由用户输入提供的L1和L2范围内的奇数和偶数相加

Question

我正在努力完成这项任务。用户正在提供L1和L2号。将L1、L2范围内的所有奇数和该范围内的所有偶数相加，并显示总和。我必须用3种方法来实现它，使用：for、while和do-while循环。我的for循环工作得很好，但while显示了一些更高的分数。

import java.util.Scanner;

public class PPO4b {

    public static void main(String[] args) {
        
        Scanner value1Check = new Scanner(System.in);
        System.out.println("Provide first value: ");
        int value1 = value1Check.nextInt();
        
        Scanner value2Check = new Scanner(System.in);
        System.out.println("Provide second value: ");
        int value2 = value2Check.nextInt();

        int sumOdd = 0;
        int sumEven = 0;
        int i = 0;

        while (i <= 0) {
            if (i % 2 == 1) {
                sumOdd = sumOdd + i;
            } else {
                sumEven = sumEven + i;
            }
            i = i + 1;
        }
        System.out.println("Sum is equal to: " + sumEven);
        System.out.println("Sum is equal to: " + sumOdd);
    }
}
 

Answer 1

通过一点数学，你可以在0（1）分钟内完成这项工作。

• 对于从1到n值的总和，n（n 1）/2将给出总和

现在有足够的信息来计算s和e之间的偶数和奇数之和。

对于给定的int e和int s；

int totalSum = ((e+s)*(e-s+1))/2;
e = (e|1);
s = (e-1)|1;
int evenCount = (e-s)/2;
int sumEven = ((e - s + 1) * (evenCount))/2;
int sumOdd = totalSum - sumEven;

下面是如何验证它。

• 使用范围的IntStream计算偶数和奇数和，根据需要过滤掉偶数和奇数值
• 接下来使用上述方法计算相同的值。
• 比较相应的总和并报告任何差异。

Random rnd = new Random();
for (int i = 0; i < 200000; i++) {
    int s = rnd.nextInt(200) +100;
    int e = rnd.nextInt(500) + s;
    int compEven = IntStream.rangeClosed(s, e)
            .filter(r -> r % 2 == 0).sum();
    int compOdd = IntStream.rangeClosed(s, e)
            .filter(r -> r % 2 == 1).sum();
    long compEvenCount = IntStream.rangeClosed(s, e)
            .filter(r -> r % 2 == 0).count();
    
    int mySum = (e + s) * (e - s + 1) / 2;
    e |= 1;
    s = (s - 1) | 1;
    int evenCount = (e - s + 1) / 2;
    int evenSum = ((e + s) * evenCount) / 2;
    int oddSum = mySum - evenSum;
    if (evenSum != compEven || oddSum != compOdd) {
        System.out.println("Oops - sums don't match");
    }
}