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{ ngx_http_upstream_keepalive_peer_data_t*Kp ngx_http_upstream_keepalive_srv_conf_t*Kcf Ngx_log_debug0 (Ngx_log_debug_http, R -Connection -Log,0,"Init keepalive peer"); Kcf=Ngx_http_conf_upstream_srv_conf (US,

decryptionFinally, let us prove that why the private key decryption, must be able to correctly get m. This is the proof of the following equation: Cd≡m (mod n) Because, according to the encryption rules Me≡c (mod n) Thus, C can be written in the following form: c = me-kn Put C into the decryption rule that we want to prove: (me-kn) d≡m (mod n) It's equivalent to proving Med≡m (mod n) Because Ed≡1 (modφ (n)) So

correctly get m. This is the proof of the following equation: Cd≡m (mod n) Because, according to the encryption rules Me≡c (mod n) Thus, C can be written in the following form: c = me-kn Put C into the decryption rule that we want to prove: (me-kn) d≡m (mod n) It's equivalent to proving Med≡m (mod n) Because Ed≡1 (modφ (n)) So ed = hφ (n) +1 Put Ed into: Mhφ (n) +1≡m (mod n) Next, the

-kn Put C into the decryption rule that we want to prove: (me-kn) d≡m (mod n) It's equivalent to proving Med≡m (mod n) Because Ed≡1 (modφ (n)) So ed = hφ (n) +1 Put Ed into: Mhφ (n) +1≡m (mod n) Next, there are two cases to prove the above equation.(1) m and N coprime.According to Euler's theorem, at this point Mφ (n) ≡1 (mod n) Get (Mφ (n)) hxm≡m (mod n) The original has been proved.(2) m and

) d≡m (mod n) It's equivalent to proving Med≡m (mod n) Because Ed≡1 (modφ (n)) So ed = hφ (n) +1 Put Ed into: Mhφ (n) +1≡m (mod n) Next, there are two cases to prove the above equation.(1) m and N coprime.According to Euler's theorem, at this point Mφ (n) ≡1 (mod n) Get (Mφ (n)) hxm≡m (mod n) The original has been proved.(2) m and n are not coprime relations.At this point, because n equals the product of the

second close is calculated at this time. If the ratio is greater than 0.8, it is ignored. This removes 90% of the error match, while only 5% of the correct match is removed. As the article says.This is the summary of the SIFT algorithm. It is highly recommended that you read the original document, which will deepen your understanding of the algorithm. Please keep in mind that this algorithm is protected by patents. So this algorithm is included in the charge module in OpenCV.The SIFT in OpenCVN

1Chcp650012 SetHome_dir=Kp_home3 Setpackage_dir=/root/Java_source4 SetConfig_home=Anydir5 Setresource_path=/home/frank/opensource6 SetExchange_dir=\exchange.90km.com\Exchange7 Setremote_host=192.168.80.848 SetLogin_user=Root9 SetRemote_passwd=123456Ten One REM1. Create a working directory A ifexist%home_dir% ( -RD/S/Q%home_dir% - ) the MD%home_dir% - REM2. Enter the working directory - CD%home_dir% - + REM3. Get the PSCP tool. -xcopy \%exchange_dir%\xf\Pscp.exe + A REM4. Use PSCP to download

private key decryptionFinally, let us prove that why the private key decryption, must be able to correctly get m. This is the proof of the following equation: Cd≡m (mod n) Because, according to the encryption rules Me≡c (mod n) Thus, C can be written in the following form: c = me-kn Put C into the decryption rule that we want to prove: (me-kn) d≡m (mod n) It's equivalent to proving Med≡m (mod n) Because Ed≡1 (modφ (n

correctly get m. This is the proof of the following equation: Cd≡m (mod n) Because, according to the encryption rules Me≡c (mod n) Thus, C can be written in the following form: c = me-kn Put C into the decryption rule that we want to prove: (me-kn) d≡m (mod n) It's equivalent to proving Med≡m (mod n) Because Ed≡1 (modφ (n)) So ed = hφ (n) +1 Put Ed into: Mhφ (n) +1≡m (mod n) Next, the

at the name of the file to know what it is ... Well, the background information is finished, give an example, is my main function to let out to do a demonstration. int _tmain (int argc, char** argv){Mat trainimg; Pictures that need to be analyzedTrainimg=imread ("1.jpg", 1); Reading picturesHogdescriptor *hog=new Hogdescriptor (cvsize (3,3), Cvsize (3,3), Cvsize (5,10), Cvsize (3,3), 9); See reference article 1,2vectorHog->compute (Trainimg, Descriptors,size (1,1), Size (0,0)); Call the calcul

DC Motor Speed Control simulation operationBlock ControllerInPort command (n=1);InPort feedback (n=1);OutPort OutPort (n=1);Real error;Real pout;Parameter Real kp=10;Parameter Real Max_output_pos = 10;Parameter Real Max_output_neg =-10;Parameter Real ki=1;AlgorithmError: = command.signal[1]-feedback.signal[1];Pout: = Kp * ERROR;If pout > Max_output_pos ThenOUTPORT.SIGNAL[1]: = Max_output_pos;ElseIf Pout OUT

#!/bin/shbackup_timestamp= ' Date +%y%m%d 'Hostname= ' HOSTNAME 'num=89################### Verification File system opt################Memuseopt= '/bin/df-kp| grep/opt | Awk-f ' {print $} ' | Awk-f '% ' {print '} 'If [$MemUseOpt-gt$num]ThenEcho${backup_timestamp},${hostname},/opt,${memuseopt}, exceeding threshold >>xunjian.csvElseEcho${backup_timestamp},${hostname},/opt,${memuseopt}, normal >>xunjian.csvFi################### Verification File system v

#-*-coding:utf-8-*-import cv2 import NumPy as NP from find_obj import Filter_matches,explore_match from Matplotlib im Port Pyplot as Plt def getsift (): ' Get and view sift feature ' img_path1 = '. /.. /data/home.jpg ' #读取图像 img = cv2.imread (img_path1) #转换为灰度图 gray= cv2.cvtcolor (img,cv2. Color_bgr2gray) #创建sift的类 sift = Cv2. SIFT () #在图像中找到关键点 can also be computed in one step #kp, des = sift.detectandcompute KP