算术编码解码！具体的数据, 假设信源符号为 00, 01, 10, 11 ,这些符号的概率分别为 0 1, 0 4, 0 2, 0 3
Arithmetic coding and decoding! For specific data, assume that the source symbols are 00, 01,10,11, and the probability of these symbols is 0 1, 0 4, 0 2, 0 3, respectively. (2018-11-21, C/C++, 5KB, 下载0次)
（1）输入并建立多项式。
（2）输出多项式，输出形式为整数序列：n，c1，e1，c1，c2，e2，…，cn，en，其中n是多项式的项数，ci，ei，分别是第i项的系数和指数，序列按指数降序排序。
（3）实现多项式a和b相加，建立多项式a+b。
（4）实现多项式a和b相减，建立多项式a-b。
（5）计算多项式在x处的值。
(1) input and establish polynomial.
(2) output polynomial, the output form is integer sequence: n, C1, E1, C1, C2, e2,... Cn, en, N, which is a polynomial number of entries, CI, EI, respectively, the coefficient and index of item I, sequence index sorted in descending order.
(3) polynomial A and B are added, and polynomial a+b is established.
(4) polynomial A and B subtraction, the establishment of polynomial a-b.
(5) calculate the value of the polynomial at X. (2017-06-20, C/C++, 1KB, 下载1次)
8×16音频矩阵控制程序，带MT8816芯片，可实现多信源，多广播区的平行广播
8 × 16 audio matrix control procedures, with MT8816 chip, multi-source, multi-parallel radio broadcast area (2014-05-26, C/C++, 10KB, 下载21次)
货位优化相关程序代码 货位在7个以下的优化设计 适合于小企业
Slotting Optimization related program code less cargo space in seven optimal design for small businesses (2013-11-08, C/C++, 13KB, 下载16次)
介绍一种结合CN-FDTD方法和UPML的新算法，它能以较高的精度解决3维电磁散射问题。
this code contains a new algorithm that combines CN-FDTD and UPML which can result in low computing dispersion in solving EM problems. (2012-11-20, C/C++, 37KB, 下载24次)
如果给定任意两个正整数x和n，快速求取x^n。题目链接：http://acm.sdibt.edu.cn/JudgeOnline/problem.php?id=1794
If given two arbitrary positive integer x and n fast strike x ^ n. Topic Link: http://acm.sdibt.edu.cn/JudgeOnline/problem.php?id=1794 (2012-08-28, C/C++, 1KB, 下载2次)
企业发放的奖金根据利润提成。利润(I)低于或等于10万元时，奖金可提10 ；利润高
于10万元，低于20万元时，低于10万元的部分按10 提成，高于10万元的部分，可可提
成7.5 ；20万到40万之间时，高于20万元的部分，可提成5 ；40万到60万之间时高于
40万元的部分，可提成3 ；60万到100万之间时，高于60万元的部分，可提成1.5 ，高于
100万元时，超过100万元的部分按1 提成，从键盘输入当月利润I，求应发放奖金总数？
Company bonus according to the percentage of profits. Profit ( I ) less than or equal to 100000yuan, the bonus is10 high profits
In100000 yuan, less than 200000 yuan, less than $100000as part of the10 Commission, more than 100000 yuan, cocoa.
7.5 200000to 400000, more than 200000 yuan, can generate 5 400000to 600000is higher than the
400000part, can generate 3 600000to 1000000, more than 600000 yuan, but percentage of 1.5 , higher than the
1000000 yuan, more than 1000000 yuan as part of the1 Commission, from the keyboard input of the I should seek profit, total bonus? (2012-04-25, C/C++, 3KB, 下载2次)
企业发放的奖金根据利润提成。利润(I)低于或等于10万元时，奖金可提10 ；利润高
于10万元，低于20万元时，低于10万元的部分按10 提成，高于10万元的部分，可可提
成7.5 ；20万到40万之间时，高于20万元的部分，可提成5 ；40万到60万之间时高于
40万元的部分，可提成3 ；60万到100万之间时，高于60万元的部分，可提成1.5 ，高于
100万元时，超过100万元的部分按1 提成，从键盘输入当月利润I，求应发放奖金总数？
Company bonus according to the percentage of profits. Profit ( I ) less than or equal to 100000yuan, the bonus is10 high profits
In100000 yuan, less than 200000 yuan, less than $100000as part of the10 Commission, more than 100000 yuan, cocoa.
7.5 200000to 400000, more than 200000 yuan, can generate 5 400000to 600000is higher than the
400000part, can generate 3 600000to 1000000, more than 600000 yuan, but percentage of 1.5 , higher than the
1000000 yuan, more than 1000000 yuan as part of the1 Commission, from the keyboard input of the I should seek profit, total bonus? (2012-04-25, C/C++, 3KB, 下载2次)
企业发放的奖金根据利润提成。利润(I)低于或等于10万元时，奖金可提10 ；利润高
于10万元，低于20万元时，低于10万元的部分按10 提成，高于10万元的部分，可可提
成7.5 ；20万到40万之间时，高于20万元的部分，可提成5 ；40万到60万之间时高于
40万元的部分，可提成3 ；60万到100万之间时，高于60万元的部分，可提成1.5 ，高于
100万元时，超过100万元的部分按1 提成，从键盘输入当月利润I，求应发放奖金总数？
Corporate bonuses based on profit commission. Profit (I) is less than or equal to 10 million, the bonus can mention 10 high profits
$ 100,000, less than 20 million, less than 10 million part of the 10 commission, higher than 10 million part of the cocoa mention
Into 7.5 200000-400000, more than 20 million part of the commission 5 between 40 million to 60 million higher than the
40 million part of the commission 3 between 60 million to 100 million, more than 60 million part, to the commission of 1.5 , higher than the
1 million yuan, more than 100 million parts of 1 commission from the keyboard, enter the month profits I seek the total number of bonuses? (2012-02-13, C/C++, 1KB, 下载2次)
动态输入节点个数、各节点数据、积分上下限及精度要求（Romberg精度要求 ）；
分别用 Lagrange 和 Newtow 插值法计算 p(x) 和 q(x) 在求积节点处的近似值；
分别用梯形公式、辛普森公式和 Romberg 算法计算：
梯形公式、辛普森公式输出T、S，Romberg 算法输出步长、等分数、Tn、Sn、Cn、Rn以及最终的计算结果
Number of dynamic input nodes, each node of data, points on the lower limit and precision (Romberg accuracy) respectively Lagrange interpolation method and Newtow p (x) and q (x) nodes in the quadrature approximation respectively trapezoidal formula, Simpson and Romberg algorithm formula: trapezoidal, Simpson formula output T, S, Romberg algorithm output step, such as scores, Tn, Sn, Cn, Rn, and the final results (2011-05-31, C/C++, 2KB, 下载4次)
optimized iFFT code for TI C64 processor
optimized iFFT code for TI C64 processor (2009-08-20, C/C++, 36KB, 下载50次)
已经知道企业发放的奖金根据利润提成。从键盘输入当月利润I，求应发放奖金总数？利用数轴来分界，定位。
Enterprises are aware of the bonuses paid royalty based on profits. Month profit from the keyboard input I, for a total prize money should be paid? Axis to make use of a few boundaries, location. (2009-05-11, C/C++, 2KB, 下载1次)
正在学数字信号处理,感觉上学期信号与系统学得不扎实，因为当时只是死记公式，这学期数信老师提倡动手实践，觉得自己在编程中对公式理解得更加深刻了。 以下是我写的FFT，欢迎指教。
is studying digital signal processing, signal feel semester with enough solid school system, because only memorizing formulas, a few letters this semester to promote hands-teachers find themselves in the right programming formula to understand more deeply. The following is what I wrote FFT welcome to enlighten. (2005-03-25, C/C++, 1KB, 下载204次)