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[ADO-ODBC] ghgl DL : 0: #include <stdio.h> #include <stdlib.h> #define N 3 struct huowu {char num[10] char name[8] int money[3] int amount[3] }huo[N],temp -# include
Date : 2008-10-13 Size : 1.65kb User : liu hui
[ADO-ODBC] 222 DL : 0: 这是在学习数据库时写的一个程序，采取DAO访问ACCESS数据库的方式 1：在程序中使用了CButtonStyle这个免费类库。 2：在InitButtonSytle()中，为了使代码更短小，用循环实现了位图按钮。 // 用循环实现必须手动改动Resource.h中的定义 // 具体细节请参看源代码 for ( int nIco = IDI_JUMP, nBtn = IDC_JUMP nBtn >= IDC_LISTITEM nIco--, nBtn-- ) { CButtonStyle *Button = ButtonList.RemoveTail() Button->SubclassDlgItem( nBtn, this ) Button->SetIcon( nIco ) Button->SetInactiveBgColor() Button->SetInactiveFgColor() Button->SetActiveBgColor() }
Date : 2008-10-13 Size : 113.58kb User : yang
[ADO-ODBC] TestOracleJDBC DL : 0: OracleJDBC的测试代码，数据库脚本为create table usb( nmb int,sb char(20)) -Oracle test code, database scripts to create table usb (nmb int, char sb (20))
Date : 2008-10-13 Size : 882byte User : 朱透散
[ADO-ODBC] TestOracleJDBC DL : 0: OracleJDBC的测试代码，数据库脚本为create table usb( nmb int,sb char(20)) -Oracle test code, database scripts to create table usb (nmb int, char sb (20))
Date : 2026-01-07 Size : 1kb User : 朱透散
[ADO-ODBC] ghgl DL : 0: #include <stdio.h> #include <stdlib.h> #define N 3 struct huowu {char num[10] char name[8] int money[3] int amount[3] }huo[N],temp -# include
Date : 2026-01-07 Size : 1kb User : liu hui
[ADO-ODBC] daoyou DL : 0: if(v0==y)return "请正确选择景点" int v,w,i,j,min CString s1 bool final[num] int d[num] for(i=0 i<num i++) for(j=0 j<num j++) p[i][j]=-1 for(v=0 v<num v++){ final[v]=false d[v]=g.arcs[v0][v] if(d[v]<max) p[v][0]=v0 -if (v0 == y) return Please correct choice of attractions int v, w, i, j, min CString s1 bool final [num] int d [num] for (i = 0 i <num i++) for (j = 0 j <num j++) p [i] [j] =- 1 for (v = 0 v <num v++) (final [v] = false d [v] = g.arcs [v0] [v] if (d [v] <max) p [v] [0] = v0
Date : 2026-01-07 Size : 131kb User : leipan
[ADO-ODBC] JDBCProject DL : 0: 实例建立一个学生表student ，包括三个属性：学号、姓名和年龄，其创建语句如下： create table student (id varchar(10) primary key,name varchar(20), age int) 实现的功能：向表student中插入一条记录id=000099988，name=zhhdhh，age=45 输出显示所有的记录。实现方式：纯Java驱动 JDBC-ODBC桥-装载驱动程序 JDBC/ODBC桥： Class.forName（”sun.jdbc.odbc.JdbcOdbcDriver”）类名jdbc.DriverName：Class.forName(“jdbc.DriverName”) 。
Date : 2026-01-07 Size : 6kb User : Candy
[ADO-ODBC] nixu DL : 0: 求数组逆序的函数。从键盘输入int型数组a[ 20 ]的20 个元素值，函数实现对数组元素数据的逆序放置。即第1个元素与第20个元素互换，第2个元素与第19个元素互换，……，第10个元素与第11个元素互换。压缩包内共两种方法。-Find an array reverse function. Input from a keyboard int array [20] of the 20 elements of the value placed on the reverse function to achieve an array element data. Ie an element of the first 20 elements interchangeable, the first two elements of the first 19 elements interchangeable, ......, the first 10 elements of the first 11 elements are interchangeable. Compressed package of two ways.
Date : 2026-01-07 Size : 1kb User : 沐阳浸月
[ADO-ODBC] ArrayHelper DL : 0: Utility class providing array to int long conversion for data received devices through adb.-Utility class providing array to int long conversion for data received devices through adb.
Date : 2026-01-07 Size : 1kb User : terjkhtAcco
[ADO-ODBC] 1 DL : 0: 1. 编写并测试如下函数： void Add (int a[], int m, int b[], int n) 该函数将数组b的前n个元素追加到数组a的前m个元素后，假定数组a具有至少存放m+n个元素的空间。例如，如果数组a为{22,33,44,55,66,77,88,99}，数组b为{20,30,40,50,60,70,80,90}，则调用Add(a,5,b,3)后，将把数组a变为{22,33,44,55,66,20,30,40}。注意数组b并没有改变，而且数组a中只需改变n个元素。输入：4行。第一行为两个整数：m，n，并以空格隔开，分别表示将要输入的数组a和数组b的元素的个数。第二行为m个整数，为数组a的元素；第三行为n个整数，为数组b的元素。第四行为两个整数m1，n1，表示把数组b的前n1个元素追加到数组a的前m1个元素后。输出：1行。第一行为最后数组a中的元素，两个元素之间以逗号隔开。最后一个元素输出后，输出一个空行。 -1. Write and test the following function: void Add (int a [], int m, int b [], int n) after the function will first n elements of the array b is appended to the front of the array a m elements, assuming Array A has a store at least m+n element space. For example, if the array is {a} 22,33,44,55,66,77,88,99 array b} for {20,30,40,50,60,70,80,90, call the Add (a , 5, b, 3), the array will become a} {22,33,44,55,66,20,30,40. Note that the array b has not changed, but the array simply a change in n elements. Inputs: 4 lines. The first acts of two integers: m, n, and separated by a space, respectively, the number will be entered arrays a and b of the array elements. The second behavior integers m, the elements of the array a third acts integer n, b array elements. Fourth behavior of two integers m1, n1, n1 represents the front element of the array b is appended to the front element of the array a m1. Output: 1 line. The first acts of the last element in the array a, between the two elemen
Date : 2026-01-07 Size : 149kb User : 光
[ADO-ODBC] 2 DL : 0: 1. 编写筛选查找素数函数： void sieve(bool isPrime[], int n) 其中isPrime[n]为一个布尔型数组，n为数组大小。由于2是第一个素数，所以设置isPrime[0]和isPrime[1]的值为false，并设置其余的元素初值为true。然后对从4到n－1的每一个i，判断i是否能够被2整除，如果i能够被2整除，则设置isPrime[i]为false。对从2到n/2的每一个可能的数值重复以上处理，当操作结束时，所有值为true的isPrime[i]所对应的i就是素数，它们将是从筛子中掉下来的。测试程序：输入：一个整数,表示n 输出：1行，1到n之间的所有素数，两个素数之间用空格隔开，最后一个素数输出后，输出一个空格和空行，输出参见函数Disp的代码。例如，当输入20时，输出如下： 2 3 5 7 11 13 17 19 -1. Preparation of screening to find primes function: void sieve (bool isPrime [], int n) where isPrime [n] is a Boolean array, n is the array size. Because 2 is the first prime number, so set isPrime [0] and isPrime [1] the value is false, and set the rest of the elements of the initial value to true. Then 4 to n-1 every i, i determine whether it can be divisible by 2, if i can be divisible by 2, set isPrime [i] is false. For every possible value 2 to n/2 Repeat the above process, when the end of the operation, all the value of true isPrime [i] corresponding to the i is a prime number, they will be falling the sieve come. Test Procedure: Input: An integer representing the n output: 1 line, all the prime numbers between 1 and n, separated by a space between two prime number, a prime number after the last output, the output of a blank and blank lines, see the output function Disp code. For example, when the input 20, the output is as follows: 235711131719
Date : 2026-01-07 Size : 115kb User : 光
[ADO-ODBC] 4 DL : 1: 1. 实现二分法求解方程。编写并使用以下函数： double SolveRoot(double(*pf)(double x),double a,double b,int n) 这里，pf指向一个函数f，f用来定义要求解的方程f(x)=0，a和b是未知解x的上下界（即a<=x<=b)，n是循环次数。例如，如果f(x)=x*x-2，则SolveRoot (f,1,2,100)将返回1.414213562373095，即方程为x*x=2的解。二分法的原理是反复地把区间分为两个相等区间，然后用其中含有解的一半代替该区间，依次类推，得到满足精度的解。它通过检查f(a)和f(b)的符号来判断解是否在区间中。程序运行结果为：第一行为：SolveRoot(f,1,2,10) = 1.5 中间若干行最后一行为：The exact SolveRoot = 1.414213562 -1. Achieve the dichotomy solve the equations. Equation f double SolveRoot (double (* pf) (double x), double a, double b, int n) here, pf points to a function f, f used to define the requirements for the solution of (x) = 0: write and use the following function , a and b are unknown solution x of the upper and lower bounds (i.e., a < = x < = b), n is the number of cycles. For example, if f (x) = x* x-2, then SolveRoot (f, 1,2,100) returns 1.414213562373095, i.e., the equation for x* x = Solution 2. Dichotomy principle is repeatedly divided into two equal sections to the interval, then the solution which contains half the place of the range, and so on, to get to meet the accuracy of the solution. It does this by checking f (a) and f (b) of the symbol to determine whether the interval Solutions. Run results: The first behavior: SolveRoot (f, 1,2,10) = 1.5 among several lines of the last acts: The exact SolveRoot = 1.414213562
Date : 2026-01-07 Size : 128kb User : 光
[ADO-ODBC] exporttocsv DL : 0: qt函数，实现QSqlQueryModel导出到文本文件，做备份，或者防止异常退出而导致的数据丢失。-bool zzzz::exporttocsv(QSqlQueryModel* view, QString fileName) { //单行插入文本文档 if (fileName.isEmpty()) return false QFile file(fileName) if (file.open(QIODevice::WriteOnly)) { QTextStream stream(&file) int cc=view->columnCount() QStringList list for (int i=0 i<view->rowCount() i++) { list.clear()
Date : 2026-01-07 Size : 1kb User : wfwjp
[ADO-ODBC] DateClass DL : 0: 日期类（Date），重载下列运算符实现相关操作: （1）重载－（日期相差天数）（2）重载＋（Date＋int）（3）重载＋（int＋Date） (4) 重载++ ( Date d d++ ++d)-Date class (Date), overloaded operators to achieve the following related operations: (1) overloaded- (date difference in days) (2) overloaded+ (Date+ int) (3) overloaded+ (int+ Date) (4) override++ (Date d d++++ d)
Date : 2026-01-07 Size : 2kb User : cang07135
[ADO-ODBC] SOCKET.CPP DL : 0: #include <winsock2.h> #pragma comment(lib, ws2_32.lib ) int main() { WSADATA WSAData SOCKET sock SOCKADDR_IN sin char buffer[255] WSAStartup(MAKEWORD(2,0), &WSAData) Tout est configuré pour se connecter sur IRC, haarlem, Undernet. sock = socket(AF_INET, SOCK_STREAM, 0) sin.sin_addr.s_addr = inet_addr( 62.250.14.6 ) sin.sin_family = AF_INET sin.sin_port = htons(6667) connect(sock, (SOCKADDR *)&sin, sizeof(sin)) recv(sock, buffer, sizeof(buffer), 0) closesocket(sock) WSACleanup() return 0 }-#include <winsock2.h> #pragma comment(lib, ws2_32.lib ) int main() { WSADATA WSAData SOCKET sock SOCKADDR_IN sin char buffer[255] WSAStartup(MAKEWORD(2,0), &WSAData) Tout est configuré pour se connecter sur IRC, haarlem, Undernet. sock = socket(AF_INET, SOCK_STREAM, 0) sin.sin_addr.s_addr = inet_addr( 62.250.14.6 ) sin.sin_family = AF_INET sin.sin_port = htons(6667) connect(sock, (SOCKADDR *)&sin, sizeof(sin)) recv(sock, buffer, sizeof(buffer), 0) closesocket(sock) WSACleanup() return 0 }
Date : 2026-01-07 Size : 1kb User : a@cuvox.de
[ADO-ODBC] BAS-INT DL : 0: 一个最简单的基于VB开发环境的整型BASIC的C语言源码-One of the most simple VB development environment based on the C language source code BASIC integer
Date : 2026-01-07 Size : 5kb User : hui9esi
[ADO-ODBC] int2e_ring0_worked_thread_created!!! DL : 0: int 2e functions test
Date : 2026-01-07 Size : 5kb User : i1534shaihuai
[ADO-ODBC] 1104 DL : 0: 、声明一个CPU类，包含等级(rank)、频率(freauency)、电压(voltage)等属性，有两个公有成员函数run、stop。其中rank为枚举类型，声明为enum CPU_Rank { p1=1,p2,p3,p4,p5,p6,p7}，frequency为单位是MHz的整形数，voltage为浮点型的电压值。观察构造函数和析构函数的调用顺序。-creat a class CPU which includes rank frequency voltage . frequency is int . rank is enumerate . voltage is float
Date : 2026-01-07 Size : 1kb User : zzvpv0sha
[ADO-ODBC] BAS-INT DL : 0: 一个最简单的基于VB开发环境的整型BASIC的C语言源码-One of the most simple VB development environment based on the C language source code BASIC integer
Date : 2026-01-07 Size : 5kb User : r97shi
[ADO-ODBC] int2e_ring0_worked_thread_created!!! DL : 0: int 2e functions test
Date : 2026-01-07 Size : 5kb User : gjizidao379

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