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how to set up a sister city links are so few routes do not intersect also the la...
动态规划法---如何设置友好城市间的航线使的航线数又最大又不相交-dynamic programming-- how to set up a sister city links are so few routes do not intersect also the largest
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- 积分：1
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Management System Status
学籍管理系统开发-Management System Status
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- 积分：1
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VC++ MFC 不可移动的窗体.rar
VC++ MFC 不可移动的窗体.rar-VC++ MFC immovable form. Rar
- 2022-09-22 02:35:03下载
- 积分：1
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Comrt_IDL.tar.gz
Comrt_IDL.tar.gz
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- 积分：1
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windows w indows w indows
windows w indows w indows-windows windows windows windows windows
- 2022-01-27 19:59:58下载
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Udp chat program
Ｕｄｐ聊天程序-Udp chat program
- 2022-03-10 03:27:24下载
- 积分：1
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When starting with protocols that work at the upper layers of the network models...
When starting with protocols that work at the upper layers of the network models, each set of data is wrapped inside the next lower layer protocol, similar to wrapping letters inside an envelope. The application creates the data, then the transport layer wraps that data inside its format, then the network layer wraps the data, and finally the link (Ethernet) layer encapsulates the data and transmits it. Each network layer either encapsulates the data stream with additional information, or manages data handling or come part of the connection
- 2022-05-23 09:07:19下载
- 积分：1
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城市公交地图编辑器源代码
城市公交地图编辑器源代码-urban transit map editor source code
- 2022-03-24 02:18:50下载
- 积分：1
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There is a infinite series, whose general term is expressed as: a [n] = 10* 10*....
有一个无穷数列，其通项表示为：
a[n]=10*10*... n=0,1,2,3.....
构成了1,10,100,1000...
把它连起来，就成了数串。
判断数串的第i位到底是0还是1.性能要求1s
-There is a infinite series, whose general term is expressed as: a [n] = 10* 10*... n = 0,1,2,3 ..... constitute 1,10100,1000 it even ... together, it becomes the number of strings. Determine the number of strings in the end of the first i-bit is 0 or 1. Performance requirements 1s
- 2022-03-22 03:32:03下载
- 积分：1
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使用解码图像从数组，我们可以做很多想想它的
#include < iostream.h >
#include < stdio.h >
#include < string.h >
结构点头
{
名称 char [10] ；
int 的时间 ；
int 等 ；
} 节点 [10]，x;
int n，m ；
双 av，w = 0 ；
void sort)
{
int i、 j ；
为 (我 = 0 ； 我 < n-1 ； + + 我）
为 (j = i + 1; j < n; + + j)
{
如果 （节点 [i] 时间 > node[j].time)
{
x = 节点 [i] ；
节点 [i] = 节点 [j] ；
节点 [j] = x;
}
}
}
void read)
{
int i;
cout n;
为 (我 = 0 ； 我 < n; + + 我）
{
cout 节点 [i].name ；
cout 时间节点 [i] ；
}
}
void print)
{
int i;
cout
- 2023-03-08 09:10:03下载
- 积分：1
