sqlite的帮助文档， This ZIP archive contains most of the static HTML files that comprise this website, including all of the SQL Syntax and the C/C++ interface specs and other miscellaneous documentation.
上传时间： 2013-12-23
上传用户：evil
全英文版的STL书籍，本人认为比较好的一本书：Design Components with the C++ STL
上传时间： 2017-05-08
上传用户：jhksyghr
将魔王的语言抽象为人类的语言：魔王语言由以下两种规则由人的语言逐步抽象上去的：α-〉β1β2β3…βm ；θδ1δ2…-〉θδnθδn-1…θδ1 设大写字母表示魔王的语言，小写字母表示人的语言B-〉tAdA，A-〉sae，eg：B(ehnxgz)B解释为tsaedsaeezegexenehetsaedsae对应的话是：“天上一只鹅地上一只鹅鹅追鹅赶鹅下鹅蛋鹅恨鹅天上一只鹅地上一只鹅”。(t-天d-地s-上a-一只e-鹅z-追g-赶x-下n-蛋h-恨)
上传时间： 2013-12-19
上传用户：aix008
《分析性写作》，介绍言简意赅： The popular, brief rhetoric that treats writing as thinking, WRITING ANALYTICALLY, Sixth Edition, offers a series of prompts that lead you through the process of analysis and synthesis and help you to generate original and well-developed ideas. The book's overall point is that learning to write well means learning to use writing as a way of thinking well. To that end, the strategies of this book describe thinking skills that employ writing. As you will see, this book treats writing as a tool of thought--a means of undertaking sustained acts of inquiry and reflection.
上传时间： 2015-08-22
上传用户：东大寺的
最小二乘法曲面拟合,包括C程序及说明文件。对于搞三维重建的有一定帮助-Least squares surface fitting, including the C procedures and documentation. For engaging in three-dimensional reconstruction to some extent help the
标签： 通信网
上传时间： 2015-11-28
上传用户：schhqq
简单命令使用grep等的使用 [zorro@isch ~]$ history 1 ifconfig 2 su 3 exit 4 ls 5 cd Desktop/ 6 ls 7 tar zxcf VMwareTools-8.4.5-324285.tar.gz 8 tar zxvf VMwareTools-8.4.5-324285.tar.gz 9 cd vmware-tools-distrib/ 10 ls 11 ./vmware-install.pl 12 su 13 ls 14 cd .. 15 ls 16 rm VMwareTools-8.4.5-324285.tar.gz 17 rm -r vmware-tools-distrib 18 ls 19 make 20 ls 21 cd redis/ 22 quit 23 ls 24 ca redis/ 25 cd redis/ 26 cd redis-2.8.17 27 make 28 cd redis-2.8.17 29 ls 30 cd redis-2.8.17 31 cd str 32 cd src 33 ls 34 ./redis-cli 35 ls 36 cd redis-2.8.17 tar.gz 37 make 38 cd src 39 ./redis-server .. /redis.conf 40 ./redis-cli 41 ./redis-server ../redis.conf 42 vi test1.sh 43 ./test1.sh 44 vi test.sh 45 ./test.sh 46 ls 47 chmod 777 test.sh 48 ./test.sh 49 vi express 50 $ grep –n ‘the’ express 51 clear 52 grep -n 'the' express 53 vi express 54 grep -n 'the' express 55 grep -vn 'the'express 56 grep -vn 'the' express 57 grep -in 'the' express 58 vi test2.c 59 grep -l 'the' *.c 60 grep -n 't[ae]st' express 61 grep -n 'oo' express 62 grep -n '[^g]oo' express 63 grep -n '[a^z]oo' express 64 grep -n '[0^9]' express 65 grep -n '^the' express 66 vi express 67 sed -e 'd' express 68 sed -e '1d' express 69 sed -e '1~7d' express 70 sed -e '$d' express 71 sed -e '1，/^$/d' express 72 ls 73 cd 74 pwd 75 history [zorro@isch ~]$
标签： 简单命令使用
上传时间： 2016-05-24
上传用户：12345678gan
实验源代码 //Warshall.cpp #include<stdio.h> void warshall(int k,int n) { int i , j, t; int temp[20][20]; for(int a=0;a<k;a++) { printf("请输入矩阵第%d 行元素:",a); for(int b=0;b<n;b++) { scanf ("%d",&temp[a][b]); } } for(i=0;i<k;i++){ for( j=0;j<k;j++){ if(temp[ j][i]==1) { for(t=0;t<n;t++) { temp[ j][t]=temp[i][t]||temp[ j][t]; } } } } printf("可传递闭包关系矩阵是:\n"); for(i=0;i<k;i++) { for( j=0;j<n;j++) { printf("%d", temp[i][ j]); } printf("\n"); } } void main() { printf("利用 Warshall 算法求二元关系的可传递闭包\n"); void warshall(int,int); int k , n; printf("请输入矩阵的行数 i: "); scanf("%d",&k); 四川大学实验报告 printf("请输入矩阵的列数 j: "); scanf("%d",&n); warshall(k,n); }
上传时间： 2016-06-27
上传用户：梁雪文以
#include "iostream" using namespace std; class Matrix { private: double** A; //矩阵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; //为向量b分配空间并初始化为0 b=new double [x]; for(int j=0;j<x;j++) b[j]=0; //为向量A分配空间并初始化为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<<"正在析构中~~~~"<<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<<"请输入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<<"请输入b:"<<endl; for(int j=0;j<size;j++){ cout<<"第"<<j+1<<"个:"<<endl; cin>>b[j]; } } double* Dooli(Matrix& A) { double *Xn=new double [A.size]; Matrix L(A.size),U(A.size); //分别求得U，L的第一行与第一列 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]; //分别求得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<<"计算U得:"<<endl; U.Disp(); cout<<"计算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; }
标签： 道理特分解法
上传时间： 2018-05-20
上传用户：Aa123456789
The aim of this book, the first of two volumes, is to present selected research that has been undertaken under COST Action IC0902 ‘‘Cognitive Radio and Net- working for Cooperative Coexistence of Heterogeneous Wireless Networks’’ (http://newyork.ing.uniroma1.it/IC0902/). COST (European Cooperation in Sci- ence and Technology) is one of the longest-running European frameworks sup- porting cooperation among scientists and researchers across Europe.
上传时间： 2020-05-26
上传用户：shancjb
The genesis for this book was my involvement with the development of the SystemView (now SystemVue) simulation program at Elanix, Inc. Over several years of development, technical support, and seminars, several issues kept recur- ring. One common question was, “How do you simulate (such and such)?” The sec- ond set of issues was based on modern communication systems, and why particular developers did what they did. This book is an attempt to gather these issues into a single comprehensive source.
标签： Communication Simulation Systems
上传时间： 2020-05-26
上传用户：shancjb
