我认识一位英语很好的程序员,他现在的薪资是2万3,之前他一直想找个线上的远程工作.
其中要注册那个远程工作网站要全英文,还要答题,是英文算法题.
虽然能看懂一大半,但是没有99%的自信说,这些英文语句我全懂.
后来他给我说可以看李靖宇的这些英文课
当时工作忙,空里有时背百词斩,也买了一段时间的百词斩会员,
但是实际要完完全全地99%说我英语很棒,还是不达标.
这次我是真的买了课程班,现在想想,除了编程课,这是我买的最贵的学习课程了.
现在把它分享出来.
1.请保证你有小学的英语水平,如果没有小学的英语水平,请努力尝试
2.视频及课程可能有不全的地方,请自行脑补
3.下载网盘数据中,请认识到自己的不足,下载完成后该如何学习
好来,上链接:
李靖瑜新思维雅思20000单词速记:
链接：https://pan.baidu.com/s/186T6kuUDCZBkh-f2zjbr7Q
提取码：1234
此文件就已够大,仅支持30天下载,这些文件我是一周内抽空就下载抽空就下载下载完成的,所以30天是足够的.如果你访问不了,无法下载,那么请留言
或扫码公众号回复"英语"获取长期下载链接
学英语很重要,尤其作为程序员
### Challenge A
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. What is the 10001st prime number?
Please implement a function/method that returns the 10001st prime number in the following code block.
/**
* @Author: zhangQi
* @Date: 2020-07-21 10:12
*/
public class PrimeTest {
//print
public static void main(String[] args) {
int the10001PrimeNumber = getThe10001PrimeNumber();
System.out.println(the10001PrimeNumber);
}
//get result
public static int getThe10001PrimeNumber(){
int the10001Index = 10001;
int j =1;
int i =1;
int result =0;
while(j<the10001Index){
if(isPrimeNumber(i)){
result =i;
j++;
}
i+=2;
}
return result;
}
//is it prime?
static boolean isPrimeNumber(int n){
if(n<2){
return false;
}
double max = Math.sqrt(n);
for(int i=2;i<=max;i++){
if(n%i==0){
return false;
}
}
return true;
}
}
### Challenge B
In this challenge we are going to estimate **the Golden Ratio** with **Fibonacci sequence**.
_**This is a coding challenge. Everything you need to solve the challenge will be explained. No extra mathematics knowledge required.**_
In mathematics, two quantities are in the Golden Ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities. The following image illustrates the geometric relationship.
The value of the Golden ratio is a irrational number: _1.6180339887..._
Fibonacci sequence is a sequence of numbers where each number is the sum of the two preceding ones, starting from 0 and 1:
- F(0) = 0
- F(1) = 1
- F(n) = F(n-1) + F(n-2)
The first 10 numbers of fibonacci sequence are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34.
**When n approaches positive infinity, the ratio F(n+1)/F(n) approaches the Golden Ratio.**
Based on the above facts, we can estimate the golden ratio with Fibonacci sequence:
Given an estimation precision parameter `t`, it should find the first `n` that matches this condition `|F(n+2)/F(n+1)-F(n+1)/F(n)| < t`, and return the value of corresponding `|F(n+1)/F(n)|` as the estimation of the golden ratio.
Your task is to write a function/method that takes `t` as the only parameter and returns the corresponding estimation of the golden ratio. `t` is the estimation precision parameter, which is a float number between 0 and 1, usually very small, `1E-6` for example.
Note: make sure you test your code with different values of `t` and at least passes the following test cases:
- `t = 1E-6`
- `t = 1E-8`
- `t = 1E-12`
- `t = 1E-16`
The smaller `t` is, the more precise the estimation should be.
/**
* @Author: zhangQi
* @Date: 2020-07-21 10:47
*/
public class FibonacciTest {
//print
public static void main(String[] args) {
double[] bds = new double[4];
bds[0] = 1E-6;
bds[1] = 1E-8;
bds[2] = 1E-12;
bds[3] = 1E-16;
//test begin
for (int i = 0; i < bds.length; i++) {
double goldenRatio = getGoldenRatio(bds[i]);
System.out.println("the "+(i+1)+" return:"+goldenRatio);
}
}
//get golden ratio
public static double getGoldenRatio(double t) {
long theNumber = 0;
//n not be zero
for (long n = 1; n < Long.MAX_VALUE; n++) {
if (Math.abs(((fibonacci(n + 2) / fibonacci(n + 1)) - (fibonacci(n + 1) / fibonacci(n)))) < t) {
theNumber = n;
break;
}
}
System.out.println("the n:"+theNumber);
return fibonacci(theNumber + 1) / fibonacci(theNumber);
}
//get Fibonacci
static long fibonacci(long n) {
if ((n == 0) || (n == 1)) {
return n;
} else {
return fibonacci(n - 1) + fibonacci(n - 2);
}
}
}