random number generator python

=shared--enable-disk_cache=shared--enable-mem_cache=shared-- enable-proxy=shared--enable-proxy_connect=shared--enable-proxy_ftp=shared--enable-proxy_http=shared-- enable-file_cache=shared--enable-charset_lite=shared--enable-case_filter=shared--enable-case_filter_in=shared- -enable-ssl=shared--WITH-APR=/USR/LOCAL/APR--with-apr-util=/usr/local/apr-utilMakeMake installNot that the first method must have a problem, I have also encountered the first installation is available, if you are the first way

\[\begin{eqnarray*}x_i=x_{i-1}+x_{i-2}\\x_i^2=x_{i-2}^2+x_{i-1}^2+2x_{i-2}x_{i-1}\\x_{i-1}x_i= (X_{i-3}+x_{i-2}) (x_{i-2}+x_{i-1}) \ \=2x_{i-3}x_{i-2}+x_{i-2}x_{i-1}+x_{i-3}^2+x_{i-2}^2\end{eqnarray*}\]Therefore, the transfer matrix can be constructed $a$ recursive.If you wish to set $n\geq m$, you can preprocess $a^0,a^1,..., a^n$ and $a^n,a^{2n},..., a^{nn}$.So the complexity of querying a number is $8^3$.Total time complexity is $o (n (8^3+\log N))

) - { - if(n1) in { -ans_matrix[0][0]=KSC (ans_matrix[0][0],matrix[0][0]); toans_matrix[1][0]= (KSC (ans_matrix[1][0],matrix[0][0]) +matrix[1][0])%m; + } -n>>=1; thell TMP1=KSC (matrix[0][0],matrix[0][0]); *ll tmp2= (KSC (matrix[1][0],matrix[0][0]) +matrix[1][0])%m; $matrix[0][0]=tmp1,matrix[1][0]=TMP2;Panax Notoginseng } - } the + voidInit () A { thescanf"%lld%lld%lld%lld%lld%lld",m,a,c,x0,n,g); +matrix[0][0]=a%m,matrix[0][1]=0, matrix[1][0]=c%m,matrix[1][1]=1;

, LL AMP;GCD) {if(!B) {x=1,y=0, Gcd=a;return; } EXGCD (B,a%b,y,x, GCD);y=y-a/b*x;} ll Get_inv (LLx, ll MOD) {ll x,y,gcd; EXGCD (x, MOD,X,Y,GCD);return(X%mod+MOD)%mod;} void Bsgs (ll a,ll b,ll C) {init (); llm=(int) Ceil (sqrt(C)); ll k=1; for(intI=0;im; i++) {intflag=1; for(intJ=head[k%mod];j!=-1; J=edge[j].Next) {if(edge[j].val==k) {flag=0; Break;} }if(flag) Edgeadd (k%mod, i,k); K=k*a%c; } ll Invk=get_inv (K,C); ll invd=1; for(intI=0; im; i++) {ll tmpb=b*INVD%c; for(intJ=head[tmpb

Question :..... Take a long look. First, since the Lexicographic Order of the sorted sequence is the smallest, You must select as few numbers as possible. Then T is 1 ~ The M * n sequence does not exist. (At the beginning, I missed this condition) Okay, this question is greedy.Select the minimum point that is not marked each time, and mark both the lower left and the upper right corner (remember to mark the duplicate break, otherwise it will be down) Note that if two int values are opened fo

1. Seeds and distribution Five types of distributions are introduced here, including the following: (a), and (). Parameters required for distribution and use are provided. 2. Test the random number generator. This section contains only one simple example, which is not described in detail. The following is the procedure for getting a

A whole bunch of borders was not known at first, and after a few random sentences, WAHelpless to download the data, and then crazy to judge all kinds of strange boundariesTo gouge out the boundary problemFirst we consider the situation of a=1X1+k*b=t (mod p)EX_GCD can be solvedConsider the situation of a>1Make s=x+b/(A-1)The original is becoming a geometric series.S1*a^k= (t+b/(A-1)) (mod p)BSGS solution after moving the itemAll other boundaries can b

Prioritize the processing of arrays in the way it describesAnd then, in order to get as many small numbers as possible in the sequence,So 1 is a must to appear, in order to make the whole sequence of the order of the dictionary after the smallest.We thought, if 2 could be in this series, it'd be better.But 2 may not be in this series, that is, 2 walk 1 is impossible to go to the place, you can not walk 2.So from small to large enumeration numbers, if the current

; - } -T/=D; the ll x, y; the EXGCD (a,b,x,y); thex= (x*t)%p; the while(x0) x+=p; - returnx+1; the } thell C=pow (A-1, P-2); thell a= (b*c+x1)%p,b=p,t= (b*c+t)%p;94 if(a0) a= (a+p)%p; the if(b0) b= (b+p)%p; thell d=gcd (A, b); the if(t%D) {98 return-1; About } -T/=D;101 ll x, y;102 EXGCD (a,b,x,y);103 while(x0) x= (x+p)%p;104x= (x*t)%p; thell ans=Work (a,x);106 if(ans!=-1)returnans+1;107 Else returnans;108 }109 intMain

Question 1-random sentence Generator-7 marksWrite a program This reads in files of words and produces random but structurally correct 中文版Sentences, according to the syntax (rules of grammar) specified on the next page.Here are the output of a demo program. These sentences is correctly structured 中文版 but (usually) makeAbsolutely no sense (but is often fun:-)It is

This article describes how to implement a random password dictionary generator in python. if you need a dictionary generator, you can refer to all the passwords you want to use. the algorithm is either nested too deep, or memory consumption (will overflow ). later, I selected an algorithm with a low probability of simp

1 #/bin/python32 ImportSYS3 Import Time4 ImportRandom5STRs = [i forIinchRange (32,128)] #产生密码的ASCII码的序列6Length =10 #生成密码长度7#下面这个函数用来生成Length长度的密码8 defrandomstr (strs,length):9Liststr = []Ten forIinchRange (length): One liststr.append (Random.choice (STRs)) A returnListstr - defstrings (): -String ="" thechar =randomstr (strs,length) - forIinchChar: -string+=chr (i) - returnstring +#上面这个函数将调用randomstr函数, returns a string password - +Start =Time.time ACount =1

The following article introduces the application of Python random number and Python random string in the form of code in practical application. The following is a detailed introduction of the article, let's share it with you! Python