一行中的星星图案 [英] Star Patterns in One Line
问题描述
我知道这是我的作业而且由于某种原因而被分配了，但是我已经在这个问题上待了一个星期但我仍然没有想到它。
所以这里的目标是使用一行C ++代码创建四种模式（行数基于用户输入）。这四种模式是：
I know this is my homework and it is assigned for a reason, but I've been sitting on the problem for a week and I still haven't figured it out.
So the goal here is to create four patterns (amount of rows is based on user input) using one line of C++ code. The four patterns are:
* *** * ***
** ** ** **
*** * *** *
我已经弄清楚如何编码每个模式，但这样做意味着我有两个/三个while循环。我们还没有在这个类中学习for循环，因此不允许使用它。但我不知道如何将这些循环压缩成一行！如果我作弊并使用for循环，它几乎适用于第一和第四种模式。但是否则......
这是我对模式的代码（n是用户输入的行数）：
模式1：
I've figured out how to code each pattern, but doing so means that I have two/three while loops. We haven't learned the for loop in this class and are therefore not allowed to use it. But I have no idea how I can even compress those loops into one single line! If I cheat and use for loops it almost works for the first and fourth patterns. But otherwise...
Here's the code I have for the patterns (n is user input of amount of rows):
Pattern 1:
int i, j, n;
i = 1;
while (i <= n) {
j = 0;
while (j < i) {
cout << "*";
j++;
}
cout << endl;
i++;
}
模式2：
Pattern 2:
int i, j, k, n;
i = 0;
while (i < n) {
k = 0;
while (k < i) {
cout << " ";
k++;
}
j = n;
while (j > i) {
cout << "*";
j--;
}
cout << endl;
i++;
}
模式3：
Pattern 3:
int i, j, k, n;
i = 1;
while (i <= n) {
k = n;
while (k > i) {
cout << " ";
k--;
}
j = 0;
while (j < i) {
cout << "*";
j++;
}
cout << endl;
i++;
}
模式4：
Pattern 4:
int i, j, n;
i = 0;
while (i < n) {
j = n;
while (j > i) {
cout << "*";
j--;
}
cout << endl;
i++;
}
我们在课堂上学到的是非常基础的，因为本课程是对计算科学的介绍。它应该触及内部的许多不同主题，而不会深入探讨。我们为C ++学到的只是基本声明，if语句和while循环。
我真的需要帮助。 :(一些提示或提示会很适合我;我正在寻求帮助，而不是有人为我做我的工作。
提前谢谢！
What we learned in class is very basic because this course is an introduction to computing science. It is supposed to touch on a lot of the different topics within in without going into too much depth. All we really learned for C++ is basic declaration, if statements, and the while loop.
I really need help. :( Some tips or hints will suit me just fine; I'm asking for help, not for someone to do my work for me.
Thank you in advance!
推荐答案
好吧，我误解了我的老师。:(非常抱歉，大家！他希望所有的图案都印在一行，不一定是一行代码。那些感兴趣的人，这是我结束的代码（并且他接受了解决方案）：
Okay, I misunderstood my teacher. :( Terribly sorry, everyone! He wanted all of the patterns printed in one line, not necessarily one line of code. For those who are interested, here's the code I ended with (and that he accepted as a solution):
int i, j, k, n;
cout << "Enter number of rows: ";
cin >> n;
i = 0;
while (i < n) {
j = 0;
while (j <= i) {
cout << "*";
j++;
}
k = n - 1;
while (k > i) {
cout << " ";
k--;
}
cout << " ";
k = 0;
while (k < i) {
cout << " ";
k++;
}
j = n;
while (j > i) {
cout << "*";
j--;
}
cout << " ";
j = n - 1;
while (j > i) {
cout << " ";
j--;
}
k = i;
while (k >= 0) {
cout << "*";
k--;
}
cout << " ";
j = n;
while (j > i) {
cout << "*";
j--;
}
cout << endl;
i++;
}
再一次，抱歉我可能引起的头痛。 ^^
Once again, sorry for the headaches I may have caused. ^^
这应该让你开始，但你需要为自己找出正确的间距值
This should get you started, but you will need to figure out the correct spacing values for yourself
int i = 0;
while (i != 3)
cout << string(i, '*') << string(4-i, ' ') << setw(4) << string(4 - i, '*') << setw(6) << string(i, '*') << string(4, ' ') << string(4 - i, '*') << string(i++, ' ') << endl;
多么大脑弯曲。我可以在这里看到一个模式，我现在还不能完全改变它。
看看你的例子，我记下了星号和空格的数量在每一行。这给了我下表：
What a brain-bender. I can see a pattern here, I just can't quite paramaterize it any better just yet.
Looking at your example, I noted down the number of asterisks and spaces on each line. This gave me the following table:
* *** * ***
** ** ** **
*** * *** *
1, 5, 3, 7, 1, 5, 3
2, 5, 2, 6, 2, 5, 2
3, 5, 1, 5, 3, 5, 1
你可以清楚地看到奇怪的关系-numbered列到最大行数，N。
您还可以看到偶数列有自己的模式 - n + 1 + 2,2 *（n + 1） -row，n + 1 + 2 N + 2,2 *（N + 1）-row，N + 2
我把一个版本砸在了一起使用for循环，然后重新阅读你的问题。所以我将它们转换为while循环。显然，我使用了超过1行代码。但是，这样做 - 我注意到每行都有一个重新出现的模式。第一行中的1,5,3重复，第二行中的2,5,2重复。什么是大脑破坏者！
这是一些值得考虑的颂歌。也许你可以找出一种我还没有完全掌握的模式。这种模式可以消除我所展示的大部分重复，并且这样做会产生（几乎/实际上）合理的代码作为单行代码。
我测试的N范围高达约25，并且获得相同的奖励，尽管规模更大。请注意，为了简化/简化每个while循环的终止条件，N的数字大于您实际想要打印的行数。
You can clearly see a relationship in the odd-numbered columns to the max number of rows, N.
You can also see that the even numbered columns have their own patterns - n+1+2, 2*(n+1)-row, n+1+2 N+2, 2*(N+1)-row, N+2
I smashed together a version using for loops, then re-read your question. So I converted them to while loops. Clearly, I'm using more than 1 line of code. But, in doing so - I noticed that there was a re-occurant pattern along each row. 1,5,3 in the first row is repeated, as is 2,5,2 in the second. What a total brain buster!!
Here's some ode to consider. Perhaps you can identify a pattern I've just not quite come to grips with yet. A patter that would eliminate much of the repetition I've exhibited and in doing so, produce code that is (almost/actually) reasonable as a single line of code.
I tested with N ranging up to about 25 and was rewarded with the same pattern, albeit scaled ever larger. Note that for the ease/brevity of the terminating condition of each while loop, N holds a number 1 larger than the number of rows you'd actually like to print.
int main()
{
int row, sp, st;
int n = 4;
row=1; while (row < n)
{
st=0; while(st++ < row) printf("*");
sp=0; while(sp++ < n+1) printf(" ");
st=0; while(st++ < n-row) printf("*");
sp=0; while(sp++ < 2*n-row) printf(" ");
//---------------
st=0; while(st++ < row) printf("*");
sp=0; while(sp++ < n+1) printf(" ");
st=0; while(st++ < n-row) printf("*");
printf("\n");
row++;
}
return 0;
}
这是（我的）n = 10的样本
Here's a sample of (my)n = 10
* ********* * *********
** ******** ** ********
*** ******* *** *******
**** ****** **** ******
***** ***** ***** *****
****** **** ****** ****
******* *** ******* ***
******** ** ******** **
********* * ********* *
我的头疼！
My head hurts!
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