a true random number generator (TRNG) in hardware which is targeted for FPGA-based crypto embedded systems. All crypto protocols require the generation and use of secret values that must be unknown to attackers.Random number generators (RNG) are required to generate public/private key pairs for asymmetric algorithm such as RSA and symmetric algorithm such as AES.
標簽： FPGA-based generator embedded hardware
上傳時間： 2014-01-09
上傳用戶：一諾88
RSA ( Rivest Shamir Adleman )is crypthograph system that used to give a secret information and digital signature . Its security based on Integer Factorization Problem (IFP). RSA uses an asymetric key. RSA was created by Rivest, Shamir, and Adleman in 1977. Every user have a pair of key, public key and private key. Public key (e) . You may choose any number for e with these requirements, 1< e <Æ (n), where Æ (n)= (p-1) (q-1) ( p and q are first-rate), gcd (e,Æ (n))=1 (gcd= greatest common divisor). Private key (d). d=(1/e) mod(Æ (n)) Encyption (C) . C=Mª mod(n), a = e (public key), n=pq Descryption (D) . D=C° mod(n), o = d (private key
標簽： crypthograph information Adleman Rivest
上傳時間： 2017-09-01
上傳用戶：chfanjiang
matlab有限元網格劃分程序 DistMesh is a simple MATLAB code for generation of unstructured triangular and tetrahedral meshes. It was developed by Per-Olof Persson (now at UC Berkeley) and Gilbert Strang in the Department of Mathematics at MIT. A detailed description of the program is provided in our SIAM Review paper, see documentation below. One reason that the code is short and simple is that the geometries are specified by Signed Distance Functions. These give the shortest distance from any point in space to the boundary of the domain. The sign is negative inside the region and positive outside. A simple example is the unit circle in 2-D, which has the distance function d=r-1, where r is the distance from the origin. For more complicated geometries the distance function can be computed by interpolation between values on a grid, a common representation for level set methods. For the actual mesh generation, DistMesh uses the Delaunay triangulation routine in MATLAB and tries to optimize the node locations by a force-based smoothing procedure. The topology is regularly updated by Delaunay. The boundary points are only allowed to move tangentially to the boundary by projections using the distance function. This iterative procedure typically results in very well-shaped meshes. Our aim with this code is simplicity, so that everyone can understand the code and modify it according to their needs. The code is not entirely robust (that is, it might not terminate and return a well-shaped mesh), and it is relatively slow. However, our current research shows that these issues can be resolved in an optimized C++ code, and we believe our simple MATLAB code is important for demonstration of the underlying principles. To use the code, simply download it from below and run it from MATLAB. For a quick demonstration, type "meshdemo2d" or "meshdemond". For more details see the documentation.
標簽： matlab有限元網格劃分程序
上傳時間： 2015-08-12
上傳用戶：凜風拂衣袖
/** * 用于在邏輯和界面間傳輸數據的bean * @version 1.0 */ public class DataBean { private int first = -1; private int second = -1; public int getFirst() { return first; } public int getSecond() { return second; } public void setFirst(int firt) { this.first = firt; } public void setSecond(int second) { this.second = second; } /** * 將1,2次均置為初始狀態
上傳時間： 2015-11-07
上傳用戶：dddhhhwww
#include <iostream> using namespace std; class Student { public: Student(int, int); int num; int grade; }; Student::Student(int n, int g) { num = n; grade = g; } int maxGradeIndex(Student* s) { int maxGrade, index = 0, i = 0; maxGrade = s[0].grade; for (i = 0; i<5; i++) { if (s[i].grade > maxGrade) { maxGrade = s[i].grade; index = i; } } return index; } int main() { Student a[5] = { Student(1, 86), Student(2, 60), Student(3, 72), Student(4, 95), Student(5, 66) }; int maxGradeStNum = maxGradeIndex(a); cout << "成績最好學生的學號是:" << a[maxGradeStNum].num << endl; cout << "成績最好學生的成績是:" << a[maxGradeStNum].grade << endl; getchar(); return 0; }
上傳時間： 2016-04-23
上傳用戶：burt1025
/*import java.util.Scanner; //主類 public class student122 { //主方法 public static void main(String[] args){ //定義7個元素的字符數組 String[] st = new String[7]; inputSt(st); //調用輸入方法 calculateSt(st); //調用計算方法 outputSt(st); //調用輸出方法 } //其他方法 //輸入方法 private static void inputSt(String st[]){ System.out.println("輸入學生的信息："); System.out.println("學號 姓名 成績1，2，3"); //創建鍵盤輸入類 Scanner ss = new Scanner(System.in); for(int i=0; i<5; i++){ st[i] = ss.next(); //鍵盤輸入1個字符串 } } //計算方法 private static void calculateSt(String[] st){ int sum = 0; //總分賦初值 int ave = 0; //平均分賦初值 for(int i=2;i<5;i++) { /計總分，字符變換成整數后進行計算 sum += Integer.parseInt(st[i]); } ave = sum/3; //計算平均分 //整數變換成字符后保存到數組里 st[5] = String.valueOf(sum); st[6] = String.valueOf(ave); } //輸出方法 private static void outputSt(String[] st){ System.out.print("學號 姓名 "); //不換行 System.out.print("成績1 成績2 成績3 "); System.out.println("總分 平均分");//換行 //輸出學生信息 for(int i=0; i<7; i++){ //按格式輸出，小于6個字符，補充空格 System.out.printf("%6s", st[i]); } System.out.println(); //輸出換行 } }*/ import java.util.Scanner; public class student122 { public static void main(String[] args) { // TODO 自動生成的方法存根 String[][] st = new String[3][8]; inputSt(st); calculateSt(st); outputSt(st); } //輸入方法 private static void inputSt(String st[][]) { System.out.println("輸入學生信息："); System.out.println("班級 學號 姓名 成績：數學 物理 化學"); //創建鍵盤輸入類 Scanner ss = new Scanner(System.in); for(int j = 0; j < 3; j++) { for(int i = 0; i < 6; i++) { st[j][i] = ss.next(); } } } //輸出方法 private static void outputSt(String st[][]) { System.out.println("序號 班級 學號 姓名 成績：數學 物理 化學 總分 平均分"); //輸出學生信息 for(int j = 0; j < 3; j++) { System.out.print(j+1 + ":"); for(int i = 0; i < 8; i++) { System.out.printf("%6s", st[j][i]); } System.out.println(); } } //計算方法 private static void calculateSt(String[][] st) { int sum1 = 0; int sum2 = 0; int sum3 = 0; int ave1 = 0; int ave2 = 0; int ave3 = 0; for(int i = 3; i < 6; i++) { sum1 += Integer.parseInt(st[0][i]); } ave1 = sum1/3; for(int i = 3; i < 6; i++) { sum2 += Integer.parseInt(st[1][i]); } ave2 = sum2/3; for(int i = 3; i < 6; i++) { sum3 += Integer.parseInt(st[2][i]); } ave3 = sum3/3; st[0][6] = String.valueOf(sum1); st[1][6] = String.valueOf(sum2); st[2][6] = String.valueOf(sum3); st[0][7] = String.valueOf(ave1); st[1][7] = String.valueOf(ave2); st[2][7] = String.valueOf(ave3); } }
上傳時間： 2017-03-17
上傳用戶：simple
CCS樣式選擇符，初學者，設計，DW，網頁制作，大一作業 部分預覽： <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>CSS樣式選擇符</title> <style type="text/css"> body { background-image:url(images/%E8%83%8C%E6%99%AF%E5%9B%BE%E7%89%87.jpg); background-repeat:repeat; } .class1 { text-align:center; font-weight:bolder; } .class2 { font-family:"仿宋"; text-indent:8em; } .class3 { font-size:18px; font-family:"宋體"; text-indent:4em; } #id1 { font-family:Zombie, Verdana, "Comic Sans MS"; font-style:oblique; font-size:64px; } #id2 { font-family:"黑體"; font-size:36px; } #id3 { color:#F69; font-weight:bolder; text-shadow:#FCC; } </style> </head> <body> <table width="780" height="1555" border="0" cellspacing="0" align="center" bgcolor="#FFFFFF"> <tr height="30"> <td align="center"><img src="images/頂部圖片.jpg" /></td> </tr>
上傳時間： 2017-12-07
上傳用戶：圈圈Ace
題目：古典問題：有一對兔子，從出生后第3個月起每個月都生一對兔子，小兔子長到第三個月后每個月又生一對兔子，假如兔子都不死，問每個月的兔子總數為多少？ //這是一個菲波拉契數列問題 public class lianxi01 { public static void main(String[] args) { System.out.println("第1個月的兔子對數: 1"); System.out.println("第2個月的兔子對數: 1"); int f1 = 1, f2 = 1, f, M=24; for(int i=3; i<=M; i++) { f = f2; f2 = f1 + f2; f1 = f; System.out.println("第" + i +"個月的兔子對數: "+f2); } } } 【程序2】 題目：判斷101-200之間有多少個素數，并輸出所有素數。 程序分析：判斷素數的方法：用一個數分別去除2到sqrt(這個數)，如果能被整除， 則表明此數不是素數，反之是素數。 public class lianxi02 { public static void main(String[] args) { int count = 0; for(int i=101; i<200; i+=2) { boolean b = false; for(int j=2; j<=Math.sqrt(i); j++) { if(i % j == 0) { b = false; break; } else { b = true; } } if(b == true) {count ++;System.out.println(i );} } System.out.println( "素數個數是: " + count); } } 【程序3】 題目：打印出所有的 "水仙花數 "，所謂 "水仙花數 "是指一個三位數，其各位數字立方和等于該數本身。例如：153是一個 "水仙花數 "，因為153=1的三次方＋5的三次方＋3的三次方。 public class lianxi03 { public static void main(String[] args) { int b1, b2, b3;
上傳時間： 2017-12-24
上傳用戶：Ariza
The 4.0 kbit/s speech codec described in this paper is based on a Frequency Domain Interpolative (FDI) coding technique, which belongs to the class of prototype waveform Interpolation (PWI) coding techniques. The codec also has an integrated voice activity detector (VAD) and a noise reduction capability. The input signal is subjected to LPC analysis and the prediction residual is separated into a slowly evolving waveform (SEW) and a rapidly evolving waveform (REW) components. The SEW magnitude component is quantized using a hierarchical predictive vector quantization approach. The REW magnitude is quantized using a gain and a sub-band based shape. SEW and REW phases are derived at the decoder using a phase model, based on a transmitted measure of voice periodicity. The spectral (LSP) parameters are quantized using a combination of scalar and vector quantizers. The 4.0 kbits/s coder has an algorithmic delay of 60 ms and an estimated floating point complexity of 21.5 MIPS. The performance of this coder has been evaluated using in-house MOS tests under various conditions such as background noise. channel errors, self-tandem. and DTX mode of operation, and has been shown to be statistically equivalent to ITU-T (3.729 8 kbps codec across all conditions tested.
標簽： frequency-domain interpolation performance Design kbit_s speech coder based and of
上傳時間： 2018-04-08
上傳用戶：kilohorse
#include "iostream" using namespace std; class Matrix { private: double** A; //矩陣A double *b; //向量b public: int size; Matrix(int ); ~Matrix(); friend double* Dooli(Matrix& ); void Input(); void Disp(); }; Matrix::Matrix(int x) { size=x; //為向量b分配空間并初始化為0 b=new double [x]; for(int j=0;j<x;j++) b[j]=0; //為向量A分配空間并初始化為0 A=new double* [x]; for(int i=0;i<x;i++) A[i]=new double [x]; for(int m=0;m<x;m++) for(int n=0;n<x;n++) A[m][n]=0; } Matrix::~Matrix() { cout<<"正在析構中~~~~"<<endl; delete b; for(int i=0;i<size;i++) delete A[i]; delete A; } void Matrix::Disp() { for(int i=0;i<size;i++) { for(int j=0;j<size;j++) cout<<A[i][j]<<" "; cout<<endl; } } void Matrix::Input() { cout<<"請輸入A:"<<endl; for(int i=0;i<size;i++) for(int j=0;j<size;j++){ cout<<"第"<<i+1<<"行"<<"第"<<j+1<<"列:"<<endl; cin>>A[i][j]; } cout<<"請輸入b:"<<endl; for(int j=0;j<size;j++){ cout<<"第"<<j+1<<"個:"<<endl; cin>>b[j]; } } double* Dooli(Matrix& A) { double *Xn=new double [A.size]; Matrix L(A.size),U(A.size); //分別求得U，L的第一行與第一列 for(int i=0;i<A.size;i++) U.A[0][i]=A.A[0][i]; for(int j=1;j<A.size;j++) L.A[j][0]=A.A[j][0]/U.A[0][0]; //分別求得U，L的第r行，第r列 double temp1=0,temp2=0; for(int r=1;r<A.size;r++){ //U for(int i=r;i<A.size;i++){ for(int k=0;k<r-1;k++) temp1=temp1+L.A[r][k]*U.A[k][i]; U.A[r][i]=A.A[r][i]-temp1; } //L for(int i=r+1;i<A.size;i++){ for(int k=0;k<r-1;k++) temp2=temp2+L.A[i][k]*U.A[k][r]; L.A[i][r]=(A.A[i][r]-temp2)/U.A[r][r]; } } cout<<"計算U得:"<<endl; U.Disp(); cout<<"計算L的:"<<endl; L.Disp(); double *Y=new double [A.size]; Y[0]=A.b[0]; for(int i=1;i<A.size;i++ ){ double temp3=0; for(int k=0;k<i-1;k++) temp3=temp3+L.A[i][k]*Y[k]; Y[i]=A.b[i]-temp3; } Xn[A.size-1]=Y[A.size-1]/U.A[A.size-1][A.size-1]; for(int i=A.size-1;i>=0;i--){ double temp4=0; for(int k=i+1;k<A.size;k++) temp4=temp4+U.A[i][k]*Xn[k]; Xn[i]=(Y[i]-temp4)/U.A[i][i]; } return Xn; } int main() { Matrix B(4); B.Input(); double *X; X=Dooli(B); cout<<"~~~~解得:"<<endl; for(int i=0;i<B.size;i++) cout<<"X["<<i<<"]:"<<X[i]<<" "; cout<<endl<<"呵呵呵呵呵"; return 0; }
標簽： 道理特分解法
上傳時間： 2018-05-20
上傳用戶：Aa123456789
