Implementation example of JavaScript printing star pyramid function
This article mainly introduces JavaScript to realize the function of printing star pyramid, and combines specific examples to analyze the principle and related implementation techniques of JavaScript for outputting any given number of rows of star pyramid graphics. Friends in need can refer to the following
The example in this article describes how to implement the printing star pyramid function in JavaScript. I share it with you for your reference. The details are as follows:
It is exactly the same for you to write in other languages.
I guess you have seen this question when you were learning C language...
That is to say, print the following ghost stuff:
#When I saw the loop structure, I felt very bored, so I disdained this question and did not pay attention to it. Think about it,
Because if it is written in JavaScript, it can be written like this. It is not even JavaScript, it is just an html:
<html> <head> <meta http-equiv="content-type" content="text/html;charset=gb2312"/> </head> <body> ＆nbsp;＆nbsp;* ＆nbsp;*** ***** ＆nbsp;*** ＆nbsp;＆nbsp;* </body> </html>
The reason why it is not used here UTF-8 encoding is because UTF-8's processing of nbsp and * font will cause typesetting confusion, that is, it does not use standard Song fonts, which makes the final result impossible to view.
Anyway, I can copy and paste the above content no matter how many lines you put in the question. However, today, when I have programmed to a certain amount and have some big data concepts, I saw this question again and carefully After thinking about it for a while, if the question maker asked to output a star-shaped pyramid with 200,000 lines of symmetry on the central axis, I would be damned.
So we still need to thoroughly understand how to implement this. Although such programming will not occur in actual programming, it is said that this question will still come up in some boring interviews. At least, latecomers will ask you C language When asked, if you output a star pyramid with 200,000 lines of symmetry on the central axis, you still have to know how to do it. If you can't do this question the first time, you can't do it.
1. Basic Goals
First, an input box pops up, allowing the user to enter an odd number. After all, axial symmetry requires an odd number,
Then for the robustness of the program, it is necessary to judge what the user inputs. If the input is not an odd number, a prompt will pop up and the subsequent program will not be executed. How to judge an odd number in JavaScript? I have already discussed it in "JavaScript's Judgment and Processing of Numbers" ” has been said before, so I won’t go into details here.
Considering the load of the browser, here, I only allow the odd number entered by the user to be 189. You can also increase it a bit. 189 feels okay in my computer, so I set this number. It has no special meaning. It was entered randomly and was not intentionally tested.
Enter a line of 189, and IE has already popped up a prompt to "abort the script", but there is no problem if you don't abort!
The running results are as follows:
If you are writing a C language program or other programs, this value can definitely be set larger!
2. Basic Idea
Once you understand this, it is very easy to write.
First of all, we only need to output spaces on the left side of *. There is no need to output spaces on the right side. After typing *, just change the line directly.
Split into two Parts, one part is when i<=n/2 in the upper half, and the other part is when i>n/2 in the lower half.
The reason why it is divided like this is because the * sign in the output of these two parts It is different from the number of spaces output.
After that, it was a common problem in junior high school to find the rules. Anyway, I found the above rules, and there was no problem when programming and running. Other math emperors found more awesome expressions, and I was defeated.
3. Production process
The code is very simple, it is the expression of conditional structure and loop structure. It goes without saying that the above ideas are understood. .
Some people may find it strange here, why should I first judge n++ and then n%2!=0, that is, judge whether n+1 is an even number to judge whether n is an odd number,
The main thing here is It is to cater to the following for loop structure...
<html>
<head>
<meta http-equiv="content-type" content="text/html;charset=gb2312"/>
</head>
<body>
</body>
</html>
<script>
var i,j,k,n;
n=window.prompt("请输入要输出的行数n，为了形成轴对称，所以你输出的必须是奇数！");
if(isNaN(n)||!n)
alert("你输入的不是数！");
else{
n++;
if(n%2!=0)
alert("你输入的不是奇数！");
else if(n>190){
alert("不要这么大嘛！臣妾做不到啊！");
}
else{
for(i=1;i<n;i++){
if(i<=n/2){
for(k=n/2-i;k>0;k--)
document.write("＆nbsp;");
for(j=0;j<2*i-1;j++)
document.write("*");
}
else{
for(k=i-n/2;k>0;k--)
document.write("＆nbsp;");
for(j=0;j<2*(n-i)-1;j++)
document.write("*");
}
document.write("<br>");
}
}
}
</script>It will make me cry if I say too much, I will experience it myself...
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