additive cipher

After the row to find the map = = The number of occurrences of the order to determine equality1#include 2#include 3#include 4#include 5 using namespacestd;6 Chars1[ the],s2[ the];7 intnum1[ -],num2[ -];8 intMain ()9 {Ten while(~SCANF ("%s%s", S1,S2)) One { A intlen=strlen (S1); - for(intI=0;i -; i++) num1[i]=num2[i]=0; - for(intI=0; i) the { -num1[s1[i]-'A']++; -num2[s2[i]-'A']++; - } +Sort (num1,num1+ -); -Sort (num2,num2+ -); + BOOLfla

(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, need to go to the heavy, here apply setCode#include #include#include#include#include#includeSet>usi

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

){ theSystem.out.print ((count_x+1) + "th:"); - for(intcount_y=0; count_y) theSystem.out.print (mid_out[count_x][count_y]+ ""); the System.out.println (); the }94System.out.println ("the" +i+ "th result as ..." +j); the the the }98 AboutSYSTEM.OUT.PRINTLN ("The result is:" +out_result); - for(inti=0; i){101 102 System.out.println (K[i]);103 }104 } the 106 Public Static BooleanCheck_succe

/*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

, *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) c[x[y[i]]]++; -Rep (I,1, M-1) C[i] + = c[i-1]; -Down (i,n-1,0) sa[--c[x[y[i] []] =Y[i]; - swap (x, y); -p =1, x[sa[0]] =0; -Rep (I,1, N-1) inX[sa[i]] = y[sa[i]] = = y[sa[i-1]] y[sa[i-1]+k] = = Y[sa[i]

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]]; - for(inti =1; I i) -Tong[i] + = tong[i-1]; - for(inti = n-1; ~i; --i) +Sa[--tong[x[i

into:WangxiMingThe second step. The ASCII values of all characters perpendicular to the same position are added to produce 6 numbers, as in the example above,It is concluded that:228 202 220 206 120 105The third step. And then the number of each digit "indent" Processing: the number of each bit is added, the figure if not aNumber, it is then indented until it becomes a digit. For example: 228 = 2+2+8=12 = 1+2=3The above number is reduced to: 344836, which is the final output of the program!Requ

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.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[m],c[m];voidBuild_sa () {intI,p,k,*x=t1,*y=t2; for(i=0; i0; for(i=0; i for(i=1; i1]; for(i=n-1;

]]--]=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; } LN*=2; m=p; }}intMain () {scanf ("%s", A +1); intN=strlen (A +1); for(inti=n+1; i1; i++) a[i]=a

\ (S[sa[i] + n-1] (s=input+input,sa[i]#include #include #include #include #include #include #include using namespaceStd#define N 200001#define REP (I, A, b) for (int i = A; I #define DRP (i, A, b) for (int i = A; I >= b; i--)#define LL Long Longinline intRead () {intx =0, flag =1;Charch = getchar (); while(!isdigit (CH)) {if(! (ch ^'-')) flag =-1; ch = getchar (); } while(IsDigit (ch)) x = (x 1) + (x 3) + CH-' 0 ', ch = getchar ();returnx * FLAG;}intNCharIn[n];intS[n];intSa[n], T1[n], t2[n], c[

]~/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 protected] ~~ ~ # ll/dev/0crw1236 June 1:Third, encrypt the whole systemHttps://wiki.archlinux.org/index.php/Dm-crypt/Encrypting_an_entire_systemThere are various ways to include boot partition encryption and so on.

First, Reason:Before because the server can only login, and limit the root account login, so ssh through all the server, to achieve the public key forward SCP free password, etc., greatly facilitate the management of the server.Recently there is a need to do offsite backup of data. The simplest use of SCP script, but script manual execution is OK, put crontab does not work properly.Second, the solution:Script execution failed, immediately thought to print verbose log debug, that scp-v then read

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