b+tree 的插入和查询操作，代码比较简练，效率高。缺少删除操作。
上传时间： 2013-12-19
上传用户：lijinchuan
求解大规模非对称线性方程组AX=b的迭代方法
上传时间： 2017-02-12
上传用户：xiaohuanhuan
我们程序设计课上要求写的b+树的c++实现，可以进行插入查找删除等操作，代码简单，适合新手
上传时间： 2013-12-16
上传用户：gxrui1991
b树的实现，包括打开关闭节点，打开节点 关闭节点，打开所有节点，关闭所有节点， 打开子节点，显示当前节点的增加节点 增加多个节点，删除节点
标签： 树
上传时间： 2014-01-13
上传用户：silenthink
DSP2810的B口IO口作50MS的中断变化输出
上传时间： 2017-02-17
上传用户：ukuk
二叉树 数据结构 忘了是第几个实验 好像是第六个 用b树做的
上传时间： 2017-02-17
上传用户：dave520l
读者要在命令行下输入tcc -B d:\tc\source\pport.c,来编译 之后需要编译server.c和clint.c
标签： csourcepport server clint tcc
上传时间： 2017-02-18
上传用户：ZJX5201314
Instead of finding the longest common subsequence, let us try to determine the length of the LCS. Then tracking back to find the LCS. Consider a1a2…am and b1b2…bn. Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. Case 2: am≠bn. Wehave to find the LCS of a1a2…am-1 and b1b2…bn, and a1a2…am and b b b b1b2…bn-1 Let A = a1 a2 … am and B = b1 b2 … bn Let Li j denote the length of the longest i,g g common subsequence of a1 a2 … ai and b1 b2 … bj. Li,j = Li-1,j-1 + 1 if ai=bj max{ L L } a≠b i-1,j, i,j-1 if ai≠j L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
标签： the subsequence determine Instead
上传时间： 2013-12-17
上传用户：evil
综合2叉树及B+树优点的能根据增删改而分裂或合并的完整程序(现在以8bit(BYTE key)为关键字,可扩充到64bit的double为key,用户数据包现在以float ton表示,可扩充到任意结构struct)
上传时间： 2017-02-19
上传用户：498732662
Reads/writes text as a character stream, buffering characters so as to provide for the efficient reading/writing of characters, arrays, and lines. The buffer size may be specified, or the default size may be used. The default is large enough for most purposes. In general, each read request made of a Reader causes a corresponding read request to be made of the underlying character or byte stream. It is therefore advisable to wrap a BufferedReader around any Reader whose read() operations may be costly, such as FileReaders and InputStreamReaders.
标签： characters character buffering efficient
上传时间： 2017-02-20
上传用户：cjl42111
