cipher

cocoa? (X,n are all positive integers in the title) input file has only one row and only a number n (112Sample Output15711HintsourceSolutionTest instructions is very simple, $x ^{2}\equiv 1\left (mod n \right) $ on $\left [1,n\right] $ on all solutionsConvert: $x ^{2}= kxn+1$$x ^{2}-1=kxn$$ (x+1) (x-1) =kxn$So make $x+1=k ' xn ', x-1=k ' xn ' $ meet $k ' xk ' =k,n ' xn ' =n$Then find the approximate number of N, enumerate more than $\sqrt n$ and judge the approximateThe answer will be repeated,

ProblemTest instructionsGive you a chemical formula that contains only Chon organic matter, such as C6H5OH, for relative molecular massAnalysis。。。Code #include UVA 1586 Ancient Cipher

[1][3]=2; k_table[3][1]=2; 111k_table[1][7]=4; K_table[7][1]=4; thek_table[1][9]=5; K_table[9][1]=5;113 //init .... thek_table[2][8]=5; K_table[8][2]=5; the //init .... thek_table[3][7]=5; K_table[7][3]=5; 117k_table[3][9]=6; K_table[9][3]=6; 118 //init ....119k_table[4][6]=5; K_table[6][4]=5; - //init ....121k_table[7][9]=8; K_table[9][7]=8;122 if(k_table[a][b]>0){123 if(r[k_table[a][b]]==1)124 return true; the E

/*Old password for Chinese questionsYou can see if you can change the location of the first character to exactly match the 26 letters of the second character.Solution: count the number of characters in two strings. If the number of characters in each string is the same, yes is output; otherwise, no is output.Difficulties: it is a little difficult to count the number of times each character appears.Key Point: SortingProblem solving person: lingnichongSolution time:Experience in solving problems:

Topic Link: Click to open the linkTitle: Give a sequence of encodings, each of which encodes the character on the I bit back to the a[i] bit. Then give a k, and the initial string, ask what the string after K-Times is.K may be very large, can not be violent, so to use the permutation group, to find the rotation of the ring, assuming that the number of m in the ring, then each code m, it represents this back to the initial state, can be used k%m, so reduce the number of encodings. If the position

#defineDown (i,j,k) for (int i = j; I >= K; i--)6 #defineMAXN 2001007 using namespacestd;8 9 CharS[MAXN];Ten intSA[MAXN], T[MAXN], T2[MAXN], C[MAXN], N; One A voidBuild_sa (intm) - { - intI, *x = t, *y =T2; theRep (I,0, M-1) C[i] =0; -Rep (I,0, N-1) C[x[i] = s[i]]++; -Rep (I,1, M-1) C[i] + = c[i-1]; -Down (i,n-1,1) Sa[--c[x[i]] =i; + for(intK =1; K 1){ - intp =0; Rep (i,n-k,n-1) y[p++] =i; +Rep (I,0, N-1)if(Sa[i] >= k) y[p++] = Sa[i]-K; ARep (I,0, M-1) C[i] =0; atRep (I,0, N-1

string for 100% does not exceed 100000.SourceSolutionThe so-called order, in fact, is to copy the string over the $sa$ value, the last character of each string corresponds to the $s[sa[i]+n-1]$1#include 2 using namespacestd;3 Chars[200005];4 intsa[200005], wv[2][350005], tong[200005];5 6 BOOLcmpint*tmp,intXintYintj)7 {8 returnTMP[X] = = Tmp[y] tmp[x + j] = = Tmp[y +j];9 }Ten One voidGetsa (intNintm) A { - intp =0, *x = wv[0], *y = wv[1]; - for(inti =0; I i) the++tong[x[i] =S[i]

:772243344836297332716652875843* End of comment on the program head*/On the code:Import Java.util.Scanner;public class Main {public static void Main (string[] args) {Scanner sc = new Scanner (system.in);int n = sc.nextint ();String arr[] = new String[n];for (int i = 0; i Arr[i] = Sc.next ();}for (int i = 0; i Z (Arr[i]);}}public static void Z (String s) {int num = 0;int arr[] = new INT[6];for (int i = 0; i ARR[I%6] + = S.charat (i);}for (int i = 0; i System.out.print (f (arr[i]));}System.out.pri

Virginia Code table650) this.width=650; "src=" Http://s1.51cto.com/wyfs02/M01/78/14/wKioL1Z1WwDBEaHWAAGY7UKfZSE837.jpg "title=" 3ac79f3df8dcd100d56d833d748b4710b9122f3c.jpg "alt=" Wkiol1z1wwdbeahwaagy7ukfzse837.jpg "/>=============================================#维吉尼亚密码 encryptionkey= ' HelloWorld 'plaintext= ' Whereisthekey '#key = ' Relations '#plaintext = ' Tobeornottobeth 'Ascii= ' ABCDEFGHIJKLMNOPQRSTUVWXYZ 'Keylen=len (Key)Ptlen=len (plaintext)ciphertext = ' 'i = 0While I j = i% KeylenK =

1.Encryption Algorithm Daquan:Rsa:RSA is made of the initial letters of the surnames of Ron Rivest, Adi Shamir, and Leonard Adleman, who first publicly desc Ribed the algorithm in 1977.Http://www.emc.com/corporate/about-rsa/index.htmHttps://en.wikipedia.org/wiki/RSA_%28cryptosystem%29DSA: Digitally Signature algorithm digital Signature AlgorithmECDSA: Elliptic Curve digital Signature algorithm Elliptic curve Signature algorithmDSS: Digital Signature StandardMD5: message-digest algorithm 5 messag

character. Equivalent to ' [^a-za-z0-9_] './xnMatch N, where n is the hexadecimal escape value. The hexadecimal escape value must be two digits long for a determination. For example, '/x41 ' matches ' A '. '/x041 ' is equivalent to '/x04 ' ' 1 '. ASCII encoding can be used in regular expressions:/numMatches num, where num is a positive integer. A reference to the obtained match. For example, ' (.) /1 ' matches two consecutive identical characters./nIdentifies an octal escape value or a backwar

Transmission DoorKind of a template.The topic says loop, then copy a bunch of stitching on it.Then we can find the suffix array, and then we'll do it.Although it is the suffix, there will be more strings in the back, but the topic is that the loop, but there is no effect.--code1#include 2#include 3#include 4 #defineN 2000055 6 intm ='Z'+1;7 intLen, Buc[n], x[n], y[n], sa[n];8 CharS[n];9 TenInlinevoidBuild_sa () One { A intI, K, p; - for(i =0; I 0; - for(i =0; i ; the for(i =1;

string are not necessarily letters, numbers, or symbols.OutputThe output line. is the encrypted string.Sample InputJSOI07Sample OutputI0o7sjHINTThe length of the data string for 100% does not exceed 100000.Idea: a template title for the suffix array.We just need to copy this string to the back, and then handle the SA, which in turn outputs the starting point in the ranking.#include #include #include #include using namespace STD;#define N n*2-1Const intm=200100;CharCH[M];intn,m=0, Sa[m],t1[m],t2

(Rsort,0,sizeof(Rsort)); for(intI=1; i; for(intI=1; i1]; for(inti=n;i>=1; i--) sa1[rsort[rank[i]]--]=i; intln=1, p=0; while(pN) {intk=0; for(inti=n-ln+1; ii; for(intI=1; iif(sa1[i]-ln>0) sa2[++k]=sa1[i]-Ln; memset (Rsort,0,sizeof(Rsort)); for(intI=1; i; for(intI=1; i1]; for(inti=n;i>=1; i--) sa1[rsort[rank[sa2[i]]]--]=Sa2[i]; for(intI=1; iRank[i]; rank[sa1[1]]=1;p =1; for(intI=2; i) { if(tt[sa1[i]]!=tt[sa1[i-1]]|| tt[sa1[i]+ln]!=tt[sa1[i-1]+LN]) p++; Rank[sa1[i]]=p;

, x[sa[1]] = ++p; Rep (I,2, N) x[sa[i]] = y[sa[i-1] [= Y[sa[i]] y[sa[i-1] + K] = = Y[sa[i] + K]? P: ++p;if((M = p) = = N) Break; }}voidGet_height () {Rep (I,1, N) rk[sa[i]] = i;intK =0; Rep (I,1, N) {if(k) k--;intp = sa[rk[i]-1]; while(S[i + K] = = s[p + K]) k++; Height[rk[i]] = k; }}intMain () {scanf ("%s", In +1); n = strlen (in +1); Rep (I,1, n) in[i + n] = In[i], s[n + i] = s[i] = (In[i] >=' A ' In[i] ' Z ') ? In[i]-' A '+1: In[i]-' A '+1+ -; n + = n; Get_sa (); Rep (I,1, N)if(Sa[i] 2

]~# Lltotal9-rw-r--r--1Root root8864May1 -:GenevaINSTALL.TXTDRWXR-xr-x2Root root6June2 on: AboutMnt[email protected]~# CD MNT [email protected]~/MNT # lltotal0[email protected]~/MNT # Touch123[email protected]~/MNT # Touch Txt[email protected]~/mnt # VIM txt [email protected]~/mnt # cat txt123456[email protected]~/mnt # ls123Txt[email protected]~/MNT # lltotal4-rw-r--r--1Root root0June2 on: $ 123-rw-r--r--1Root root7June2 on: $Txt[email protected]~/MNT # CD.5. Uninstall shutdown[Email protec

= ' HOSTNAME ' # HOSTNAME not set some machinesIf [-x/usr/bin/keychain-a-F $HOME/.keychain/${hostname}-sh]; Then/usr/bin/keychain--clear $HOME/.ssh/id_rsaSOURCE $HOME/.keychain/${hostname}-shFiThen add in the script:SOURCE $HOME/.keychain/${hostname}-shThird, reference:http://www.snowfrog.net/2007/11/15/ssh-ssh-agent-keychain-and-cron-notes/Https://wiki.gentoo.org/wiki/Keychain/zh-cnHttps://serverfault.com/questions/92683/execute-rsync-command-over-ssh-with-an-ssh-agent-via-crontabSolution to th

http://www.lydsy.com/JudgeOnline/problem.php?id=1031It is easy to think of this as copying the string to its end and then the suffix array out of the SA and output by the interval.Then change the template and put the cardinal sort outside#include Suffix array of the original: #include 　　 DescriptionLike to delve into the problem of JS classmate, and recently fascinated by the encryption method of thinking. One day, he suddenly came up with what he thought was the ultimate encryption: to

Because it is a ring, it can be reproduced in the back again.#include 　　bzoj1031: [JSOI2007] character encryption cipher suffix array

($ password). $salt), that is, after MD5 the user's password, add a salt, and then MD5, saved in the password field. Therefore, if a different system of data conversion, according to this principle, the other system's user name and password calculation, import ucenter uc_members table, to achieve user migration. For example, if the original system uses an algorithm such as MD5 (password) to save the password, then the program randomly generated salt, and then calculate the cumulative MD5, so it