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Description: 此程序实现的是一个运算器，是在一个程序的基础上进行改进的，计算器中需要的运算以及操作对象：
IF,AND,OR,<,<=,>,>=,<>,=,==
数字0-9,(,),+,-,*,/,
SIN,COS,TAN,POW,EXP,ABS,SQRT,INT,LN,LOG,MIN,MAX.-this program is a computing device that is in a process of improvement based on the needs of calculator operation and the operation of Object : IF, AND, OR, lt; , Lt; =, Gt; , Gt; =, Lt; Gt; , =, == figures 0-9 ,(,), ,-,*,/, SIN, COS, TAN, POW, EXP, ABS, SQRT, INT, LN, LOG, MIN, MAX.
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Size: 46278 |
Author: 天空 |
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Description: 能运算的函数： sin,cos,tg,ctg,e^,pow(x,y),cosh,sinh,tgh,log10,ln,sqrt,arcsin,arccos,
运算方式： +,-,*,/,绝对值（“[ ]”）,^,!,
输入规则： 用键盘或按钮都可，输入完按回车运算，(光标要在最后)
sin(21-32)/(12-43)
4(323-4343) 4*(323-4343) e^2-sin3-3^4,(不要输入pow(3,4)) //有无*都可
2*3^4是（2*3）^4 而不是2*(3^4) 也就是要用x^y就要一定要（x^y）加上一个括号
[3-4]是求3-4的绝对值不是中括号 -performance computing functions : sin, cos, tg, rays, e ^, pow (x, y), cosh, sinh, tgh, log10, ln, sqrt, Simulation, arccos computational methods : absolute ,-,*,/, ( "[]"),^,!, importation rules : a keyboard or buttons which can be imported by the transport operator End (cursor to the final) sin (21-32) / (12-43) 4 (323-4343) 4 * ( 323-4343), e ^ 2 - KF-3 ^ 4 (do not enter pow (3,4)) / / * there can be 2 * 3 ^ 4 is (2 * 3) ^ 4 rather than 2 * (3 ^ 4) was needed x ^ y must have (x ^ y) with a brackets [3-4] is the pursuit of 3-4 is not the absolute value of the brackets
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Size: 111005 |
Author: qwq |
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Description: 注意事项：
1>在进行DFT时，强烈建议选用64*64的图像，选用256*256的图像，**时间太长，中间可能导致死机**。
2>程序中在显示频谱图像时有两种方式，即移至中心和未移至中心，可以通过将/** pow(-1, x+y)*/的注释去掉来进行转换。
由于本程序使用了VFW组件，所以在Project－－Setting中－－Object/libray modules中添加vfw32.lib。
-Note : 1
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Size: 151959 |
Author: pudn_jin |
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Description: ANNUAL REPORT TO CONGRESS
The Military Power of the People’s Republic of China
2005-ANNUAL REPORT TO CONGRESS The Military Pow er of the People's Republic of China 2005
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Size: 1250307 |
Author: zhybb |
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Description: 1.产生(0-1)分布随机数的方法:z i =(16807zi-1 +1)mod(pow(2,32)).被主函数调用.
2.产生U(0-1)分布利用公式:x1=sqrt(-2lnU1)cos(2PiU2), x2=sqrt(-2lnU1)sin(2PiU2)
3.产生瑞利分布,用两个独立同分布的正态分布值,用公式 x= sqrt(Y1*Y1+Y2*Y2)产生瑞利随机数.
4泊松分布-1. 20 (0-1) random number distribution methods : i = z (16807zi-1 1) mod (pow (2,32)). The main function was called. 2. Have U (0-1) distribution using the formula : x1 = sqrt (-2lnU1) cos (2PiU2), x2 = sqrt (-2lnU1) sin (2PiU2) 3. Rayleigh distribution, using two independent normal distribution with the distribution of values with the formula x = sqrt (Y1 - Y2 Y1 - Y2) - Health Rayleigh random number. 4 Poisson distribution
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Size: 52813 |
Author: 吕军红 |
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Description: 一、RSA基本原理
对明文分组M和密文分组C，加密与解密过程如下：
C = POW (M , e) mod n
M = POW(C , d) mod n = POW(POW( M ,e), d) mod n=POW( M,e*d)
其中POW是指数函数，mod是求余数函数。
其中收发双方均已知n，发送放已知e，只有接受方已知d，因此公钥加密算法的公钥为
KU={ e , n},私钥为KR={d , n}。该算法要能用做公钥加密，必须满足下列条件：
1. 可以找到e ，d和n，使得对所有M<n ,POW(M ,e*d)=M mod n .
2. 对所有 M<n,计算POW (M , e)和POW(C , d)是比较容易的。
3. 由e 和n确定d是不可行的
-one, the basic tenets of RSA expressly group M and cipher block C, encryption and decryption process is as follows : C = POW (M, e) mod n = M POW (C, d) mod n = POW (POW (M, e), d) mod n = POW (M, e * d), which is an exponential function POW, mod is the pursuit of the remaining functions. Transceivers which both known n, send e Fang known, the only known recipient d, therefore the public key encryption algorithm for public key e KU = (n), private key for KR = (d, n). The algorithm could be used to be a public key encryption, must meet the following conditions : 1. E can be found, and d n, making all the right M
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Size: 5523 |
Author: 烟翔 |
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Description: 功能说明：
次组件提供的功能有二：
1、解析出数学表达式中的参数
2、计算数学表达式
此组件不但可以解析和计算数学表达式，还提供了一系列的数学基本函数，可应用于表达式中一块处理。提供的数学基本函数有：将值舍入到最接近的整数 rount(x)、向下取整函数 int(x)、向上取整 ceiling(x)、求两个数中最大数的函数 max(a,b)、求两个数中最小数的函数 min(a,b)、幂函数 pow(x,n)、求平方根的函数 sqrt(x)。
下载地址：http://www.qiusuo365.com/qiusuo365/viewthread.php?tid=2422&extra=page％3D1-Functional Description : sub-components provided by the two functions : one, to figure out the mathematical expression of two parameters. calculate mathematical expression of this component will not only analytical and computational mathematical expression, but also provided a series of basic math functions, expression can be applied to a handle. The basic function of mathematics : the value of rounding to the nearest integer rount (x), downward for the entire function int (x), up from the entire ceiling (x), 2 for the largest number of the number of function max (a, b), 2 for the smallest number of the number of function min (a, b), power function pow (x, n), for the square root function sqrt (x). Download Address : http : / / www.qiusuo365.com/qiusuo365/viewthread . php tid = 24
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Size: 3687 |
Author: jason |
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Description: double pow(double,double)
省去了libm的麻烦。可以自由使用-double pow (double, double) save a libm trouble. Free use
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Size: 1336 |
Author: 段云 |
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Description: 本科生毕业设计
基于PowerBuider的图书管理系统
本文介绍了在PowerBuider环境下采用“自上而下地总体规划，自下而上地应用开发”的策略开发一个管理信息系统的过程。通过分析传统的人工管理图书馆的不足，创建了一套行之有效的计算机管理图书馆的方案。文章详细介绍了图书管理信息系统的系统分析部分，包括可行性分析、组织机构分析、管理职能分析、业务流程分析、数据流程分析、数据字典、处理描述等等；系统设计部分主要介绍了系统功能设计和数据库设计；系统实现部分列出了几个主要的程序框图，并附带了一些主要的窗口和程序。-Graduates'Clinical POWERBUIDER based on the design of library management system described in this paper Pow erBuider environment for the adoption of the "top down overall planning, Application to the development of bottom-up "strategy to develop a management information system process. By analyzing the traditional library management artificial shortage created a set of effective management of library computer programs. The article detailed the library management information systems analysis, including feasibility analysis, organizational analysis, Analysis of management functions, business process analysis, data flow analysis, data dictionary, handling description, and so on. system design some of the major functions of the system design and database design; System lists some
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Size: 1073632 |
Author: wu |
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Description: Learning Kernel Classifiers: Theory and Algorithms, Introduction This chapter introduces the general problem of machine learning and how it relates to statistical inference. 1.1 The Learning Problem and (Statistical) Inference It was only a few years after the introduction of the first computer that one of man’s greatest dreams seemed to be realizable—artificial intelligence. Bearing in mind that in the early days the most powerful computers had much less computational power than a cell phone today, it comes as no surprise that much theoretical research on the potential of machines’ capabilities to learn took place at this time. This becomes a computational problem as soon as the dataset gets larger than a few hundred examples.-Learning Kernel Classifiers : Theory and Algorithms. Introduction This chapter introduces the gene the acidic problem of machine learning and how it relat es to statistical inference. 1.1 The Learning P roblem and (Statistical) It was only inference a few years after the introduction of the first c omputer that one of man's greatest dreams seeme d to be realizable-artificial intelligence. B earing in mind that in the early days the most pow erful computers had much less computational po wer than a cell phone today, it comes as no surprise that much theoretical're search on the potential of machines' capabilit ies to learn took place at this time. This become 's a computational problem as soon as the dataset gets larger than a few hundred examples.
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Size: 2537081 |
Author: google2000 |
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Description: 1.实现科学计算器功能，对出错的输入计算式能进行识别
2.可以使用+,-,*,/,(,),&,|,~,sin,cos,tan,pow,log,exp,asin,acos,atan,sqrt以及0～9,pi的数字构成运算符；
3.负号的使用具有特殊性，可以使用2--2；也可以使用sin--2(表示sin2)；
4.10进制支持小数；支持10、16、8、2进制运算；对于16、8、2进制，仅支持整型输入操作；
5、8、2进制的计算结果仍然用原进制表示，若结果为负则用10进制表示结果；
6.sin 之类的函数可用角度或弧度运算，sin之类的函数支持嵌套使用，如sinsinsin1；
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Size: 12529 |
Author: 程梁 |
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Description: double pow(double,double)
省去了libm的麻烦。可以自由使用-double pow (double, double) save a libm trouble. Free use
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Size: 1024 |
Author: 段云 |
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Description: 由于小波变换转换过程中出现浮点数,使得256*256的图像压缩后不能为128*128.采用小波提升算法可以解决这个问题,由于小波提升算法可以实现整数到整数的变换,且可以完全重构信号.并且采用提升算法可以提高算法执行的速度.-Wavelet transform as a result of the conversion process appear in floating-point numbers, making the image 256* 256 can not be compressed to 128* 128. Wavelet lifting scheme can solve this problem, as the wavelet lifting scheme can realize integer to integer transform, and can be completely remodeling signal. and using lifting scheme algorithm can improve the speed of implementation.
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Size: 1024 |
Author: desertwing |
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Description: 计算 pow (m,n) 的精确值。 C语言环境-precise calculation of pow(m,n) using C
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Size: 2048 |
Author: bobo |
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Description: 这是C++语言中pow函数的源代码，通过学习此代码，可以了解C语言中是如何实现指数运算的。-pow function for C/C++
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Size: 1024 |
Author: Frank |
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Description: 利用vc++实现RSA加密解密算法源代码
[VC_RSA.rar] - 一、RSA基本原理 对明文分组M和密文分组C，加密与解密过程如下： C = POW (M , e) mod n M = POW(C , d) mod n = POW(POW( M ,e), d) mod n=POW( M,e*d) 其中POW是指数函数，mod是求余数函数。-Vc++ to achieve the use of RSA encryption and decryption algorithm source code [VC_RSA.rar]- a, RSA express the basic principles of M and the ciphertext packet division C, encryption and decryption process is as follows: C = POW (M, e) mod n M = POW (C, d) mod n = POW (POW (M, e), d) mod n = POW (M, e* d) which is a POW exponential function, mod is to strive for balance function.
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Size: 17408 |
Author: 易天行 |
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Description: pow为幂法计算，求解矩阵的特征值和特征向量，本文提供了算例和程序-pow method for the power, solve the matrix eigenvalue and eigenvector, the paper provides examples and procedures
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Size: 10240 |
Author: 陈岩 |
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Description: cvmx pow bist stat for Linux v2.13.6.
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Size: 3072 |
Author: doufanse |
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Description: cvmx pow bist stat for Linux v2.13.6.
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Size: 3072 |
Author: ghsnisabnr |
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Description: mpi-pow.c - MPI functions for Linux v2.13.6.
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Size: 3072 |
Author: knkjby |
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