这条Java流的这一行是什么意思？

发布于 2025-01-21 05:47:18 字数 247 浏览 2 评论 0原文

因此，我有这条代码。它给了我输出[6,28]。你们知道为什么吗？我不知道有人试图打印什么数字。

System.out.println( IntStream.range(1,30).filter(n -> IntStream.range(1,n).filter(i->n%i == 0).sum() ==n)
               .boxed().collect(Collectors.toList()));

原文

So I have this line of code. It gives me output [6,28].
Do you guys know why? I dont know what kind of numbers was someone trying to print.

System.out.println( IntStream.range(1,30).filter(n -> IntStream.range(1,n).filter(i->n%i == 0).sum() ==n)
               .boxed().collect(Collectors.toList()));

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何以畏孤独 2025-01-28 05:47:18

IntStream.range(1,n).filter(i -> n % i == 0).sum() == n

通过检查n的所有除数，通过检查n除以潜在的除数分隔（i-＆gt; n％i == 0 ）。然后将所有除数求和与n本身进行比较。

IntStream.range(1,30).filter(n -> ...).boxed().collect(Collectors.toList())

我上面描述的是1到30之间的数字，并且只保留了比较为true的数字。因此，它计算出1到30之间的所有数字，其中除数总和到数字本身。这些称为完美的数字。前两个是6和28，其次是496，...

IntStream.range(1,n).filter(i -> n % i == 0).sum() == n

calculates all the divisors of n by checking if n divided by the potential divisor has no remainder (i -> n % i == 0). All divisors are then summed and compared with n itself.

IntStream.range(1,30).filter(n -> ...).boxed().collect(Collectors.toList())

Just does what I described above for the numbers between 1 and 30 and only keeps those where the comparison is true. It therefore calculates all the numbers between 1 and 30 where the divisors sum to the number itself. Those are called perfect numbers. The first two are 6 and 28, followed by 496, ...

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