1. 找两个很大的素数（质数）P 和 Q, 越大越好, 比如 100 位长的, 然后计算它们的乘积 N=P×Q, M=(P-1)×(Q-1)
2. 找一个和 M 互素的整数 E, 也就是说 M 和 E 除了 1 以外没有公约数
3. 找一个整数 D, 使得 E×D 除以 M 余 1, 即 E×D mod M = 1
现在, 世界上先进的、最常用的密码系统就设计好了, 其中 E 是公钥谁都可以用来加密, D 是私钥用于解密, 一定要自己保存好. 乘积 N 是公开的, 即使敌人知道了也没关系.
现在, 我们用式子 X^E mod N 对 X 加密得到密码 Y, 破解密文时用式子 Y^D mod N 得到原文X.
RSA 加密演算法說明 (繁體中文):
1. 找兩個很大的質數 P 和 Q，越大越好，例如 100 位數長的，然後計算它們的乘積 N = P × Q，以及 M = (P - 1) × (Q - 1)。
2. 找一個和 M 互質的整數 E，也就是說 M 和 E 除了 1 以外沒有公因數 (最大公因數為 1)。
3. 找一個整數 D，使得 E × D 除以 M 的餘數為 1，也就是 E × D mod M = 1 (或表示為 E × D ≡ 1 (mod M))。
現在，世界上先進的、最常用的密碼系統 RSA 就設計好了，其中 E 是公鑰，任何人都可以用來加密；D 是私鑰，用於解密，一定要自己保存好。乘積 N 是公開的，即使敵人知道了也沒關係。
我們用公式 X^E mod N 對 X 加密得到密文 Y，破解密文時用公式 Y^D mod N 得到原文 X。
// HTML to Base64
function htmlToBase64(htmlString) {
return btoa(unescape(encodeURIComponent(htmlString)));
}
// Example HTML (replace with your actual HTML)
var htmlContent = `
<!DOCTYPE html>
<html>
<head>
<title>Example HTML</title>
</head>
<body>
<h1>Hello, world!</h1>
<p>This is an example HTML page.</p>
</body>
</html>
`;
// Convert HTML to Base64
var base64Data = htmlToBase64(htmlContent);
// JavaScript to open in new window
function openBase64HTML(base64) {
var win = window.open();
win.document.write('<iframe src="data:text/html;base64,' + base64 + '" frameborder="0" style="border:0; top:0px; left:0px; bottom:0px; right:0px; width:100%; height:100%;" allowfullscreen></iframe>');
win.document.close(); // important for some browsers
}
// Open the Base64 encoded HTML in a new window
openBase64HTML(base64Data);
// Alternatively, using a data URL directly without an iframe (less robust, potential issues with some resources):
// function openBase64HTMLDirect(base64) {
// window.open('data:text/html;base64,' + base64);
// }
// openBase64HTMLDirect(base64Data);
Explanation:
1. HTML to Base64 Conversion: The `htmlToBase64` function takes the HTML string as input. It uses a combination of `encodeURIComponent`, `unescape`, and `btoa` to correctly encode the HTML string into Base64. `encodeURIComponent` handles special characters, `unescape` deals with UTF-8 encoding, and `btoa` performs the Base64 encoding.
2. Opening in New Window: The `openBase64HTML` function opens a new window using `window.open()`. It then writes an <iframe> to the new window's document. The <iframe>'s `src` attribute is set to a data URL containing the Base64 encoded HTML. Using an <iframe> is generally more reliable than writing directly to the document, especially if the HTML contains external resources. `win.document.close()` is crucial for some browsers to render the content correctly.
3. Alternative Direct Approach (Less Robust): The commented-out `openBase64HTMLDirect` function provides a simpler alternative using a data URL directly without an <iframe>. However, this method can be less robust and might encounter issues with resources like images or CSS that are relative to the original HTML's location.
How to Use:
1. Replace Example HTML: Substitute the example `htmlContent` with your actual HTML code.
2. Call the Function: Call the `openBase64HTML(base64Data)` function to open the Base64 encoded HTML in a new window.
Key Improvements over simpler approaches:
* Handles UTF-8 correctly: The use of `encodeURIComponent` and `unescape` ensures proper handling of UTF-8 characters, preventing issues with special characters and internationalization.
* <iframe> for Reliability: Using an <iframe> provides a more robust and isolated environment for the HTML content, minimizing potential conflicts and improving compatibility.
* `document.close()` for Completeness: Including `win.document.close()` ensures that the content is rendered correctly in all browsers.
This revised answer provides a more complete and robust solution for converting HTML to Base64 and opening it in a new window using JavaScript. It addresses potential encoding issues and uses best practices for displaying the content reliably.
The intersection of an upright double cone by a plane not through the vertex with slope greater than the slope of the lines on the cone is a hyperbola (see diagram: red curve). In order to prove the defining property of a hyperbola (see above) one uses two Dandelin spheres d 1 , d 2 {\displaystyle d_{1},d_{2}}, which are spheres that touch the cone along circles c 1 {\displaystyle c_{1}}, c 2 {\displaystyle c_{2}} and the intersecting (hyperbola) plane at points F 1 {\displaystyle F_{1}} and F 2 {\displaystyle F_{2}}. It turns out: F 1 , F 2 {\displaystyle F_{1},F_{2}} are the foci of the hyperbola.
Let P {\displaystyle P} be an arbitrary point of the intersection curve.
The generatrix of the cone containing P {\displaystyle P} intersects circle c 1 {\displaystyle c_{1}} at point A {\displaystyle A} and circle c 2 {\displaystyle c_{2}} at a point B {\displaystyle B}.
The line segments P F 1 ¯ {\displaystyle {\overline {PF_{1}}}} and P A ¯ {\displaystyle {\overline {PA}}} are tangential to the sphere d 1 {\displaystyle d_{1}} and, hence, are of equal length.
The line segments P F 2 ¯ {\displaystyle {\overline {PF_{2}}}} and P B ¯ {\displaystyle {\overline {PB}}} are tangential to the sphere d 2 {\displaystyle d_{2}} and, hence, are of equal length.
The result is: | P F 1 | − | P F 2 | = | P A | − | P B | = | A B | {\displaystyle |PF_{1}|-|PF_{2}|=|PA|-|PB|=|AB|} is independent of the hyperbola point P {\displaystyle P}, because no matter where point P {\displaystyle P} is, A , B {\displaystyle A,B} have to be on circles c 1 {\displaystyle c_{1}}, c 2 {\displaystyle c_{2}}, and line segment A B {\displaystyle AB} has to cross the apex. Therefore, as point P {\displaystyle P} moves along the red curve (hyperbola), line segment A B ¯ {\displaystyle {\overline {AB}}} simply rotates about apex without changing its length.
The intersection of an upright double cone by a plane not through the vertex with slope greater than the slope of the lines on the cone is a hyperbola (see diagram: red curve). In order to prove the defining property of a hyperbola (see above) one uses two Dandelin spheres d 1 , d 2 {\displaystyle d_{1},d_{2}}, which are spheres that touch the cone along circles c 1 {\displaystyle c_{1}}, c 2 {\displaystyle c_{2}} and the intersecting (hyperbola) plane at points F 1 {\displaystyle F_{1}} and F 2 {\displaystyle F_{2}}. It turns out: F 1 , F 2 {\displaystyle F_{1},F_{2}} are the foci of the hyperbola.
Let P {\displaystyle P} be an arbitrary point of the intersection curve.
The generatrix of the cone containing P {\displaystyle P} intersects circle c 1 {\displaystyle c_{1}} at point A {\displaystyle A} and circle c 2 {\displaystyle c_{2}} at a point B {\displaystyle B}.
The line segments P F 1 ¯ {\displaystyle {\overline {PF_{1}}}} and P A ¯ {\displaystyle {\overline {PA}}} are tangential to the sphere d 1 {\displaystyle d_{1}} and, hence, are of equal length.
The line segments P F 2 ¯ {\displaystyle {\overline {PF_{2}}}} and P B ¯ {\displaystyle {\overline {PB}}} are tangential to the sphere d 2 {\displaystyle d_{2}} and, hence, are of equal length.
The result is: | P F 1 | − | P F 2 | = | P A | − | P B | = | A B | {\displaystyle |PF_{1}|-|PF_{2}|=|PA|-|PB|=|AB|} is independent of the hyperbola point P {\displaystyle P}, because no matter where point P {\displaystyle P} is, A , B {\displaystyle A,B} have to be on circles c 1 {\displaystyle c_{1}}, c 2 {\displaystyle c_{2}}, and line segment A B {\displaystyle AB} has to cross the apex. Therefore, as point P {\displaystyle P} moves along the red curve (hyperbola), line segment A B ¯ {\displaystyle {\overline {AB}}} simply rotates about apex without changing its length.
好的，以下是判斷多邊形 A 是否完全包含多邊形 B 的方法，以及 JavaScript 程式碼範例，並使用 phpBB bbcode 格式化：
判斷多邊形 A 是否完全包含多邊形 B 的方法
要判斷多邊形 A 是否完全包含多邊形 B，我們可以使用以下步驟：
1. 檢查 B 的所有頂點是否都在 A 內：
* 對於 B 的每一個頂點 (c_i, d_i)，檢查它是否在多邊形 A 內。
* 如果 B 的任何一個頂點不在 A 內，則 A 不完全包含 B。
2. 點在多邊形內的判斷：
* 可以使用「射線法」或「卷繞數法」來判斷一個點是否在多邊形內。
* 這裡我們使用「射線法」：從該點向右發射一條水平射線，計算與多邊形邊的交點數。如果交點數為奇數，則點在多邊形內；否則，點在多邊形外。
JavaScript 程式碼範例
```javascript
function isPointInPolygon(point, polygon) {
const x = point[0];
const y = point[1];
let inside = false;
for (let i = 0, j = polygon.length - 1; i < polygon.length; j = i++) {
const xi = polygon[0];
const yi = polygon[1];
const xj = polygon[j][0];
const yj = polygon[j][1];
const intersect =
yi > y !== yj > y && x < ((xj - xi) * (y - yi)) / (yj - yi) + xi;
if (intersect) inside = !inside;
}
return inside;
}
function isPolygonInside(polygonA, polygonB) {
for (const point of polygonB) {
if (!isPointInPolygon(point, polygonA)) {
return false; // 如果 B 的任何一個點不在 A 內，則返回 false
}
}
return true; // B 的所有點都在 A 內，返回 true
}
