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Practice of dynamic arrays (Second Edition) a large amount of data management is...
实习性
动态数组(第二版)
大量数据的管理是很多程序员的心病,很难找到一个速度快、效率高、支持超大规模数据的表动态数组是一个功能强大的列表形数据管理链表,利用它可以轻松实现超大数据量的随机插入、删除、修改等操作,它另外一个特点就是速度极快,内存利用率高。 -Practice of dynamic arrays (Second Edition) a large amount of data management is the heart of many programmers, it is difficult to find a fast, high efficiency, support for ultra-large-scale data table dynamic array is a powerful list of shape data management list using it can easily realize large amount of data random insert, delete, modify, such as operation, it is another feature of this approach is extremely fast, high memory utilization.
- 2023-05-25 14:35:03下载
- 积分:1
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算法课程代码LAB8
CS101 是关于使用计算机和解决问题。学生将会学习如何计算机功能以及如何使用它们作为一种工具来做有益的事情。主要的重点,但是,是的设计与实现的自定义程序。本课程强调软件工程原则,在整个。采用一种设计方法有助于产生程序是"一次成功",此外,也可维护性。假定以前没有计算机或者编程的知识。
- 2022-01-27 18:54:50下载
- 积分:1
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a program for get the shortest path of two points.
距离最近的点对计算方法-a program for get the shortest path of two points.
- 2023-04-02 07:30:03下载
- 积分:1
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Code in java for the Traveling salesman problem. This code is very simple becaus...
Code in java for the Traveling salesman problem. This code is very simple because has only three pages. It s works same.
- 2022-03-04 06:11:45下载
- 积分:1
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csimple2d CFD的C源程序
CSIMPLE2d CFD C源程序-CSIMPLE2d CFD C source
- 2022-06-27 03:32:30下载
- 积分:1
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决策树ID3算法预测鲍鱼年龄
用了abalone 数据集,是通过鲍鱼的性别,大小,总量等等来判断鲍鱼的年龄。作业用了ID3决策树算法C++
- 2022-07-06 22:57:19下载
- 积分:1
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C language Multiplier
用C语言实现的乘法器-C language Multiplier
- 2022-01-27 19:14:22下载
- 积分:1
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unknownexercice
#include < stdlib.h >
#include < stdio.h >
#include < sys/types.h >
#include < sys/stat.h >
#include < fcntl.h >
int 主要 (int argc、 char *argv[])
{
int fd、 fd2、 fd3 ;
int i;
浅黄色 char [10] ;
fd=open(argv[1],O_RDONLY) ;
if(fd2=open(argv[2],O_WRONLY) = =-1)
{
fd3 = 共创 (argv [2],S_IRUSR |S_IWUSR) ;
read(fd,buff,10) ;
write(fd3,buff,10) ;
}
其他
{
fd2=open(argv[2],O_TRUNC) ;
read(fd,buff,10) ;
write(fd2,buff,10) ;
}
返回 0 ;
}
- 2023-05-04 19:30:03下载
- 积分:1
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下相当的问题说明:2个可装2个八个玻璃…
倒酒问题描述: 设有两个能装8两的酒杯(称为1号,2号)装满了酒, 和1个能装3两的空酒杯(称为3号), 问怎样用这3个酒杯向4个人 敬酒, 使得每个人都喝4两酒. 要求: 用程序计算出可行方案。 输入: 无 输出: 每一步决策.-down quite Problem description : two can be loaded with two of the eight glasses (known as 1, 2) filled with the wine, and one can hold three two empty glasses (known as the 3rd) and asked how to use it three to four glasses and toast the individual, so that each person maximum servings of both types 4 2 liquor. Request : procedures calculated options. Input : None output : each step of the decision-making.
- 2023-06-15 16:35:03下载
- 积分:1
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GNU的数值运算库,内容丰富!
GNU的数值运算库,内容丰富!-GNU numerical arithmetic library, a rich content!
- 2022-05-29 13:04:47下载
- 积分:1