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RSS订阅原创 个人关于chrome浏览器的一些看法
我本身是一个忠实的google迷，邮箱用gmail，手机用android，等等，当然浏览器也不例外，一直使用chrome浏览器，下面我将从两方面谈一下我个人对使用chrome浏览器的看法。第一，对于普通用户来说，普通用户平时用浏览器无非就是上网，而上网时浏览器打开页面的速度尤为重要，如果浏览器打开网页的速度很快，会在用户体验上加不少分，而chrome浏览器就是以速度而著名的，所以首先说一下ch
2011-11-22 16:39:52
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原创 eclipse修改workspace的方法（转）
eclipse中改变默然的workspace的方法可以有：1.在创建project的时候，手动选择使用新的workspace，如创建一个web project，在向导中的Location选项，取消使用"Use default location",同时在下面选择新的worksp
2011-08-22 08:31:04
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原创 css学习笔记（四）
三 文本的精细排版1，调整字符间距letter-spacing字符间距letter-spacing属性用来控制字符之间的间距，这个间距实际就是在浏览器中所显示字符间的空格间距，同时间距的取值必须符合长度标准基本语法：letter-spacing:normal|长度normal表示
2011-07-22 10:55:36
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原创 css学习笔记（三）
二 字体的设置1，设置字体在HTML中设置字体使用的标记的face属性，而在css中使用font-family属性来设置字体基本语法：font-family:字体1, 字体2, ....;应用font-family属性可以一次定义多个字体，而在浏览器读取字体时，会按照定义的先后顺
2011-07-19 20:42:40
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原创 css学习笔记（二）
13，编写头部的css因为要将css文件定义在HTML文件头部的方法为内部样式表方法，所以下面举例说明怎么应用内部样式表方法在HTML文件的头部编写css（1）打开记事本，在记事本中输入如下一段普通的HTML代码，然后将代码文件以扩展名.html的形式保存实例代码： 编写头部
2011-07-17 11:19:30
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原创 css学习笔记（一）
一 css样式表基础1，css基本概念css（cascading style sheet）即层叠样式表，简称样式表，要理解样式表的概念先要理解样式的概念，样式是对网页中的元素（字体，段落图像，列表等）属性的整体概括，即描述所有网页对象的显示形式（例如文字的大小，字体，背景及图像的
2011-07-17 09:37:55
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原创 HTML学习笔记（四）
八 层的应用1，图层的创建图层也是网页制作中用于定位元素或者布局的一种技术，但是图层表格的布局更加的灵活，它能够将图层的内容摆放到浏览器的任意位置，同时放入到层中的HTML元素包括：文字，图像，动画甚至是图层。一个网页文件中可以使用多个层，
2011-07-16 08:54:47
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原创 HTML学习笔记（二）
15，忽略HTML标记 在HTML文件中利用成对的标记可以忽略HTML标记的作用，让标记之间的包括标记的所有内容直接显示在页面上 基本语法： 实例代码： 忽略HTML标记 哈哈哈
2011-07-10 18:27:20
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原创 html学习笔记（一）
最近开始学习网站制作。首先开始学习最基础的html语言。一 html基本语法1，标记语法html用于描述功能的符号称为标记。在html中标记的大小写作用相同。标记分为单标记和双标记单标记语法：双标记语法：双标记是由“始标记”和“尾标记”两部分构成的，必须成对使用。始标记告诉浏览器
2011-07-10 18:21:30
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原创 poj 2689解题报告（区间筛素数，经典）
<br /> Prime DistanceTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4814 Accepted: 1282<br />DescriptionThe branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theo
2011-02-04 12:32:00
3650
转载 【转】素数总结
<br /><br />定义：除了1和其本身，没有其他约数的数。<br />测试：用n分别试除2到sqrt(n)的数，如果中间有一个能整除，即<br />为合数，否则即为素数<br />bool is_prime(int n)//判断n是否为素数，是素数返回1<br />{<br /> int i;<br /> bool flag = 1;<br /> for(i = 2; i <= sqrt(n); i++)<br /> {<br /> if(n % i == 0){
2011-02-04 11:07:00
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原创 poj 1811解题报告
<br /> Prime TestTime Limit: 6000MS Memory Limit: 65536KTotal Submissions: 17747 Accepted: 3644Case Time Limit: 4000MS<br />DescriptionGiven a big integer number, you are required to find out whether it's a prime number.<br />InputThe first line contains
2011-02-02 17:42:00
1517
原创 poj 3641解题报告
Pseudoprime numbersTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3611 Accepted: 1285DescriptionFermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by
2011-02-01 14:01:00
3076
原创 zoj 1312解题报告
<br /> Prime CutsTime Limit: 1 Second Memory Limit: 32768 KB<br />A prime number is a counting number (1, 2, 3, ...) that is evenly divisible only by 1 and itself. In this problem you are to write a program that will cut some number of prime numbers
2011-02-01 12:33:00
1238
原创 poj3518解题报告
<br /> Prime GapTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 5323 Accepted: 2988<br />Description<br />The sequence of n − 1 consecutive composite numbers (positive integers that are not prime and not equal to 1) lying between two successive p
2011-01-30 21:08:00
2007
1
原创 poj 2739解题报告
<br /> Sum of Consecutive Prime NumbersTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10154 Accepted: 5767<br />DescriptionSome positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations
2011-01-30 20:36:00
1390
原创 poj 1365
<br /> Prime LandTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 1246 Accepted: 556<br />DescriptionEverybody in the Prime Land is using a prime base number system. In this system, each positive integer x is represented as follows: Let {pi}i=0,1,
2011-01-30 14:37:00
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原创 poj 1286解题报告
Necklace of BeadsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 3349 Accepted: 1405DescriptionBeads of red, blue or green colors are connected together into a circular necklace of n beads ( n InputThe input has several lines, and each line cont
2011-01-29 17:13:00
2018
原创 poj 2409解题报告
Let it BeadTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2190 Accepted: 1348Description"Let it Bead" company is located upstairs at 700 Cannery Row in Monterey, CA. As you can deduce from the company name, their business is beads. Their PR dep
2011-01-29 16:35:00
1652
原创 poj 1026解题报告
<br /> CipherTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 13633 Accepted: 3438<br />DescriptionBob and Alice started to use a brand-new encoding scheme. Surprisingly it is not a Public Key Cryptosystem, but their encoding and decoding is based
2011-01-29 15:16:00
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转载 转（acm数学）
view plaincopy to clipboardprint?·········10········20········30········40········50········60········70········80········90········100·······110·······120·······130·······140·······150 从放暑假前周sir给我讲了一个用polya计数法和burnside定理做的题目（pku2409）后，突然觉得组合数学挺有意思，然后从那时起
2011-01-28 18:37:00
4358
原创 joj 1031 Hanoi Tower Troubles Again!解题报告
<br /> 1031: Hanoi Tower Troubles Again!ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K1034606StandardPeople stopped moving discs from peg to peg after they know the number of steps needed to complete the entire task. But on the other hand, they
2011-01-28 18:25:00
1732
原创 joj 1026解题报告
<br /> 1026: The StaircasesResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K1590530Standard<br />One curious child has a set of N little bricks. From these bricks he builds different staircases. Staircase consists of steps of different sizes in a
2011-01-28 14:39:00
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原创 zoj 1133||joj 1027解题报告
<br /> Smith NumbersTime Limit: 1 Second Memory Limit: 32768 KB<br />While skimming his phone directory in 1982, Albert Wilansky, a mathematician of Lehigh University, noticed that the telephone number of his brother-in-law H. Smith had the followin
2011-01-27 17:57:00
1582
原创 joj 1019解题报告
<br /> 1019: Do the UntwistResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K992556Standard<br />Cryptography deals with methods of secret communication that transform a message (theplaintext) into a disguised form (the ciphertext) so that no one s
2011-01-24 19:06:00
816
原创 zoj1828||joj1949 解题报告
<br /> Fibonacci NumbersTime Limit: 1 Second Memory Limit: 32768 KB<br />A Fibonacci sequence is calculated by adding the previous two members of the sequence, with the first two members being both 1.<br />f(1) = 1, f(2) = 1, f(n > 2) = f(n - 1) + f
2011-01-23 12:13:00
1578
原创 poj1001解题报告
<br /> ExponentiationTime Limit: 500MS Memory Limit: 10000KTotal Submissions: 80517 Accepted: 19098<br />DescriptionProblems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the nat
2011-01-23 11:05:00
9152
原创 pku 2389解题报告
<br /> Bull MathTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7923 Accepted: 4094<br />DescriptionBulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... or so they say. Far
2011-01-22 22:56:00
835
原创 joj 1920解题报告
<br /> 1920: JojerResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K733231Standard<br />Mr Jojer is a teacher, and he is crazy in programming, he wants the computer to do every thing for him, one day, one of his colleague Mr NEtiger asks him to wri
2011-01-22 21:36:00
703
原创 joj 1386解题报告
<br /> 1386: 500!ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K2898518Standard<br />In these days you can more and more often happen to see programs which perform some useful calculations being executed rather then trivial screen savers. Some o
2011-01-22 21:31:00
514
原创 joj 1386解题报告
<br /> 1386: 500!ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K2898518Standard<br />In these days you can more and more often happen to see programs which perform some useful calculations being executed rather then trivial screen savers. Some of
2011-01-21 21:20:00
521
原创 zju1086 （高精度）解题报告
Octal FractionsTime Limit: 1 Second Memory Limit: 32768 KBFractions in octal (base 8) notation can be expressed exactly in decimal notation. For example, 0.75 in octal is 0.963125 (7/8 + 5/64) in decimal. All octal numbers of n digits to the right
2011-01-20 23:16:00
775
原创 zju1078题解
<br />题目：<br />Palindrom NumbersTime Limit: 1 Second Memory Limit: 32768 KBStatement of the Problem<br />We say that a number is a palindrom if it is the sane when read from left to right or from right to left. For example, the number 75457 is a pal
2011-01-02 14:03:00
773
原创 zju1067题解
题目：Color Me LessTime Limit: 1 Second Memory Limit: 32768 KBProblemA color reduction is a mapping from a set of discrete colors to a smaller one. The solution to this problem requires that you perform just such a mapping in a standard twenty-four bit
2011-01-02 13:48:00
724
原创 zju1058题解
<br />题目：<br />Currency ExchangeTime Limit: 1 Second Memory Limit: 32768 KBWhen Issac Bernand Miller takes a trip to another country, say to France, he exchanges his US dollars for French francs. The exchange rate is a real number such that when m
2011-01-02 13:33:00
790
原创 zju1049题解
<br />题目：<br /> I Think I Need a HouseboatTime Limit: 1 Second Memory Limit: 32768 KB<br /> <br /> <br />Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the land, he learned that th
2011-01-02 11:37:00
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原创 zju1045题解
题目：HangOverTime Limit: 1 Second Memory Limit: 32768 KBHow far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the
2011-01-02 11:12:00
2834
原创 zju1037题解
题目： GridlandTime Limit: 1 Second Memory Limit: 32768 KBBackgroundFor years, computer scientists have been trying to find efficient solutions to different computing problems. For some of them efficient algorithms are already available, these are the
2011-01-02 10:46:00
1300
原创 辗转相除法
辗转相除法是一个很基础的用递归思想的算法：#includeint main(){ int x,y; int gcd(int x,int y); scanf("%d%d",&x,&y); printf("%d",gcd(x,y)); system("pause"); return 0;}int gcd(int x,int y){ if(y==0)return x; else return gcd(y,x%y);}
2010-11-15 23:49:00
653
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