酱油爆鱼
问题描述
我尽可能简化任务，所以我可以将它应用到我的算法中.
I tied to simplify the task as much as possible, so I could apply it to my algorithm.
这是数学家和程序员面临的挑战:
And here is the challenge for mathematicians and programmers:
我需要创建一个传递参数 int n 的方法:
I need to create a method where I pass parameter int n:
public void optionality_generator(int n){
//some kind of loops, or recursions...to make it workable
System.out.println("current combination: ...");
}
输出应显示真假的所有可能组合.
The output should show all possible combinations of true's and false's.
这里是 N=1 的例子；N=2；N=3；N=4；N=5，其中 x=false，0=true；请注意，空折线只是为了让您更容易识别模式.希望我包括了所有可能的组合):
Here is examples where N=1; N=2; N=3; N=4; N=5 where x=false and 0=true; Please note, empty break lines is just for you to recognise easier the patterns. Hopefully, I included all possible combinations):
Combination of 1: 0 x Combination of 2: 00 x0 0x xx Combination of 3: 000 X00 0X0 00X XX0 0XX XXX Combination of 4: 0000 X000 0X00 00X0 000X XX00 X0X0 X00X 0XX0 0X0X 00XX XXX0 XX0X X0XX 0XXX XXXX Combination of 5: 00000 X0000 0X000 00X00 000X0 0000X XX000 X0X00 X00X0 X000X X0X00 X00X0 X000X 0XX00 0X0X0 0X00X 00XX0 00X0X 000XX XXX00 XX0X0 XX00X X0XX0 X0X0X X00XX 0XXX0 0XX0X 00XXX XXXX0 XXX0X XX0XX X0XXX 0XXXX XXXXX
另外，如果您看到输出，这是我识别的模式，所有组合都反转了一半(例如，第一个组合是 00000，最后一个是 XXXXX，第二个是 X0000，最后一个之前的一个是 0XXXX 等..).也许，这种模式将有助于使整个算法更高效，对此不确定.提前谢谢！
Also, If you see the output, here is the pattern I recognized, that all combinations are inverted on half (e.g first combination is 00000 last one will be XXXXX, second one X0000, one before the last one will be 0XXXX etc..). Maybe, this pattern will help to make the whole algorithm more efficient, not sure about this. Thank you in advance!
推荐答案
这是一个仅使用 Java API 的非常基本的方法:
Here is a really basic way using only Java APIs:
final int n = 3;
for (int i = 0; i < Math.pow(2, n); i++) {
String bin = Integer.toBinaryString(i);
while (bin.length() < n)
bin = "0" + bin;
System.out.println(bin);
}
结果:
000 001 010 011 100 101 110 111
当然，您可以将 n 设置为任何您喜欢的值.并且，有了这个结果，您可以从字符串中选择第 n 个字符作为真/假.
Of course, you can set n to whatever you like. And, with this result, you can pick the nth character from the string as true/false.
如果只需要检查某个位是否为真，则不需要将其转换为字符串.这只是为了说明输出值.
If you only need to check if a bit is true, you don't need to convert it to a string. This is just to illustrate the output values.
