如何通过使用一个循环来编写相同的函数 [英] How can I write the same function by using exactly one loop

问题描述

<pre>static void fun(int x) {
int a, b, c, d;
d = 1;
for (a = 1; a <= x; a++) {
b = a;
c = 1;
while (b > 0) {
c &= b % 2;
b /= 2;
}
d += c;
}
 System.out.println(d);
}

我的尝试：



我试图将while循环更改为if循环，然后在循环将新值赋给变量之后，我尝试使用continue;但结果却截然不同。我也试图摆脱for循环并在没有循环的情况下递增a值，但结果也不同。

What I have tried:

I have tried to change the while loop into if loop and then after the loop assigns the new values to the variables, I have tried to use continue; but results has been completely different. I have also attempted to get rid of the for loop and increment the a value without a loop but result has been different also.

推荐答案

static void very_fun(int x)
{
  int b = 1, c = 1;
  for (int a = 1; a <= x; a += b)
  {
    ++c;
    b *= 2;
  }
  System.out.println(c);
}

你需要的是一个递归函数。



DuckDuckGo的java递归函数 [ ^ ]

What you need is a recursive function.

java recursive functions at DuckDuckGo[^]

引用：

我试过继续使用;但结果完全不同。

I have tried to use continue; but results has been completely different.

尝试随机更改代码是一个坏主意。



学会正确缩进你的代码，它显示其结构，它有助于阅读和理解。它还有助于发现结构错误。

Trying random changes in code is a bad idea.

Learn to indent properly your code, it show its structure and it helps reading and understanding. It also helps spotting structures mistakes.

static void fun(int x) {
	int a, b, c, d;
	d = 1;
	for (a = 1; a <= x; a++) {
		b = a;
		c = 1;
		while (b > 0) {
			c &= b % 2;
			b /= 2;
		}
		d += c;
	}
	System.out.println(d);
}

缩进风格 - 维基百科 [ ^ ]



专业程序员的编辑有这个功能和其他功能，如括号匹配和语法高亮。

Notepad++主页 [ ^ ]

ultraedit [ ^ ]

Indentation style - Wikipedia[^]

Professional programmer's editors have this feature and others ones such as parenthesis matching and syntax highlighting.
Notepad++ Home[^]
ultraedit[^]

引用：

怎么能我通过使用一个循环来编写相同的函数

How can I write the same function by using exactly one loop

在第一次阅读时，看起来需要2个循环，所以如果你必须将它保存在1个函数中，你需要完全理解这段代码是做什么的。



这个

On first reading, it looks like the 2 loops are needed, so if you have to keep it in 1 function, you need to fully understand what is doing this code.

This

c &= b % 2;
b /= 2;

告诉代码正在弄乱<$的二进制形式c $ c> b



首先要做的是运行x = 1,2,3,4 ... 16的函数

备注匹配结果和二进制x

tells that the code is messing with the binary form of b

first thing to do is run the function with x=1, 2, 3, 4 ... 16
note matching results and binary of x

x binary result
  1    1
  2   10
  3   11
...

然后你必须找到结果改变的原因，这种理解可能导致简化和循环移除。

否则，CPallini的解决方案只是神奇的，你什么都不会学到。

then you have to find the reason why result change, this understanding may lead to simplify and a loop removal.
Otherwise, CPallini's solution is just magic and you will learn nothing.

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