求第一个 N 奇数和偶数之和的Java程序

当任何以 0、2、4、6、8 结尾的数除以 2 时，即为偶数。而当任何以 1、3、5、7、9 结尾的数不除以二时，就是奇数。

例子：

Input : 8
Output: Sum of First 8 Even numbers = 72
        Sum of First 8 Odd numbers = 64

方法#1：迭代

1. 创建两个变量 evenSum 和oddSum 并将它们初始化为 0。
2. 从 1 到 2*n 开始 For 循环。
3. 如果我是 even Add i with evenSum。
4. 否则用oddSum添加i。
5. 在循环结束时打印evenSum 和oddSum。

下面是Java程序的实现：

Java

// Calculate the Sum of First N Odd & Even Numbers in Java
import java.io.*;
  
public class GFG {
  
    // Driver function
    public static void main(String[] args)
    {
        int n = 8;
        int evenSum = 0;
        int oddSum = 0;
  
        for (int i = 1; i <= 2 * n; i++) {
            // check even & odd using Bitwise AND operator
            if ((i & 1) == 0)
                evenSum += i;
            else
                oddSum += i;
        }
        // Sum of even numbers less then 17
        System.out.println("Sum of First " + n
                           + " Even numbers = " + evenSum);
  
        // sum of odd numbers less then 17
        System.out.println("Sum of First " + n
                           + " Odd numbers = " + oddSum);
    }
}

输出

Sum of First 8 Even numbers = 72
Sum of First 8 Odd numbers = 64

时间复杂度： O(N)，其中 N 是前 N 个偶数/奇数的数量。

方法 2：使用 AP 公式。

• 前 N 个偶数之和 = n * (n+1)
• 前 N 个奇数之和 = n * n

下面是上述方法的实现：

Java

// Calculate the Sum of First N Odd & Even Numbers in Java
import java.io.*;
  
public class GFG {
  
    // Function to find the sum of even numbers
    static int sumOfEvenNums(int n) { return n * (n + 1); }
  
    // Function to find the sum of odd numbers.
    static int sumOfOddNums(int n) { return n * n; }
  
    // Driver function
    public static void main(String[] args)
    {
        int n = 10;
        int evenSum = sumOfEvenNums(n);
        int oddSum = sumOfOddNums(n);
  
        // Sum of even numbers
        System.out.println("Sum of First " + n
                           + " Even numbers = " + evenSum);
  
        // sum of odd numbers
        System.out.println("Sum of First " + n
                           + " Odd numbers = " + oddSum);
    }
}

输出

Sum of First 10 Even numbers = 110
Sum of First 10 Odd numbers = 100

时间复杂度： O(1)