英文:
How to print an additional * after every 5th and following 3rd number in a positive integer sequence?
问题
我正在尝试打印如下内容：
1, 2, 3, 4, 5, *, 6, 7, 8, *, 9, 10, 11, 12, 13, *, 14, 15, 16, *, 17, ...
其中，在每第5个和随后的第3个整数之后，都应该重复打印一个额外的 *。
以下是我目前的代码：
for (int i = 1; i <= 100; i++) {
if (i % 5 == 0) {
System.out.print("* ");
} else {
System.out.print(i + " ");
}
}
但这段代码打印出来的结果是：
1, 2, 3, 4, *, 6, 7, 8, 9, *, 11, 12, 13, 14, *, 16, 17, ...
我在这里做错了什么？
英文:
I'm trying to print
1, 2, 3, 4, 5, *, 6, 7, 8, *, 9, 10, 11, 12, 13, *, 14, 15, 16, *, 17, ...
where an additional * should be printed after every 5th and following 3rd integer repeatedly.
Here's my current code
for (int i = 1; i <= 100; i++) {
if (i % 5 == 0) {
System.out.print("* ");
} else {
System.out.print(i + " ");
}
}
which prints
1, 2, 3, 4, *, 6, 7, 8, 9, *, 11, 12, 13, 14, *, 16, 17, ...
What am I doing wrong here?
答案1
得分: 1
这产生了预期的输出：
class Main {
public static void main(String[] args) {
for (int i = 1; i <= 100; i++) {
System.out.print(i + " ");
if ((i-5) % 8 == 0 || i % 8 == 0) {
System.out.print("* ");
}
}
}
}
// 输出：
// 1 2 3 4 5 * 6 7 8 * 9 10 11 12 13 * 14 15 16 * 17 18 19 20 21 * 22
// 23 24 * 25 26 27 28 29 * 30 31 32 * 33 34 35 36 37 * 38 39 40 * 41
// 42 43 44 45 * 46 47 48 * 49 50 51 52 53 * 54 55 56 * 57 58 59 60 61
// * 62 63 64 * 65 66 67 68 69 * 70 71 72 * 73 74 75 76 77 * 78 79 80
// * 81 82 83 84 85 * 86 87 88 * 89 90 91 92 93 * 94 95 96 * 97 98 99 100
Dry run 链接：https://repl.it/repls/MarriedPreemptiveLeadership
同时，感谢 @Rogue 提供的数学公式。
这是 @Rogue 提供的解释：
你有 5、8、13、16、21 等。因此，这些索引位于间隔为 5、5+3（8）或 5+(8n) 的位置。因此，你可以检查是否满足 ((i-5) % 8 == 0 || i % 8 == 0)，如果满足，则在打印数字后在模式中打印一个星号。毕竟，数字总是被打印出来的。
英文:
This is giving the expected output:
class Main {
public static void main(String[] args) {
for (int i = 1; i <= 100; i++) {
System.out.print(i + " ");
if ((i-5) % 8 == 0 || i % 8 == 0) {
System.out.print("* ");
}
}
}
}
// Output:
// 1 2 3 4 5 * 6 7 8 * 9 10 11 12 13 * 14 15 16 * 17 18 19 20 21 * 22
// 23 24 * 25 26 27 28 29 * 30 31 32 * 33 34 35 36 37 * 38 39 40 * 41
// 42 43 44 45 * 46 47 48 * 49 50 51 52 53 * 54 55 56 * 57 58 59 60 61
// * 62 63 64 * 65 66 67 68 69 * 70 71 72 * 73 74 75 76 77 * 78 79 80
// * 81 82 83 84 85 * 86 87 88 * 89 90 91 92 93 * 94 95 96 * 97 98 99 100
Dry run link: https://repl.it/repls/MarriedPreemptiveLeadership
Also, thanks to @Rogue for coming up with the mathematical formula.
Here's the explanation provided by @Rogue
> You have 5, 8, 13, 16, 21, etc. So it follows that these indices lie
> after a gap of 5, or 5+3 (8), or 5+(8n). So from there, you could
> check if ((i-5) % 8 == 0 || i % 8 == 0), and if so, print an asterisk
> in your pattern after printing your number. After all, the numbers are
> always printed.
答案2
得分: 0
我不能说如何使其显示您想要的内容，但它之所以在每个5（5、10、15等）处显示*，是因为％是一个“模数”运算符，它在除法后获得余数。
因此，5％5的结果为== 0（因此打印出“*”）因为在将5 / 5相除后没有余数。同样，10 / 5也是如此，等等。6％5会给出余数1`。
英文:
I can't say how to make it show what you want, but the reason it's showing * at every 5 (5, 10, 15, etc.) is that % is a "modulus" operator which gets the remainder after dividing.
So 5 % 5 does == 0 (and thus prints out "* ") as there is no remainder after dividing 5 / 5. Same as 10 / 5, etc. 6 % 5 would give 1 as the remainder.
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