A hard puzzle

Problem Description
lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.

Input
There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)

Output
For each test case, you should output the a^b's last digit number.

Sample Input
7 66
8 800

Sample Output
9
6

4个回答

Yes of cause the upstair solution is OK, but if you want to know why the number like that in array "sam" like that, you can reference my LastBitNumber function.
This function is applied to any modulo not only 10, I've test my program just now, and fix some bug, you can press enter and press controll + Z to end the input. here is the program

#include "stdafx.h"
#include
#include

using namespace std;

// actually it can calculate any modulo last bit, here called this function with modulo = 10
unsigned int Fun_LastBitNumber(unsigned int a, long int exponet, unsigned int modulo)
{
long int i = 0;
unsigned int remainder = 0, temp = 1, remainder_a = 0;

remainder_a = a % modulo;

for (i = 0; i < exponet; i++)
{
    temp *= remainder_a;
    remainder = temp % modulo;
    temp = remainder;
}

return remainder;

}

int _tmain(int argc, _TCHAR* argv[])
{
unsigned int a = 0, lastbitNumber = 0, modulo = 10;// binary octal hexadecimal are also OK;
long int exponet = 0;
std::vector vecLastBitNum;
std::vector::iterator itervecLastBitNum;

while(cin >> a >> exponet)
{
    lastbitNumber = Fun_LastBitNumber(a, exponet, modulo);
    vecLastBitNum.push_back(lastbitNumber);
}

//Output
itervecLastBitNum = vecLastBitNum.begin();
while( itervecLastBitNum != vecLastBitNum.end() )
{
    cout << *itervecLastBitNum << endl;
    itervecLastBitNum++;
}
return 0;

}

You'd better to understand knowledge about polynomial or number theory first.
Here is the program for your reference.
#include "stdafx.h"
#include
#include

using namespace std;

// actually it can calculate any modulo last bit, here called this function with modulo = 10
unsigned int Fun_LastBitNumber(unsigned int a, long int exponet, unsigned int modulo)
{
long int i = 0;
unsigned int remainder = 0, temp = 1, remainder_a = 0;

remainder_a = a % modulo;

for (i = 0; i < exponet; i++)
{
    temp *= remainder_a;
    remainder = temp % modulo;
    temp = remainder;
}

return remainder;

}

int _tmain(int argc, _TCHAR* argv[])
{
unsigned int a = 0, lastbitNumber = 0, modulo = 10;// binary octal hexadecimal are also OK;
long int exponet = 0;
std::vector vecLastBitNum;
std::vector::iterator itervecLastBitNum;

while(1)
{
    cin >> a;
    cin >> exponet;
    lastbitNumber = Fun_LastBitNumber(a, exponet, modulo);
    vecLastBitNum.push_back(lastbitNumber);
}

//Output
itervecLastBitNum = vecLastBitNum.begin();
while( itervecLastBitNum != vecLastBitNum.end() )
    cout << *itervecLastBitNum;

cout << endl;
return 0;

}

you can add any break condition in the while(1) to end of the input,
and obviously the 2 include files are iotream and vector because "<>"can not display

Csdn user default icon
上传中...
上传图片
插入图片
抄袭、复制答案,以达到刷声望分或其他目的的行为,在CSDN问答是严格禁止的,一经发现立刻封号。是时候展现真正的技术了!
其他相关推荐
A hard puzzle 是怎么来写得
Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin. this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise. Input There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30) Output For each test case, you should output the a^b's last digit number. Sample Input 7 66 8 800 Sample Output 9 6
Equation of XOR 异或的问题
Problem Description Recently, Jimmy is learning about linear algebra from Blue Mary while having the course of Boolean algebra in class offered by Prof. Z. Since Jimmy has been thoroughly bored by the boring homework assigned by two teachers, evil Jimmy plans to set a hard question to baffle them as revenge for their heavy tasks. As a result, Jimmy comes up with an idea that merging the knowledge from both the two classes and constructs a complicate problem: the XOR equation system. Let’s consider the following equations: (a11 . x1) ^ (a12 . x2) ^ … ^ (a1m . xm) = 0 (a21 . x1) ^ (a22 . x2) ^ … ^ (a2m . xm) = 0 … (an1 . x1) ^ (an2 . x2) ^ … ^ (anm . xm) = 0 which satisfies the following conditions: 1. aij in {0,1} for 1 ≤ i ≤ n and 1 ≤ j ≤ m; 2. xi in Si where Si is a subset of {0,1,2,3}, 1 ≤ i ≤ m; 3. |Si| ≤ 3, 1 ≤ i ≤ m; 4. 1≤n ≤ 30, 1 ≤ m ≤ 22. In the system of equations, operation “ . “ denotes the multiplication operation while “ ^ ” is for bitwise XOR. Moreover, the bitwise XOR takes two bit patterns of equal length and performs the logical XOR operation on each pair of corresponding bits. The result in each position is 1 if the two bits are different, and 0 if they are the same. Rather than expecting a solution of a specified equation system, Jimmy would like to ask the teachers to calculate that how many distinct solutions can satisfy a given equation system. What a confusing puzzle! Help Jimmy’s teachers please! Input There are several test cases. The first line of input is a single positive integer T (<= 15) indicating the number of test cases, then T cases follow. For each test case, the first line contains two integers N and M giving the two dimensions of the equation system respectively where N is the number of rows and M for columns. Then N lines are following, each line contains m integers. Item at line i and column j represents aij. The next m lines are descriptions of Si that the leading integer K denotes the number of elements in Si and the following K integers represent elements. Output For each test case, output one line containing a single integer which represents the answer of this case. Sample Input 1 1 2 1 1 3 0 1 2 2 3 1 Sample Output 1
Very Hard Problem
It was in the ancient world. ZOJ, the greatest treasure hunter in the world has been lost in the forest for more than 100 hours. Being with no food and no water for such a long time, he was really exhausted this night. He then took out an old map, which was a very strange map. There are many stared positions on the map and it seems those stars are connected by some roads. "I'm sure the treasure is near me, but where is it?", said ZOJ, "If I can't find it, I will be laughed by others." Suddenly, ZOJ noticed a slight light. "It was unsual.", said ZOJ and he started looking for the treasure again. After three hours' search, he finally found an entrance. But to enter the entrance, a puzzle should be solved. The puzzle was described like this. Every time, you were given a character in the set {'-', '!', '~'} and a b-based number. You should take the character as an operator (i.e. '-' changes a number to its opposite number, '!' changes zero to one and non-zero value to zero, '~' takes bitwise operation NOT on all 64 bits of a number) and operates on the number. All you need to do is to print out the result. The puzzle seemed quite simple, but ZOJ was only good at working out the output of the programs written by others, and had no idea about how to solve such a problem. So he turned to you for help. Input There are multiple test cases. In each test case, a character ch, a number b and a number n in b-based (2 ≤ b ≤ 16, when b is no less than 10, 'a'..'f' or 'A'..'F' are used) are given in order in one line. ch is assured to be in the set {'-', '!', '~'}, the number (n)b is assured in the range of a signed 64-bit integer. ch, b, n are seperated by one or more spaces. Leading or trailing spaces may also be added to the lines. Output For each test case, print one line, the 10-based result. Sample Input ~ 10 4 ! 10 0 - 16 -F Sample Output -5 1 15
Jugs
Description In the movie "Die Hard 3", Bruce Willis and Samuel L. Jackson were confronted with the following puzzle. They were given a 3-gallon jug and a 5-gallon jug and were asked to fill the 5-gallon jug with exactly 4 gallons. This problem generalizes that puzzle. You have two jugs, A and B, and an infinite supply of water. There are three types of actions that you can use: (1) you can fill a jug, (2) you can empty a jug, and (3) you can pour from one jug to the other. Pouring from one jug to the other stops when the first jug is empty or the second jug is full, whichever comes first. For example, if A has 5 gallons and B has 6 gallons and a capacity of 8, then pouring from A to B leaves B full and 3 gallons in A. A problem is given by a triple (Ca,Cb,N), where Ca and Cb are the capacities of the jugs A and B, respectively, and N is the goal. A solution is a sequence of steps that leaves exactly N gallons in jug B. The possible steps are fill A fill B empty A empty B pour A B pour B A success where "pour A B" means "pour the contents of jug A into jug B", and "success" means that the goal has been accomplished. You may assume that the input you are given does have a solution. Input Input to your program consists of a series of input lines each defining one puzzle. Input for each puzzle is a single line of three positive integers: Ca, Cb, and N. Ca and Cb are the capacities of jugs A and B, and N is the goal. You can assume 0 < Ca <= Cb and N <= Cb <=1000 and that A and B are relatively prime to one another. Output Output from your program will consist of a series of instructions from the list of the potential output lines which will result in either of the jugs containing exactly N gallons of water. The last line of output for each puzzle should be the line "success". Output lines start in column 1 and there should be no empty lines nor any trailing spaces. Sample Input 3 5 4 5 7 3 Sample Output fill B pour B A empty A pour B A fill B pour B A success fill A pour A B fill A pour A B empty B pour A B success
Crazy Search
Many people like to solve hard puzzles some of which may lead them to madness. One such puzzle could be finding a hidden prime number in a given text. Such number could be the number of different substrings of a given size that exist in the text. As you soon will discover, you really need the help of a computer and a good algorithm to solve such a puzzle. Your task is to write a program that given the size, N, of the substring, the number of different characters that may occur in the text, NC, and the text itself, determines the number of different substrings of size N that appear in the text. As an example, consider N=3, NC=4 and the text "daababac". The different substrings of size 3 that can be found in this text are: "daa", "aab", "aba", "bab", "bac". Therefore, the answer should be 5. Input The first line of input consists of two numbers, N and NC, separated by exactly one space. This is followed by the text where the search takes place. You may assume that the maximum number of substrings formed by the possible set of characters does not exceed 16 Millions. Output The program should output just an integer corresponding to the number of different substrings of size N found in the given text. This problem contains multiple test cases! The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks. The output format consists of N output blocks. There is a blank line between output blocks. Sample Input 1 3 4 daababac Sample Output 5
Word-Search Wonder
Description The Pyrates Restaurant was starting to fill up as Valentine McKee walked in. She scanned the crowd for her sister, brother-in-law, and nephew. Seeing her sister waving from the far end of the restaurant, she made her way back to their booth. ``Hi, Valentine,'' her sister and brother-in-law, Niki and Dennis Chapman, greeted her. ``Hi, guys,'' she replied. ``What are you doing, Wade?'' she asked her nephew. He was busy working on one of the restaurant's activity sheets with a crayon. ``I'm doing a word search game,'' Wade explained. ``I have to find all of these words in this big mess of letters. This is really hard.'' Wade looked intently at the paper in front of him. ``Can I help?'' asked Valentine, looking across the table at the activity sheet. ``Sure. These are the words we're looking for. They're the names of different kinds of Planes, Trains, and Automobiles.'' Input The first line of input will specify the length (in characters) of the sides of the letter matrix (the matrix of letters will be square). The length, l, will be in the range 1 <= l <= 100. The next l lines of input will be the matrix itself, each line will contain l uppercase letters. A list of words will follow. Each word will be on a line by itself; there will be 100 or fewer words. Each word will be 100 or fewer characters long, and will only contain uppercase letters. The final line of input will contain a single zero character. Output Your program should attempt to find each word from the word list in the puzzle. A word is ``found'' if all the characters in the word can be traced in a single (unidirectional) horizontal, vertical, or diagonal line in the letter matrix. Words may not ``wrap around'' rows or columns, but horizontal and diagonal words may proceed from right to left (``backwards''). For each word that is found, your program should print the coordinates of its first and last letters in the matrix on a single line, separated by a single space. Coordinates are pairs of comma-separated integers (indexed from 1), where the first integer specifies the row number and the second integer specifies the column number. If a word is not found, the string ``Not found'' should be output instead of a pair of coordinates. Each word from the input can be ``found'' at most once in the puzzle. Sample Input 5 EDEEE DISKE ESEEE ECEEE EEEEE DISC DISK DISP 0 Sample Output 1,2 4,2 2,1 2,4 Not found
一个拖动拼图游戏的代码小问题想不通
html代码: <!doctype html> <html> <head> <title>Sliding Puzzle</title> <style> .picture { border: 1px solid black; } </style> </head> <body> <div id="title"> <h2>Sliding Puzzle</h2> </div> <div id="slider"> <form> <label>Easy</label> <input type="range" id="scale" value="4" min="3" max="5" step="1"> <label>Hard</label> </form> <br/> </div> <div id="main" class="main"> <canvas id="puzzle" width="480px" height="480px"></canvas> </div> <script src="test04.js"></script> </body> </html> js代码:var context = document.getElementById('puzzle').getContext('2d'); var img = new Image(); img.src = './img/dimetrodon.jpg'; img.addEventListener('load', drawTiles, false); var boardSize = document.getElementById('puzzle').width; var tileCount = document.getElementById('scale').value; var tileSize = boardSize / tileCount; var clickLoc = new Object; clickLoc.x = 0; clickLoc.y = 0; var emptyLoc = new Object; emptyLoc.x = 0; emptyLoc.y = 0; var solved = false; var boardParts = new Object; setBoard(); document.getElementById('scale').onchange = function() { tileCount = this.value; tileSize = boardSize / tileCount; setBoard(); drawTiles(); }; document.getElementById('puzzle').onmousemove = function(e) { clickLoc.x = Math.floor((e.pageX - this.offsetLeft) / tileSize); clickLoc.y = Math.floor((e.pageY - this.offsetTop) / tileSize); }; document.getElementById('puzzle').onclick = function() { if (distance(clickLoc.x, clickLoc.y, emptyLoc.x, emptyLoc.y) == 1) { slideTile(emptyLoc, clickLoc); drawTiles(); } if (solved) { setTimeout(function() {alert("You solved it!");}, 500); } }; function setBoard() { boardParts = new Array(tileCount); for (var i = 0; i < tileCount; ++i) { boardParts[i] = new Array(tileCount); for (var j = 0; j < tileCount; ++j) { boardParts[i][j] = new Object; boardParts[i][j].x = (tileCount - 1) - i; boardParts[i][j].y = (tileCount - 1) - j; } } emptyLoc.x = boardParts[tileCount - 1][tileCount - 1].x; emptyLoc.y = boardParts[tileCount - 1][tileCount - 1].y; solved = false; } function drawTiles() { context.clearRect ( 0 , 0 , boardSize , boardSize ); for (var i = 0; i < tileCount; ++i) { for (var j = 0; j < tileCount; ++j) { var x = boardParts[i][j].x; var y = boardParts[i][j].y; if(i != emptyLoc.x || j != emptyLoc.y || solved == true) { context.drawImage(img, x * tileSize, y * tileSize, tileSize, tileSize, i * tileSize, j * tileSize, tileSize, tileSize); } } } } function distance(x1, y1, x2, y2) { return Math.abs(x1 - x2) + Math.abs(y1 - y2); } function slideTile(toLoc, fromLoc) { if (!solved) { boardParts[toLoc.x][toLoc.y].x = boardParts[fromLoc.x][fromLoc.y].x; boardParts[toLoc.x][toLoc.y].y = boardParts[fromLoc.x][fromLoc.y].y; boardParts[fromLoc.x][fromLoc.y].x = tileCount - 1; boardParts[fromLoc.x][fromLoc.y].y = tileCount - 1; // var a = toLoc.x; // var b = toLoc.y; toLoc.x = fromLoc.x; toLoc.y = fromLoc.y; // fromLoc.x = a; // fromLoc.y = b;//这边为什么不用给emptyLoc改变坐标? checkSolved(); } } function checkSolved() { var flag = true; for (var i = 0; i < tileCount; ++i) { for (var j = 0; j < tileCount; ++j) { if (boardParts[i][j].x != i || boardParts[i][j].y != j) { flag = false; } } } solved = flag; } 请问为什么js中的emptyLoc的x和y不需要更新,或者他的x和y是怎么更新的,他代码里面emptyLoc.x和emptyLoc.y的值貌似都是2,没有变化,不更新的话没办法继续判断距离distance==1啊,为什么它程序可以正确运行,还是我理解错了?
爬虫福利二 之 妹子图网MM批量下载
爬虫福利一:27报网MM批量下载 点击 看了本文,相信大家对爬虫一定会产生强烈的兴趣,激励自己去学习爬虫,在这里提前祝:大家学有所成! 目标网站:妹子图网 环境:Python3.x 相关第三方模块:requests、beautifulsoup4 Re:各位在测试时只需要将代码里的变量path 指定为你当前系统要保存的路径,使用 python xxx.py 或IDE运行即可。 ...
Java学习的正确打开方式
在博主认为,对于入门级学习java的最佳学习方法莫过于视频+博客+书籍+总结,前三者博主将淋漓尽致地挥毫于这篇博客文章中,至于总结在于个人,实际上越到后面你会发现学习的最好方式就是阅读参考官方文档其次就是国内的书籍,博客次之,这又是一个层次了,这里暂时不提后面再谈。博主将为各位入门java保驾护航,各位只管冲鸭!!!上天是公平的,只要不辜负时间,时间自然不会辜负你。 何谓学习?博主所理解的学习,它是一个过程,是一个不断累积、不断沉淀、不断总结、善于传达自己的个人见解以及乐于分享的过程。
大学四年自学走来,这些私藏的实用工具/学习网站我贡献出来了
大学四年,看课本是不可能一直看课本的了,对于学习,特别是自学,善于搜索网上的一些资源来辅助,还是非常有必要的,下面我就把这几年私藏的各种资源,网站贡献出来给你们。主要有:电子书搜索、实用工具、在线视频学习网站、非视频学习网站、软件下载、面试/求职必备网站。 注意:文中提到的所有资源,文末我都给你整理好了,你们只管拿去,如果觉得不错,转发、分享就是最大的支持了。 一、电子书搜索 对于大部分程序员...
linux系列之常用运维命令整理笔录
本博客记录工作中需要的linux运维命令,大学时候开始接触linux,会一些基本操作,可是都没有整理起来,加上是做开发,不做运维,有些命令忘记了,所以现在整理成博客,当然vi,文件操作等就不介绍了,慢慢积累一些其它拓展的命令,博客不定时更新 free -m 其中:m表示兆,也可以用g,注意都要小写 Men:表示物理内存统计 total:表示物理内存总数(total=used+free) use...
比特币原理详解
一、什么是比特币 比特币是一种电子货币,是一种基于密码学的货币,在2008年11月1日由中本聪发表比特币白皮书,文中提出了一种去中心化的电子记账系统,我们平时的电子现金是银行来记账,因为银行的背后是国家信用。去中心化电子记账系统是参与者共同记账。比特币可以防止主权危机、信用风险。其好处不多做赘述,这一层面介绍的文章很多,本文主要从更深层的技术原理角度进行介绍。 二、问题引入 假设现有4个人...
程序员接私活怎样防止做完了不给钱?
首先跟大家说明一点,我们做 IT 类的外包开发,是非标品开发,所以很有可能在开发过程中会有这样那样的需求修改,而这种需求修改很容易造成扯皮,进而影响到费用支付,甚至出现做完了项目收不到钱的情况。 那么,怎么保证自己的薪酬安全呢? 我们在开工前,一定要做好一些证据方面的准备(也就是“讨薪”的理论依据),这其中最重要的就是需求文档和验收标准。一定要让需求方提供这两个文档资料作为开发的基础。之后开发...
网页实现一个简单的音乐播放器(大佬别看。(⊙﹏⊙))
今天闲着无事,就想写点东西。然后听了下歌,就打算写个播放器。 于是乎用h5 audio的加上js简单的播放器完工了。 演示地点演示 html代码如下` music 这个年纪 七月的风 音乐 ` 然后就是css`*{ margin: 0; padding: 0; text-decoration: none; list-...
Python十大装B语法
Python 是一种代表简单思想的语言,其语法相对简单,很容易上手。不过,如果就此小视 Python 语法的精妙和深邃,那就大错特错了。本文精心筛选了最能展现 Python 语法之精妙的十个知识点,并附上详细的实例代码。如能在实战中融会贯通、灵活使用,必将使代码更为精炼、高效,同时也会极大提升代码B格,使之看上去更老练,读起来更优雅。
数据库优化 - SQL优化
以实际SQL入手,带你一步一步走上SQL优化之路!
2019年11月中国大陆编程语言排行榜
2019年11月2日,我统计了某招聘网站,获得有效程序员招聘数据9万条。针对招聘信息,提取编程语言关键字,并统计如下: 编程语言比例 rank pl_ percentage 1 java 33.62% 2 cpp 16.42% 3 c_sharp 12.82% 4 javascript 12.31% 5 python 7.93% 6 go 7.25% 7 p...
通俗易懂地给女朋友讲:线程池的内部原理
餐盘在灯光的照耀下格外晶莹洁白,女朋友拿起红酒杯轻轻地抿了一小口,对我说:“经常听你说线程池,到底线程池到底是个什么原理?”
经典算法(5)杨辉三角
杨辉三角 是经典算法,这篇博客对它的算法思想进行了讲解,并有完整的代码实现。
腾讯算法面试题:64匹马8个跑道需要多少轮才能选出最快的四匹?
昨天,有网友私信我,说去阿里面试,彻底的被打击到了。问了为什么网上大量使用ThreadLocal的源码都会加上private static?他被难住了,因为他从来都没有考虑过这个问题。无独有偶,今天笔者又发现有网友吐槽了一道腾讯的面试题,我们一起来看看。 腾讯算法面试题:64匹马8个跑道需要多少轮才能选出最快的四匹? 在互联网职场论坛,一名程序员发帖求助到。二面腾讯,其中一个算法题:64匹...
面试官:你连RESTful都不知道我怎么敢要你?
干货,2019 RESTful最贱实践
JDK12 Collectors.teeing 你真的需要了解一下
前言 在 Java 12 里面有个非常好用但在官方 JEP 没有公布的功能,因为它只是 Collector 中的一个小改动,它的作用是 merge 两个 collector 的结果,这句话显得很抽象,老规矩,我们先来看个图(这真是一个不和谐的图????): 管道改造经常会用这个小东西,通常我们叫它「三通」,它的主要作用就是将 downstream1 和 downstre...
为啥国人偏爱Mybatis,而老外喜欢Hibernate/JPA呢?
关于SQL和ORM的争论,永远都不会终止,我也一直在思考这个问题。昨天又跟群里的小伙伴进行了一番讨论,感触还是有一些,于是就有了今天这篇文。 声明:本文不会下关于Mybatis和JPA两个持久层框架哪个更好这样的结论。只是摆事实,讲道理,所以,请各位看官勿喷。 一、事件起因 关于Mybatis和JPA孰优孰劣的问题,争论已经很多年了。一直也没有结论,毕竟每个人的喜好和习惯是大不相同的。我也看...
SQL-小白最佳入门sql查询一
不要偷偷的查询我的个人资料,即使你再喜欢我,也不要这样,真的不好;
项目中的if else太多了,该怎么重构?
介绍 最近跟着公司的大佬开发了一款IM系统,类似QQ和微信哈,就是聊天软件。我们有一部分业务逻辑是这样的 if (msgType = "文本") { // dosomething } else if(msgType = "图片") { // doshomething } else if(msgType = "视频") { // doshomething } else { // doshom...
【图解经典算法题】如何用一行代码解决约瑟夫环问题
约瑟夫环问题算是很经典的题了,估计大家都听说过,然后我就在一次笔试中遇到了,下面我就用 3 种方法来详细讲解一下这道题,最后一种方法学了之后保证让你可以让你装逼。 问题描述:编号为 1-N 的 N 个士兵围坐在一起形成一个圆圈,从编号为 1 的士兵开始依次报数(1,2,3…这样依次报),数到 m 的 士兵会被杀死出列,之后的士兵再从 1 开始报数。直到最后剩下一士兵,求这个士兵的编号。 1、方...
致 Python 初学者
欢迎来到“Python进阶”专栏!来到这里的每一位同学,应该大致上学习了很多 Python 的基础知识,正在努力成长的过程中。在此期间,一定遇到了很多的困惑,对未来的学习方向感到迷茫。我非常理解你们所面临的处境。我从2007年开始接触 python 这门编程语言,从2009年开始单一使用 python 应对所有的开发工作,直至今天。回顾自己的学习过程,也曾经遇到过无数的困难,也曾经迷茫过、困惑过。开办这个专栏,正是为了帮助像我当年一样困惑的 Python 初学者走出困境、快速成长。希望我的经验能真正帮到你
“狗屁不通文章生成器”登顶GitHub热榜,分分钟写出万字形式主义大作
一、垃圾文字生成器介绍 最近在浏览GitHub的时候,发现了这样一个骨骼清奇的雷人项目,而且热度还特别高。 项目中文名:狗屁不通文章生成器 项目英文名:BullshitGenerator 根据作者的介绍,他是偶尔需要一些中文文字用于GUI开发时测试文本渲染,因此开发了这个废话生成器。但由于生成的废话实在是太过富于哲理,所以最近已经被小伙伴们给玩坏了。 他的文风可能是这样的: 你发现,...
程序员:我终于知道post和get的区别
是一个老生常谈的话题,然而随着不断的学习,对于以前的认识有很多误区,所以还是需要不断地总结的,学而时习之,不亦说乎
GitHub标星近1万:只需5秒音源,这个网络就能实时“克隆”你的声音
作者 | Google团队 译者 | 凯隐 编辑 | Jane 出品 | AI科技大本营(ID:rgznai100) 本文中,Google 团队提出了一种文本语音合成(text to speech)神经系统,能通过少量样本学习到多个不同说话者(speaker)的语音特征,并合成他们的讲话音频。此外,对于训练时网络没有接触过的说话者,也能在不重新训练的情况下,仅通过未知...
《程序人生》系列-这个程序员只用了20行代码就拿了冠军
你知道的越多,你不知道的越多 点赞再看,养成习惯GitHub上已经开源https://github.com/JavaFamily,有一线大厂面试点脑图,欢迎Star和完善 前言 这一期不算《吊打面试官》系列的,所有没前言我直接开始。 絮叨 本来应该是没有这期的,看过我上期的小伙伴应该是知道的嘛,双十一比较忙嘛,要值班又要去帮忙拍摄年会的视频素材,还得搞个程序员一天的Vlog,还要写BU...
加快推动区块链技术和产业创新发展,2019可信区块链峰会在京召开
11月8日,由中国信息通信研究院、中国通信标准化协会、中国互联网协会、可信区块链推进计划联合主办,科技行者协办的2019可信区块链峰会将在北京悠唐皇冠假日酒店开幕。   区块链技术被认为是继蒸汽机、电力、互联网之后,下一代颠覆性的核心技术。如果说蒸汽机释放了人类的生产力,电力解决了人类基本的生活需求,互联网彻底改变了信息传递的方式,区块链作为构造信任的技术有重要的价值。   1...
程序员把地府后台管理系统做出来了,还有3.0版本!12月7号最新消息:已在开发中有github地址
第一幕:缘起 听说阎王爷要做个生死簿后台管理系统,我们派去了一个程序员…… 996程序员做的梦: 第一场:团队招募 为了应对地府管理危机,阎王打算找“人”开发一套地府后台管理系统,于是就在地府总经办群中发了项目需求。 话说还是中国电信的信号好,地府都是满格,哈哈!!! 经常会有外行朋友问:看某网站做的不错,功能也简单,你帮忙做一下? 而这次,面对这样的需求,这个程序员...
网易云6亿用户音乐推荐算法
网易云音乐是音乐爱好者的集聚地,云音乐推荐系统致力于通过 AI 算法的落地,实现用户千人千面的个性化推荐,为用户带来不一样的听歌体验。 本次分享重点介绍 AI 算法在音乐推荐中的应用实践,以及在算法落地过程中遇到的挑战和解决方案。 将从如下两个部分展开: AI算法在音乐推荐中的应用 音乐场景下的 AI 思考 从 2013 年 4 月正式上线至今,网易云音乐平台持续提供着:乐屏社区、UGC...
【技巧总结】位运算装逼指南
位算法的效率有多快我就不说,不信你可以去用 10 亿个数据模拟一下,今天给大家讲一讲位运算的一些经典例子。不过,最重要的不是看懂了这些例子就好,而是要在以后多去运用位运算这些技巧,当然,采用位运算,也是可以装逼的,不信,你往下看。我会从最简单的讲起,一道比一道难度递增,不过居然是讲技巧,那么也不会太难,相信你分分钟看懂。 判断奇偶数 判断一个数是基于还是偶数,相信很多人都做过,一般的做法的代码如下...
【管理系统课程设计】美少女手把手教你后台管理
【文章后台管理系统】URL设计与建模分析+项目源码+运行界面 栏目管理、文章列表、用户管理、角色管理、权限管理模块(文章最后附有源码) 1. 这是一个什么系统? 1.1 学习后台管理系统的原因 随着时代的变迁,现如今各大云服务平台横空出世,市面上有许多如学生信息系统、图书阅读系统、停车场管理系统等的管理系统,而本人家里就有人在用烟草销售系统,直接在网上完成挑选、购买与提交收货点,方便又快捷。 试想,若没有烟草销售系统,本人家人想要购买烟草,还要独自前往药...
4G EPS 第四代移动通信系统
目录 文章目录目录4G 与 LTE/EPCLTE/EPC 的架构E-UTRANE-UTRAN 协议栈eNodeBEPCMMES-GWP-GWHSSLTE/EPC 协议栈概览 4G 与 LTE/EPC 4G,即第四代移动通信系统,提供了 3G 不能满足的无线网络宽带化,主要提供数据(上网)业务。而 LTE(Long Term Evolution,长期演进技术)是电信领域用于手机及数据终端的高速无线通...
日均350000亿接入量,腾讯TubeMQ性能超过Kafka
整理 | 夕颜出品 | AI科技大本营(ID:rgznai100)【导读】近日,腾讯开源动作不断,相继开源了分布式消息中间件TubeMQ,基于最主流的 OpenJDK8开发的Tencent Kona JDK,分布式HTAP数据库 TBase,企业级容器平台TKEStack,以及高性能图计算框架Plato。短短一周之内,腾讯开源了五大重点项目。其中,TubeMQ是腾讯大数据平台部门应用的核心组件,...
8年经验面试官详解 Java 面试秘诀
作者 |胡书敏 责编 | 刘静 出品 | CSDN(ID:CSDNnews) 本人目前在一家知名外企担任架构师,而且最近八年来,在多家外企和互联网公司担任Java技术面试官,前后累计面试了有两三百位候选人。在本文里,就将结合本人的面试经验,针对Java初学者、Java初级开发和Java开发,给出若干准备简历和准备面试的建议。 Java程序员准备和投递简历的实...
面试官如何考察你的思维方式?
1.两种思维方式在求职面试中,经常会考察这种问题:北京有多少量特斯拉汽车?某胡同口的煎饼摊一年能卖出多少个煎饼?深圳有多少个产品经理?一辆公交车里能装下多少个乒乓球?一个正常成年人有多少根头发?这类估算问题,被称为费米问题,是以科学家费米命名的。为什么面试会问这种问题呢?这类问题能把两类人清楚地区分出来。一类是具有文科思维的人,擅长赞叹和模糊想象,它主要依靠的是人的第一反应和直觉,比如小孩...
so easy! 10行代码写个"狗屁不通"文章生成器
前几天,GitHub 有个开源项目特别火,只要输入标题就可以生成一篇长长的文章。 背后实现代码一定很复杂吧,里面一定有很多高深莫测的机器学习等复杂算法 不过,当我看了源代码之后 这程序不到50行 尽管我有多年的Python经验,但我竟然一时也没有看懂 当然啦,原作者也说了,这个代码也是在无聊中诞生的,平时撸码是不写中文变量名的, 中文...
知乎高赞:中国有什么拿得出手的开源软件产品?(整理自本人原创回答)
知乎高赞:中国有什么拿得出手的开源软件产品? 在知乎上,有个问题问“中国有什么拿得出手的开源软件产品(在 GitHub 等社区受欢迎度较好的)?” 事实上,还不少呢~ 本人于2019.7.6进行了较为全面的回答,对这些受欢迎的 Github 开源项目分类整理如下: 分布式计算、云平台相关工具类 1.SkyWalking,作者吴晟、刘浩杨 等等 仓库地址: apache/skywalking 更...
MySQL数据库总结
一、数据库简介 数据库(Database,DB)是按照数据结构来组织,存储和管理数据的仓库。 典型特征:数据的结构化、数据间的共享、减少数据的冗余度,数据的独立性。 关系型数据库:使用关系模型把数据组织到数据表(table)中。现实世界可以用数据来描述。 主流的关系型数据库产品:Oracle(Oracle)、DB2(IBM)、SQL Server(MS)、MySQL(Oracle)。 数据表:数...
20行Python代码爬取王者荣耀全英雄皮肤
引言 王者荣耀大家都玩过吧,没玩过的也应该听说过,作为时下最火的手机MOBA游戏,咳咳,好像跑题了。我们今天的重点是爬取王者荣耀所有英雄的所有皮肤,而且仅仅使用20行Python代码即可完成。 准备工作 爬取皮肤本身并不难,难点在于分析,我们首先得得到皮肤图片的url地址,话不多说,我们马上来到王者荣耀的官网: 我们点击英雄资料,然后随意地选择一位英雄,接着F12打开调试台,找到英雄原皮肤的图片...
张小龙-年薪近3亿的微信之父,他是如何做到的?
张小龙生于湖南邵东魏家桥镇, 家庭主要特点:穷。 不仅自己穷,亲戚也都很穷,可以说穷以类聚。爷爷做过铜匠,总的来说,标准的劳动阶级出身。 家有兄弟两人, 一个小龙,一个小虎。 小虎好动,与邻里打成一片, 小龙好静,喜好读书。 “文静的像个妹子。”张小龙的表哥如是说。 穷文富武,做个读书郎是个不错的选择。 87年至94年, 华中科技大学本硕连读。 本科就读电信系, 不喜欢上课...
阿里靠什么武功秘籍渡过“双十一“的天量冲击
双十一大概会产生多大的数据量呢,可能大家没概念,举个例子央视拍了这么多年电视新闻节目,几十年下来他存了大概80P的数据。而今年双11一天,阿里要处理970P的数据,做为一个IT人,笔者认为今年”双十一“阿里最大的技术看点有有以下两个: 阿里的数据库,也就是刚刚拿下TPC冠军的OcceanBase,处理峰值也达到了骇人听闻的6100万次/秒, 阿里核心系统百分百上云了。 如果把信息系统比做一个武...
西游记团队中如果需要裁掉一个人,会先裁掉谁?
2019年互联网寒冬,大批企业开始裁员,下图是网上流传的一张截图: 裁员不可避免,那如何才能做到不管大环境如何变化,自身不受影响呢? 我们先来看一个有意思的故事,如果西游记取经团队需要裁员一名,会裁掉谁呢,为什么? 西游记团队组成: 1.唐僧 作为团队teamleader,有很坚韧的品性和极高的原则性,不达目的不罢休,遇到任何问题,都没有退缩过,又很得上司支持和赏识(直接得到唐太宗的任命,既给袈...
iOS Bug 太多,苹果终于坐不住了!
开源的 Android 和闭源的 iOS,作为用户的你,更偏向哪一个呢? 整理 | 屠敏 出品 | CSDN(ID:CSDNnews) 毋庸置疑,当前移动设备操作系统市场中,Android 和 iOS 作为两大阵营,在相互竞争的同时不断演进。不过一直以来,开源的 Android 吸引了无数的手机厂商涌入其中,为其生态带来了百花齐放的盛景,但和神秘且闭源的 iOS 系统相比,不少网友...
究竟你适不适合买Mac?
我清晰的记得,刚买的macbook pro回到家,开机后第一件事情,就是上了淘宝网,花了500元钱,找了一个上门维修电脑的师傅,上门给我装了一个windows系统。。。。。。 表砍我。。。 当时买mac的初衷,只是想要个固态硬盘的笔记本,用来运行一些复杂的扑克软件。而看了当时所有的SSD笔记本后,最终决定,还是买个好(xiong)看(da)的。 已经有好几个朋友问我mba怎么样了,所以今天尽量客观...
相关热词 c#处理浮点数 c# 生成字母数字随机数 c# 动态曲线 控件 c# oracle 开发 c#选择字体大小的控件 c# usb 批量传输 c#10进制转8进制 c#转base64 c# 科学计算 c#下拉列表获取串口
立即提问