對于方程組

• x = a (mod p)
• x = b (mod q)

其中p, q互素。

可以采用中國剩余定理，x = q * Eq * a + p * Ep * b (mod pq ) , 其中 Eq * q + Ep * p = 1;

而模不互素的情況，卻有類似的形式：

• x = a (mod pd)
• x = b (mod qd)

其中p, q互素, d > 1。

如果d 不整除 a - b, 則無解, 否則
x = q * Eq * a + p * Ep * b ( mod pqd ) , 其中 Eq * q + Ep * p = 1;


可以驗算這個構(gòu)造解是適合上面兩個方程的。

比如驗算第一個方程：
首先變形得到 x = (1 - Ep * p ) * a + Ep * p * b  (mod pd);
又有：x = a + Ep * p *( b - a )   (mod pd);
又有：d | (b - a)  所以 pd | p*(b - a)
所以 x = a ( mod pd ) 

也可以證明x 模上 pqd 具有唯一解
     摘要: 待續(xù)  閱讀全文      摘要: 一個多項式的差分的等價形式---棋盤上放車的種數(shù)  閱讀全文      摘要:
感覺以前很少接觸到這種劃分的問題，但是它又好像很經(jīng)典的樣子  閱讀全文 ZOJ

題號	摘要	提交次數(shù) / coding耗時
2313	模板的弊端,具體優(yōu)化	13    / ---
2317	走道鋪磚	3     / 60"
2318	環(huán)顧法判點在多邊形內(nèi)，搜索樹，所有回路	---   / ---

PKU
            
             

題號	分類	注釋	鏈接
1012	遞歸    recursion	joseph問題，joseph是經(jīng)典的遞歸問題
1186	雙向枚舉	現(xiàn)枚舉前一半，再二分查找后一半是否有對應(yīng)的值
1285	組合 & 計數(shù)	有限制的可重復(fù)排列    dp （pku 的 G++不識 unsigned long long 尷尬）
1286	burnside	2154的簡化版
1316	質(zhì)因數(shù)分解  Prime- factor	有點進制轉(zhuǎn)換的感覺	:D
1351	組合 & 計數(shù)	有相鄰問題可重復(fù)的排列   dfs
1430	stirling數(shù)	很考察觀察能力
1715	組合 & 計數(shù)	詢問第n位上是哪個數(shù)，比較常見的一類題
1718	joseph	計算倒數(shù)第二個被殺的人是誰
1737	遞歸 recursion	其實不是很復(fù)雜
1809	奇偶性	奇偶性
1811	miller-rabin + pollard rho	很適合初學這兩種算法
1831	枚舉 構(gòu)造	枚舉幾項小的，再用S= 2*P+2(p/2 + 1/2 = 1) 和 S = 2*P + 9(p/2 +　１＋1/3 + 1/6 = 1)構(gòu)造
1845	積性函數(shù)	積性函數(shù)
2034	反素數(shù)  antiprime	dfs	:D
2142	解不定方程	解不定整數(shù)方程ax + by = c 其中a，b，c ，x，y為整數(shù)
2154	burnside  歐拉數(shù)  觀察	想法不算繞彎，只要知道這些知識點完全能解出來	:D
2282	數(shù)字游戲	統(tǒng)計[a，b]中0,1,2...9的個數(shù)
2429	質(zhì)因數(shù)分解   pollard rho	pollard rho
2689	素數(shù)    prime	刷表	:)
2739	素數(shù)    prime	暴力
2769	同余	刷表
2891	合并同余方程	合并同余方程
2917	質(zhì)因數(shù)	分解質(zhì)因數(shù)
2992	約數(shù) divisor	分解連續(xù)的數(shù)的質(zhì)因數(shù) 水題
3126	素數(shù)    prime	其實重點不是prime。。。 bfs關(guān)鍵
3128	循環(huán)節(jié)	找規(guī)律
3132	素數(shù)    prime	其實重點不是prime。。。 dp關(guān)鍵 -_-!
3252	數(shù)字游戲	算[a,b]里有多少數(shù)的二進制0比1多
3324	大數(shù) +針對該題目的一些優(yōu)化	mod （2^p-1）可以優(yōu)化
3508	大數(shù)加法	大數(shù)加法
3518	素數(shù)    prime	二分
3641	素數(shù)    prime	miller-rabin   注意 a^p%p=a 不等價與 a^(p-1)%p=1
3725	數(shù)字游戲	分各位十位百位。。。統(tǒng)計， 也可以通過二分做，注意不要溢出這題不順

要有激情
剩下的就是提高實力了，首先是想法，其次是代碼。看大量的書，看大量的論文。做大量的題
要了解自己的隊友，要熟悉現(xiàn)在那些人是牛人，多關(guān)注牛人，見賢思齊


     摘要: 奇跡只會發(fā)生在不言放棄的人身上  閱讀全文 Why XAML Needed?

Since WPF applications can be developed entirely in code, you may ask a
perfectly natural question – why do we need XAML in the first place? The
reason can be traced back to the question of efficiently implementing complex,
graphically rich applications. A long time ago, developers realized that the most
efficient way to develop these kinds of applications was to separate the graphics
portion from the underlying code. In this way, the designers could work on the
graphics, while the developers could work on the code behind the graphics. Both
parts could be designed and refined separately, without any versioning
headaches.

Before WPF, it was impossible to separate the graphics content from the code.
For example, when you work with Windows Forms, you define every form
entirely in C# code or any other language. As you add controls to the UI and
configure them, the program needs to adjust the code in corresponding form
classes. If you want to decorate your forms, buttons, and other controls with
graphics developed by designers, you must extract the graphic content and
export it to a bitmap format. This approach works for simple applications;
however, it is very limited for complex, dynamic applications. Plus, graphics in
bitmap format can lose their quality when they get resized.

The XAML technology introduced in WPF resolves these issues. When you
develop a WPF application in Visual Studio, the window you are creating isn’t
translated into code. Instead, it is serialized into a set of XAML tags. When you
run the application, these tags are used to generate the objects that compose the
UI.

XAML isn’t a must in order to develop WPF applications. You can implement
your WPF applications entirely in code. However, the windows and controls
created in code will be locked into the Visual Studio environment and available
only to programmers; there is no way to separate the graphics portion from the
code.

In orther words, WPF doesn’t require XAML. However, XAML opens up world
of possibilities for collaboration, because many design tools understand the
XAML format.



刷表就是一種預(yù)處理

Cubic-free numbers II

要求[ L，R )上的不是Cubic數(shù)的個數(shù)，發(fā)現(xiàn)求區(qū)間上有多少Cubic數(shù)更清晰，求這種區(qū)間問題有一種比較經(jīng)典的處理技巧，求出[1，L)和[1,R)
[L , R) = [1, R) - [1, L);

我們可以用容斥來求區(qū)間[1，k）上有多少Cubic數(shù)，這里刷表表示容斥就很方便了
唯一注意一點，就是先把含有i*i的數(shù)標記成無效，因為我們的容斥不會去判一個集合自己和自己的關(guān)系，我們都是比較一個集合和其他集合的關(guān)系

Coprimes

這也是一道容斥題，刷表

     摘要:   閱讀全文