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E-ruler
- 电子尺源程序说明 本程序使用ADO访问Access2000的数据库。 请适当修改stdafx.h中 #import \"g:\\program files\\common files\\system\\ado\\msado15.dll\" no_namespace rename(\"EOF\",\"adoEOF\") msado15.dll的路径-source of the present procedures for the use of ADO Access 2000 database. P
魔方阵的问题求解
- #include"stdio.h" #include"stdlib.h" int main() { int i ,j,a[100][100],n,k,flag=1; while(scanf("%d",&n)!=EOF&&(n%2!=0)) { for(i=0;i<n;i++) for(j=0;j<n;j++) a[i][j]=0; j=n/2; a[0][j]=1; i=0
读写SQL中的image字段.zip
- Private Declare Function DeleteFile Lib "kernel32" Alias "DeleteFileA" (ByVal lpFileName As String) As Long '************************************************************* Public Function ImportPhoto(sPicFile As String, sEmpID As
气象常用fortran程序集魏凤英.rar
- 各种突变检验方法、EOF等程序。这些程序是魏凤英老师编写的,我只是把我下载到的内容发布在这里,网上可以直接下载这些程序,所以应该不会牵扯到版权问题。 希望我可以获得积分,然后下载其他程序。
经验正交分析的fortran程序
- 经验正交分析的fortran程序,适合科研使用.,The analysis of empirical orthogonal fortran program for research use.
tianpingwenti.rar
- 输入: 多组测试数据,每行只有一个正整数n,表示要称的物体质量, n在32bit有符号整数所能表示的数的范围内, 最后请使用EOF结束本程序 输出: 怎么称质量n出来 ,Input: multiple sets of test data, each line there is only one positive integer n, said that the quality of the object, n in 32bit signed integer
CEOF.rar
- 气象上常用的负经验正交分解,用fortran写的 跟eof方法不同,Weather on the commonly used orthogonal decomposition negative experience with fortran written with different eof
fortran
- Fortran程序集,用于气象中的各种程序计算,比如EOF,SVD,滑动T检验,累计距平。-Fortran program set in a variety of procedures for meteorological calculations, such as EOF, SVD, sliding T-test, the cumulative anomaly
iostream1
- 编写程序,使用输入流对象cin的成员函数eof()和get()让用户逐个读取输入的字符,并根据eof()判断是否结束,然后使用输出流对象cout的成员函数put()将读取的内容输出。 -Programming, using input stream object cin the member function eof () and get () allows users to read input characters one by one, and under eof () to dete
REOF
- fortran 编写的旋转EOF程序,比普通EOF更能刻画局部特征。作者李建平,用于气象海洋研究-EOF program written in fortran rotation, better than the average EOF characterize local features. Of Li Jianping, marine research for the Meteorological
20085841710354
- 价值千元的E灵通在线考试系统商业版,人数限制已破解。 关于程序中人数的限制,修改方法: if rsuser.recordcount>=20 then rsuser.move 20 do while not rsuser.eof rsuser.delete rsuser.update rsuser.movenext loop 找到涉及文件的以上代码,修改20为你所定义的人数即可,看似很简单,但原来的代码加密了,没有办法修改! ID:admin
shell
- #include<stdio.h> #include<string.h> #include<limits.h> #include<unistd.h> #include<sys/types.h> #define PROMPT_STRING "[myshell]$" #define QUIT_STRING "exit\n" static char inbuf[MAX_CANON] char *
BatDelay
- A Windows console app to delay a specified number of seconds and exit if a key is pressed. Handy for automation batch files. For example: :start batdelay 10 if ERRORLEVEL 1 goto :EOF ...do something goto start
cangkuguanlixitong
- 仓库管理系统 含源代码- #import "c:\Program Files\Common Files\System\ado\msado15.dll" no_namespace rename("EOF","adoEOF") rename("BOF","adoBOF") class ADOConn { public: _ConnectionPtr m_pConnection _RecordsetPtr m_pRecordset public:
eofstain
- 站点分布式EOF分析源代码,用于分析数据的空间时间构型,相关性和主要形式,全代码以77格式编写,可以使用。-EOF analysis of sites distributed source code, used to analyze the data structure of space time, the relevance and the main form, the format code to 77, you can use.
jtxor
- 程序使用C++展示相当强的XOR加密。 当它unencrypting文件时,包括EOF验错和自动侦查。程序很好被提供并且做在串类建立的好用法。-Program Uses C++ to Demonstrate A Fairly Strong XOR Encryption. Includes EOF error checking and automatic detection of when it is unencrypting a file. Program Is Documented
explore
- String Pointers to explore how the pointer work -The program takes lines from standard input and keeps the last n of them in memory as it goes through standard input. When it gets to an EOF, it prints the last n lines out. You may assume n is les
pc
- 1.开启档案 ( 来源档 、 目的档 ) 2.逐一检查来源档每个字元,若不是EOF则到3. ,否则结束 3. (1)若发现字元为 a~z 或 A~Z 或 0 ~ 9 或 _ 丢给procId()处理 (2)若发现字元为 (字元) 时,丢给procChar() 处理 (3)若发现字元为 " (字串) 时,丢给procString()处理 (4)若发现字元为 \ (注解) 时,丢给procComment()处理 (5)若发现字元为空白或跳格时,丢给printp
chaofan
- #include<iostream> using namespace std int s[100005] int main() { int i,j,k,sum,n while(scanf(" d",&n)!=EOF) { for(i=0 i<n i++) scanf(" d",&s[i]) sort(s,s+n) for(i=sum=0 i<n i++) { sum += s[i] * (n - i)