cipher

Test Instructions:LinkMethod:Permutation groupparsing:Quite a simple permutation group? But the first time was too two to think wrong. I thought I was going to just follow that sequence and get the power of the permutation group OK. The results found that I did not affect this many loops-it is not funny. So hurry back and ask for a loop. For each of the intra-cycle permutations, the water problem is followed.Code:#include #include #include #include #define Nusing namespace STD;intN,k;intA[n];int

#include #include #include #include using namespace Std;int main (){int ce[26], co[26];string encrypted, orginal;String::size_type I, Len;Ios::sync_with_stdio (FALSE);while (CIN >> encrypted >> orginal){memset (CE, 0, sizeof (CE));memset (Co, 0, sizeof (CO));Len = Orginal.length ();for (i = 0; i {++ce[encrypted[i]-' A '];++co[orginal[i]-' A '];}Sort (CE, ce+26);Sort (Co, co+26);if (Equal (CE, ce+26, CO))cout Elsecout }return 0;}UVa1339 Ancient Cipher

The program task is to convert the genetic phonetic alphabet to 6 digits, we can use any well-remembered pinyin string, the transformation process is as follows, the first step, the string six a group of folding up, such as wangximing into: WangxiMingIn the second step, we want to add the ASCII value of all the characters perpendicular to the same position, and draw out 6 numbers, as the example above, then draw 228 202 206Then the number of each digit processing, that is, the number of each bit

Cipher Time limit:1000 ms Memory limit:10000 K Total submissions:19228 Accepted:5148 DescriptionBob and Alice started to use a brand-new encoding scheme. surprisingly it is not a public key cryptosystem, but their encoding and decoding is based on secret keys. they chose the secret key at their last meeting in Philadelphia on February 16th, 1996. they chose as a secret key a sequence of N distinct integers, A1 ;.

Poj 1026 Cipher (replacement group) Link: poj 1026 Given n integers (1-n) as keys, a string is given, Evaluate the k-encoded string Analysis: brute-force solving times out and can be solved using the knowledge of the replacement group. Replacement group: a one-to-one transformation of a finite set is called replacement. A pair of replicas form a replacement group. For a set a (a [1], a [2], a [3]... a [n ]), (B [a [1], B [a [2], B [a [3]... B [a [n])

(inti = 0;i )TenA[i% 6] = a[i% 6] +S.charat (i); One for(inti = 0;i ) { A while(A[i] >= 10) { -String temp = "" +A[i]; - intA = 0; the for(intj = 0;j ) -A = A + (Temp.charat (j)-' 0 '); -A[i] =A; - } + } -StringBuffer T =NewStringBuffer (""); + for(inti = 0;i ) A t.append (A[i]); at returnt.tostring (); - } - - Public Static voidMain (string[] args) { -Main test =NewMain (); -Scanner in =New

/////////////////////////////////////////////////////////////Author:stardicky//e-mail: [email protected]//qqnumber:9531511//Companyname:ezone International//class:hbs-0308//Title: Using dotnet Password system to ensure data security///////////////////////////////////////////////////////////////Note: A des symmetric encryption algorithm using one of the dotnet cipher Systems ensures data security/////////////////////////////////////////////////////////

; for(inti =0; I 1]; for(inti = n1; I >=0; i--) sa[--top[x[y[i] []] =Y[i]; Swap (x, y); P=1; x[sa[0]] =0; //Calculate Rank Value for(inti =1; I ) X[sa[i]]= (y[sa[i-1] [= Y[sa[i]] y[sa[i-1]+k] = = Y[sa[i]+k])? P1: p++; if(P >= N) Break; M=p; }}voidbuild_height () {//Rep (i,0,n) rank[sa[i]] = i; intK =0; for(intI=0; ii; for(intI=0; i){ if(k) k--; intj = Sa[ra[i]-1]; while(S[i+k] = = S[j+k]) k++; Height[ra[i]]=K; }}intMain () {scanf ("%s", s); intLen =strlen (s); fo

{ - for(intI=0; i -; i++) About { $ if(ch==C[i]) - { - if(i== -) - { Afout' '; + } the Else - { $chout=Char('a'+i); thefoutChout; the } the } the } - } in}Input file file1_1.in// file1_1.inWe'll attack tomorrow morningOutput File File1_1.out// file1_1.outfu flbb dssdkw sovoaaof VOANLNCSingle-table substitution

Topic: Given a string, the string of the last column after the string is sorted to the beginning of each character.Legend of the suffix array 0.0 last night DC3 did not understand, so wrote a multiplier 0.0 Mr. Luo's 25 lines of code is really abstract Qaq Konjac Konjac Express Understanding can not qaq so I wrote a relatively clear version QaqFirst, this is the ring, so we add the first n-1 character of the string to the end of the string and then the suffix array.After this, enumerate each beg

Topic Link Topic 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 an encryption method he thought was the ultimate: In a circle of information that needs to be encrypted, it is clear that there are many different ways of reading them. For example, it can be read as:JSOI07 soi07j oi07js i07jso 07JSOI 7jsoi0 sort them by the size of the string: 07JSOI 7jsoi0 i07jso JSOI07 oi07js soi07j read th

This paper illustrates the method of generating random cipher by Python3, which has a wide range of practical value in Python programming. The specific methods are as follows: This article mainly realizes the creation of 8-bit random password (uppercase and lowercase + digits), and uses Python3 to generate the random password of the primary algorithm. The main functional code is as follows: __author__ = ' Goopand ' import string import random

keyboard to locate success, must be recognized through the image, Identify the location coordinate information of the keyboard The following function is to identify the coordinates of the cipher keyboard by means of an image recognition method. The password keyboard encountered is shown in the following figure There is also the contents of the input box, I have to intercept. Solution idea: 1, image recognition, and then click the coordinates 2, dev

Portal: http://www.lydsy.com/JudgeOnline/problem.php?id=1031Save the suffix array template.In fact, if the data range is smaller, or the space limit is a little larger, or even as long as the alphabet is a little bit more can use the suffix automaton, but unfortunately the space is not allowed, just use the suffix array!In fact, I still do not understand the code, whether it should be a black box code back down?#include 　　_bzoj1031 [JSOI2007] character encryption

http://www.lydsy.com/JudgeOnline/problem.php?id=1406Test instructions: Given n, ask for x^2==1%nX^2-1=k*n(x+1) * (x-1) ==k*n(x+1) (x-1)%n=0N=a*b(x+1) |a and (x-1) |b or (x-1) |a and (x+1) |b1#include 2#include 3#include 4#include 5#include string>6#include 7#include Set>8 #definell Long Long9 using namespacestd;Ten intN; One Setint>ans; A intMain () { -scanf"%d",n); - intLen= (int) sqrt (n); theAns.insert (1); - for(intI=1; i) - if(n%i==0){ - inta=n/i,b=i,x; + for

) for (int i = (r); i > (l); i--)#defineRep_1 (i,r,l) for (int i = (r); I >= (l); i--)#defineMS0 (a) memset (A,0,sizeof (a))#defineMS1 (a) memset (A,-1,sizeof (a))#defineMSi (a) memset (A,0x3f,sizeof (a))#defineINF 0x3f3f3f3f#defineLson L, M, RT #defineRson m+1, R, RT #definePB Push_backtypedef __int64 Ll;templatevoidRead1 (T m) {T x=0, f=1;CharCh=GetChar (); while(ch'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} while(ch>='0'ch'9') {x=x*Ten+ch-'0'; ch=GetChar ();} M= x*F;} TemplatevoidRead2

First of all, for each position, it starts the hash of the length of Y, and then for the hash value of the vertical column of the hash value, sorted after the outstanding number.Time Complexity $o (n^2\log N) $.#include 　　BZOJ1170: [Balkan2007]cipher

Sometimes we do not log on to the server for a long time will forget the password, want to retrieve the password, you can try to use the following tools.The purpose of this article is not to give you the solution, just for the convenience of work to retrieve the password. In fact, you can also change the password through rescue mode. If someone has done something bad through this article, it is irrelevant to the author of this article. Technology is endless, use caution.wget--no-check-certificat

Test instructions: Was killed by test instructions ... orz ..... That substitution is not an ASCII code plus a few ... is to randomly change to another character ...Solution: Just count the number of occurrences of each letter and then sort the array teller.Code:#include 　　POJ 2159 Ancient 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