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No Internet access, the use of CD-ROM installation of GCC and gcc-c++ method is as follows: First put the CD into the optical drive, the virtual machine in the words on the ISO CD image can be. And then the following: [Root@localhost ~]# more gcc_install.sh #!/bin/bashMount/dev/cdrom/mntcd/mnt/centos/RPM-IVH glibc-common-2.5-42.i386.rpmRPM-IVH kernel-headers-2.6.18-164.el5.i386.rpmRPM-IVH libgcc-4.1.2-46.el5.i386.rpmRPM-IVH glibc-2.5-42.i686.rpmRPM-IVH cpp-4.1.2-46.el5.i386.rpmRPM-IVH libgomp-4.

an authentication mechanism on the SNMP agent. SNMP supports different authentication mechanisms, depending on the different versions of the SNMP protocol, monitoring Bao currently supports V2C and V3 two versions, where the V2C version of the authentication mechanism is relatively simple, it is based on plaintext password and authorized IP to authenticate, While the V3 version is authenticated by encrypti

(1) The figure has 6 vertex v1-v6, each edge of the Benquan value is on the graph, in the prim algorithm, I randomly select a vertex as the starting point, of course, we generally choose V1 as the starting point, OK, now we set the U set for the currently found in the smallest spanning tree vertex, TE set as the found edge, Now the status is as follows:U={V1}; te={};(2) Now find a vertex in the U set, the other vertex in the V-u collection of the minimum weight, such as, the Red Line intersectio

array Anv from subscript s to the part of T 3 int temp; 4 while (S Second: N (n>=2) stage race chart construction Hamiltonian pathwayN-Order Race chart: A forward graph with n vertices with an edge between each pair of vertices. There must be Hamiltonian paths for the N-Order race chart.The mathematical inductive method proves that the competition chart must exist in N >= 2 o'clock Hamilton Road:(1) n = 2 o'clock conclusion clearly established;(2) Assuming n = k, the conclusion is also establi

print (new_a) print (NEW_A1) print (NEW_A2)Output He xin He xinhe Xin he xin Segmentation User_info = "Ex|in s123 9" v1 = user_info.split (' | ') V2 = user_info.split (' | ', 1) v3 = User_info.rsplit (", 1) print (v1) print (v2) print (v3)Output[' Ex ', ' in s123 9 '] [' Ex ', ' in s123 9 '] [' Ex|in s123 ', ' 9 '] Length Len () User_info = "Ex|in s123 9" v1 = Len (user_in

Notes CUDNN v5 Runtime Library for Ubuntu14.04 (DEB) CUDNN v5 Developer Library for Ubuntu14.04 (DEB) CUDNN v5 Code Samples and User Guide (DEB)Download CUDNN V4 (FEB, 2016), for CUDA 7.0 and later. CUDNN User Guide CUDNN Install Guide CUDNN v4 Library for Linux CUDNN v4 Library for Linux (IBM Power8) CUDNN v4 Library for L4T (ARMV7) CUDNN v4 Library for L4T (ARM64) CUDNN v4 Library for Android (ARMV7) CUDNN v4 Library for Android (ARM64) CUDNN v4 Library for Windows CUDNN v4 Library for OS X C

for the first line segment (starting from p1 and p2), p3 and p4 must be on both sides. at the same time, for the second line segment (from p3 and p4 as the endpoint), p1 and p2 must be on both sides of the second line segment. to prove the intersection of two line segments, you only need to prove the two conditions behind them. This is a necessary and sufficient condition. How can we prove that the two points are on both sides of a straight line? Use the vector method to see the figure: In

http://poj.org/problem?id=1755Test instructions: Triathlon, everyone has their own speed at each stage, is there a 3 route length not 0 design so that someone can strictly win?We enumerate everyone wins and get the inequality group: s1/v1+s2/v2+s3/v3We divide both sides by S3.(S1/S3) * (1/V1) + (S2/S3) * (1/V2) + (1/V3) So S1/s3 and S2/s3 become the only two variables, so that the inequality group becomes two yuan an inequality group, with a semi-plan

==================================== NFS server can be considered as a file server. It allows your PC to mount the files shared by the remote NFS server to its own system through the network, in the client's view, remote files using NFS are like local files.The NFS protocol has been available in multiple versions since its birth, such as NFS V2 (rfc1094) and NFS V3 (rfc1813) (the latest version is V4 (rfc3010 ).Ii. Main differences between NFS Protoc

Problem description: In a connected graph G with several vertices, if the subgraph G' contains all vertices and some edges in G and does not form a loop, G' is called the Spanning Tree of graph G, the spanning tree with the minimum cost is called the least spanning tree. For example, if G is set, find the edge connecting all vertices (V1, V2, V3, V4, V5, V6) of graph G, and the sum of the weights of these edges is the minimum. How can we generate the

: any point in a triangle can be determined by the U, V, and the coordinates of the three vertices, where 0 Principle 2: any point on the Ray can be expressed by multiplying the direction vector (formatted) of the ray by its modulo (length of the vector), as follows: vpoint = originpoint + dir * LenIf the triangle and triangle intersect, the above two conditions must be met at the same time, so there are:(-DIR) * Len + (V2-V1) * u + (V3-V1) * V = orig

it is the shortest distance from the V1 vertex to the rest of the vertices, find a vertex nearest to the 1th vertex. It is known from the array dis that the V1 vertex is the V3 vertex recently. When the number 2nd vertex is selected, the value of dis[2] (the subscript starting from 0) has changed from "Estimated value" to "fixed value", i.e. the shortest distance from the V1 vertex to the V3 vertex is the

tree.The V0 (=V3) is also incorporated into the set U. The value of the secondary array is then modified. 1) will closedge[2].lowcost = 0;//to represent vertex V3 Three has been merged into U 2) because the weight of the Edge (V2,V3) is less than closedge[1].lowcost, it is necessary to modify closedge[1] for the Edge (V2,V