l);
    System.out.print(num1 - num2 +"\n");
}
  
// Driver Code
public static void main(String[] args)
{
    int l1 = 1, r1 = 400;
    leapNum(l1, r1);
  
    int l2 = 400, r2 = 2000;
    leapNum(l2, r2);
}
}
  
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation to find the 
# count of leap years in given 
# range of the year 
  
# Function to calculate the number 
# of leap years in range of (1, year) 
def calNum(year) : 
  
    return (year // 4) - (year // 100) + (year // 400); 
  
# Function to calculate the number 
# of leap years in given range 
def leapNum(l, r) :
  
    l -= 1; 
    num1 = calNum(r); 
    num2 = calNum(l); 
    print(num1 - num2); 
  
# Driver Code 
if __name__ == "__main__" : 
  
    l1 = 1; r1 = 400; 
    leapNum(l1, r1); 
  
    l2 = 400; r2 = 2000; 
    leapNum(l2, r2); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation to find the 
// count of leap years in given
// range of the year 
using System;
  
class GFG
{
  
// Function to calculate the number
// of leap years in range of (1, year)
static int calNum(int year)
{
    return (year / 4) - (year / 100) +
                        (year / 400);
}
  
// Function to calculate the number
// of leap years in given range
static void leapNum(int l, int r)
{
    l--;
    int num1 = calNum(r);
    int num2 = calNum(l);
    Console.Write(num1 - num2 +"\n");
}
  
// Driver Code
public static void Main(String[] args)
{
    int l1 = 1, r1 = 400;
    leapNum(l1, r1);
  
    int l2 = 400, r2 = 2000;
    leapNum(l2, r2);
}
}
  
// This code is contributed by PrinciRaj1992

输出：

97
389

时间复杂度： O(1)