 用java,c#,c++,golong最少的代码,求出结果 5C

请问小明要去上海,第一天步行了10000米,从第二天开始小明换了交通工具,以后的行程每天都是前一天的2倍加1000米,请各位大捞们用最简单的代码方式实现小明n天的行程(也就是n天走了多少米),提示到此,伙伴们发挥你们的想象吧,加油,面包会有的,java,c#,c++,golong各种语言都行
代码最简单，而且速度最快就是套公式
result = 10000*(2^(n1))+1000*(2^(n1)1)
化简
result = 11000*(2^(n1))1000
又因为2的n次方在一些语言中可以用向左移位表示。所以给一个C/C++语言最简单的（不考虑溢出问题，n超过30就要用大数计算了）
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
printf("%d", 11000*(1 << (n  1))1000);
}
public int Travel(int num){
int Alllength=10000;
for(int i=1;i<num;i++){
Alllength+=Alllength*2+1000
}
return Alllength;
}
public int Num(int N){
int DayNum=10000; //第一天走了多少米
int Num=DayNum;//N天走了多少米
for(int i=1;i<N;i++){
DayNum=DayNum*2+1000;// 第N天走了多少米
Num+=DayNum;//N天走了多少米
}
return Num;
}
import java.util.Scanner;
public class Hello {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
System.out.print("请输入行走天数：");
int countday=sc.nextInt();//行走天数
int i;//循环变量
long alllength=10000;//总的行走距离
long todaylength=alllength;//第几天走的距离,第1天走了10000米
System.out.println(countday);
for (i=2;i<=countday;i++) {
todaylength=2*todaylength+1000;
alllength+=todaylength;
System.out.println(i+";"+todaylength+";"+alllength);
}
System.out.println("第"+countday+"天行走了"+todaylength+"米；一共走了"+alllength+"米");
}
}
public static void main(String[] args) {
int x = 10000;
int t = 3;//设置天数
for(int i=1;i<t;i++)
{x=x*2+1000;}
System.out.println(t+"天走了"+x+"米");
}
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20190118Problem Description 输入一个百分制的成绩t，将其转换成对应的等级，具体转换规则如下： 90~100为A; 80~89为B; 70~79为C; 60~69为D; 0~59为E; Input 输入数据有多组，每组占一行，由一个整数组成。 Output 对于每组输入数据，输出一行。如果输入数据不在0~100范围内，请输出一行：“Score is error!”。 Sample Input 56 67 100 123 Sample Output E D A Score is error!
带有斜线方向的单词的连线的问题，怎么使用C语言的程序的编写之后的计算方式来实现查找_course
20190531Problem Description Anna Graham is a puzzle maker who prides herself in the quality and complexity of her work. She makes puzzles of all kinds  crosswords, logic problems, acrostics, and word search puzzles, to name but a few. For each puzzle, she has developed a set of rules which she constrains herself to follow. For word search puzzles, she insists not only that all the words be connected to one another (as in most word search puzzles), but also that removing any word from the word list will not cause one or more words to become disconnected from the rest. (Two words are connected if they contain a common letter in the grid.) The example word search puzzle on the left satisfies this condition, but the one on the right does not (removing the word Pascal from the word list disconnects Java from the other two words). Your job is to write a program that checks to see if Anna’s word search problems are up to snuff. Input Input will consist of multiple test cases. The first line of each test case contains 3 integers n m l, where n and m are the number of rows and columns in the puzzle and l is the number of words. Following this are n lines containing m uppercase characters each (the puzzle) followed by l lines containing one word each (the word list, in mixed case). Each word in the word list will appear in the puzzle exactly once. There will be no more than 100 rows and 100 columns in the puzzle and no more than 100 words to search for. There will be no spaces in the input words. Output For each problem instance, output the word Yes or No depending on whether the puzzle satisfies Anna’s constraints. Sample Input 5 6 3 PBROGR PASCAL ASMMIN GIICON TCELST BASIC LISP Pascal 5 6 4 PBROJR PASCAL ASMMVN GIICAN TCELST BASIC Java LISP Pascal 0 0 0 Sample Output Yes No
如何才能选择最小难度的路径的计算，具体如何使用C语言的程序的设计的代码的编写去解决这个问题_course
20190702Problem Description Little Timmy likes playgrounds, especially those with complicated constructions of wooden towers interconnected with plank bridges or ropes, with slides and ropeladders. He could play on his favourite playground for days, if only his parents let him. Sooner or later they decide that it is time to go home. For their next trip to a playground Timmy has a plan: he will not simply let his father grab him, but climb to the highest platform of the most complex structure and hide there. His father will never be able to reach him there or at least it will give Timmy some extra time. An adventure playground consists of several platforms. The difficulty of reaching a platform directly from ground level varies. In addition the different platforms are interconnected by “bridges” of different difficulty. There are connections that can be used a lot more easily in one direction than in the other, e.g. slides. Given a plan of an adventure playground you need to help Timmy find the platform that is most difficult to reach from ground level. The difficulty for a path in the adventure playground can be estimated as a sum of the difficulties of the connections used. The difficulty of reaching a platform is the difficulty of the least difficult path from ground level to the platform. Input The input starts with a line containing a single integer n, the number of test cases. Each test case starts with a line containing two integers 1 <= p, c <= 10000, the number of platforms and connections respectively (we allow for large playgrounds). A line with p integers pi ∈ [0, 1000] follows, where pi is the difficulty of reaching platform i directly from groundlevel. The next c lines each describe a connection between two platforms. They consist of four integers i, j, a, b; i < j, the zerobased index of the two connected platforms and the difficulties a ∈[0, 1000] and b ∈ [0, 1000] of using the connection to get from pi to pj and from pj to pi respectively. There can be multiple connections between any two platforms. Output The output for every test case begins with a line containing “Scenario #i:”, where i is the number of the test case counting from 1. Then, output a single line containing the zerobased index of the platform that is most difficult to reach from ground level. If two or more platforms are equally difficult to reach, output the smallest index of among those platforms. Terminate each test case with an empty line. Sample Input 2 4 3 1 10 10 40 0 3 3 2 1 3 4 3 2 3 5 4 2 2 11 10 0 1 2 2 0 1 1 0 Sample Output Scenario #1: 2 Scenario #2: 0
解码base64并将其编码为十六进制_course
20190708<div class="posttext" itemprop="text"> <p>I have following code which decodes base64 and then encodes it to hex.</p> <pre><code>doc_id := "Can35qPeFkm9Xgmp9+aj3g==" base64_decode, err := base64.StdEncoding.DecodeString(doc_id) if err != nil { log.Fatal("error:", err) } base64_decoded := fmt.Sprintf("%q", base64_decode) fmt.Printf("base_decoded %v ", base64_decoded) src := []byte(base64_decoded) fmt.Println(src) hex_encode := make([]byte, hex.EncodedLen(len(src))) hex.Encode(hex_encode, src) hex_encoded := fmt.Sprintf("%s", hex_encode) fmt.Printf("hex_encoded %v", hex_encoded) </code></pre> <p>where doc_id is base64 format. base64_decoded is its decoded value. I have to encode it to hex, so i pass it to src.</p> <p>The problem is when i pass the identifier base64_decoded to src i get wrong when i pass in the value that base64_decoded is holding i get correct answer. for example: if i get base64_decoded value as "\x11z\xc0[d~\xfcK\xb1\xf8\x11z\xc0[d~" if i pass its value which is "\x11z\xc0[d~\xfcK\xb1\xf8\x11z\xc0[d~", i get correct answer 117ac05b647efc4bb1f8117ac05b647e</p> <p>if i pass the variable holding "\x11z\xc0[d~\xfcK\xb1\xf8\x11z\xc0[d~" i get wrong answer 225c7831317a5c7863305b647e5c7866634b5c7862315c7866385c7831317a5c7863305b647e22dn</p> <p>Has it got something with this assignment <strong>base64_decoded := fmt.Sprintf("%q", base64_decode)</strong></p> <p>what am i doing wrong</p> </div>
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20160626课程主要面向嵌入式Linux初学者、工程师、学生 主要从一下几方面进行讲解： 1.linux学习路线、基本命令、高级命令 2.shell、vi及vim入门讲解 3.软件安装下载、NFS、Samba、FTP等服务器配置及使用
玩转Linux：常用命令实例指南
20190928人工智能、物联网、大数据时代，Linux正有着一统天下的趋势，几乎每个程序员岗位，都要求掌握Linux。本课程零基础也能轻松入门。 本课程以简洁易懂的语言手把手教你系统掌握日常所需的Linux知识，每个知识点都会配合案例实战让你融汇贯通。课程通俗易懂，简洁流畅，适合0基础以及对Linux掌握不熟练的人学习； 【限时福利】 1）购课后按提示添加小助手，进答疑群，还可获得价值300元的编程大礼包！ 2）本月购买此套餐加入老师答疑交流群，可参加老师的免费分享活动，学习最新技术项目经验。  29元=掌握Linux必修知识+社群答疑+讲师社群分享会+700元编程礼包。
150讲轻松搞定Python网络爬虫
20190516【为什么学爬虫？】 1、爬虫入手容易，但是深入较难，如何写出高效率的爬虫，如何写出灵活性高可扩展的爬虫都是一项技术活。另外在爬虫过程中，经常容易遇到被反爬虫，比如字体反爬、IP识别、验证码等，如何层层攻克难点拿到想要的数据，这门课程，你都能学到！ 2、如果是作为一个其他行业的开发者，比如app开发，web开发，学习爬虫能让你加强对技术的认知，能够开发出更加安全的软件和网站 【课程设计】 一个完整的爬虫程序，无论大小，总体来说可以分成三个步骤，分别是： 网络请求：模拟浏览器的行为从网上抓取数据。 数据解析：将请求下来的数据进行过滤，提取我们想要的数据。 数据存储：将提取到的数据存储到硬盘或者内存中。比如用mysql数据库或者redis等。 那么本课程也是按照这几个步骤循序渐进的进行讲解，带领学生完整的掌握每个步骤的技术。另外，因为爬虫的多样性，在爬取的过程中可能会发生被反爬、效率低下等。因此我们又增加了两个章节用来提高爬虫程序的灵活性，分别是： 爬虫进阶：包括IP代理，多线程爬虫，图形验证码识别、JS加密解密、动态网页爬虫、字体反爬识别等。 Scrapy和分布式爬虫：Scrapy框架、Scrapyredis组件、分布式爬虫等。 通过爬虫进阶的知识点我们能应付大量的反爬网站，而Scrapy框架作为一个专业的爬虫框架，使用他可以快速提高我们编写爬虫程序的效率和速度。另外如果一台机器不能满足你的需求，我们可以用分布式爬虫让多台机器帮助你快速爬取数据。 从基础爬虫到商业化应用爬虫，本套课程满足您的所有需求！ 【课程服务】 专属付费社群+每周三讨论会+1v1答疑
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