neon text generator

[BZOJ1030] [JSOI2007] Text generatorQuestion DescriptionJsoi to the players zyx a task, the production of a "text generator" computer software: The user of the software is some low-young people, they are now using the GW Text Generator V6 version. The software can generate s

DescriptionJsoi to the players zyx a task, the production of a "text generator" computer software: The user of the software is some low-young people, they are now using the GW Text Generator V6 version. The software can generate some random articles----always generate a fixed length and completely random article--that

http://www.lydsy.com/JudgeOnline/problem.php?id=1030DescriptionJsoi to the players zyx a task, the production of a "text generator" computer software: The user of the software is some low-young people,They are now using the GW Text Generator version V6. The software can generate a number of random articles---always gen

Title DescriptionJsoi to the players zyx a task, the production of a "text generator" computer software: The user of the software is some low-young people, they are now using the GW Text Generator V6 version.The software can generate some random articles----always generate a fixed length and completely random article--

]; A Inc (h); the forI: =1 to - Do begin +Y: =tr[x, I]; - ifY >0 Then begin $ Inc (T); $Q[t]: =y; -X1: =Fai[x]; - whileTrue Do begin the ifTR[X1, I] >0 Then begin -Fai[y]: =tr[x1, I];Wuyi ifTAIL[TR[X1, I]] ThenTail[tr[x, I]]: =true; the Break - End; WuX1: =fai[x1]; - End; About End; $ End; - End; - End; - A procedureDP (t:longint); + var the I, J, X:longint; - $ begin the forI: =1 toTot Do begin the ifTail[

++) { intson=T[x].c[i]; if(son==-1)Continue; if(x==0) t[son].fail=0; Else { intj=T[x].fail; while(j!=0t[j].c[i]==-1) j=T[j].fail; T[son].fail=max (0, T[j].c[i]); if(t[t[son].fail].s==1) t[son].s=1; } list[++tail]=Son; } head++; }}intf[ the][11000];intMain () {intn,m; scanf ("%d%d",n,m); Tot=0; Clean (0); for(intI=1; i) {scanf ("%s", st+1); BT (); } BFS (); Memset (F,0,sizeof(f)); intsum=1; for(intI=1; i -)%Mod; f[0][0]=1; for(intu=1; u) {

PHP implementation of the text-based Moss code generator Introduced I recently encountered a requirement to generate Morse code audio files based on input text. After a few searches, I decided to write a generator myself. Download source Codes –2.63 KB Because I wanted to access my Morse code audio files through th

This article mainly introduces related information about PHP implementation of text-based Moss code generator. For more information, see a recent requirement for generating Moss code audio files based on input text. After several unsuccessful searches, I decided to write a generator myself. I decided to use PHP as my

Recently encountered a requirement to generate Morse code audio files based on input text. After a few searches, I decided to write a generator myself. Because I wanted to access my Morse code audio files through the web, I decided to use PHP as my main programming language. The above shows a webpage that starts to generate Moss code. In the downloaded zip file, contains the Web page for submitting the

']; } t[nw].end=1;}voidGetfail () {t[0].fail=0; for(intI=0;i -; i++) if(t[0].son[i]) {t[t[0].son[i]].fail=0; Q.push (t[0].son[i]); } while(Q.size ()) {intx=Q.front (); Q.pop (); for(intI=0;i -; i++) { if(T[x].son[i]) {T[t[x].son[i]].fail=T[t[x].fail].son[i];//T[t[x].son[i]].end|=t[t[t[x].son[i]].fail].end;Q.push (T[x].son[i]); } Elset[x].son[i]=T[t[x].fail].son[i]; } if(t[t[x].fail].end) t[x].end=1; } }intMain () {scanf ("%d%d",n,m); for(intI=

AC automata.F[I][J] is expressed as the number of scenarios in the text string that match the J node in the dictionary tree for length i.1#include 2 using namespacestd;3typedefLong Longll;4 Const intn=6005, mod=10007;5ll ans1,ans2,f[ the][n];6 Chars[ the];intcnt,n,m;7queueint>Q;8 structnode9 {Ten ints,v[ -],f; One }t[n]; A voidbuild () - { - intN=strlen (s);intnow=0; the for(intI=0; ii) - { - if(t[now].v[s[i]-'A']) now=t[now].

CharS[maxlen]; One A intf[2][maxn*Maxlen],pre,now; - intI,j,k,n,m,ans; - theInlinevoidInsertintLen) { -RegisterintI,now; - for(i=1; i'A'; - for(now=0, i=1; i) +now=ch[now][s[i]]?ch[now][s[i]]:ch[now][s[i]]=++tot; -tag[now]=1; + } AInlinevoidGetfail () { atRegisterinti,j,tmp; -L=0, r=1, dl[1]=0; - while(lR) - for(j=0,i=dl[++l];j -; j + +)if(Ch[i][j]) { - for(tmp=fail[i];tmp!ch[tmp][j];tmp=fail[tmp]); -FAIL[CH[I][J]]=I?CH[TMP][J]:0, intag[ch[i][j]]|=Tag[ch[t

(I,0, len-1) the { + intc=s[i]-'A'; A if(!ch[temp][c]) ch[temp][c]=++Ed; thetemp=Ch[temp][c]; + } -v[temp]=x; $ } $ voidGetfail () - { -queueint>h;intK; theH.push (1); - while(!h.empty ())Wuyi { the intx=H.front (); H.pop (); -Re (I,0, -) Wu { - if(!ch[x][i])Continue; About intvv=Ch[x][i]; $ H.push (vv); -k=Pre[x]; - while(!ch[k][i]) k=Pre[k]; -pre[vv]=Ch[k][i]; A if(V[ch[k][i]]) v[ch[x][i]]=V[ch[k

=strlen (s); -Foru (I,0, L-1){ + intid=s[i]-'A'; A if(!k->Nxt[id]) { atp= (trie*)malloc(sizeof(Trie)); -P->init (); -k->nxt[id]=p; -k=p; -ind[++cnt]=p; -}Else{ inK=k->Nxt[id]; - } to } +k->v=1; - } the * voidSetfail () { $Trie *k=r,*p;Panax Notoginseng ints=1, t=0; -q[++t]=R; the while(st) { +k=q[s++]; AForu (I,0, -){ the if(k->Nxt[i]) { +P=k->fail; - while(P!p->nxt[i]) p=p->fail; $K->nxt[i]->fail= (p?p->nxt[i]:r); $Q[++t]=k->Nxt

]-'A']=++tot; Anow=ac[now].nxt[s[i]-'A']; at } -Ac[now].e=1; - } - voidGet_fail () { -queueint>Q; - for(intI=0; i -; i++){ in if(ac[0].nxt[i]) { - //cout toQ.push (ac[0].nxt[i]); + } - } the while(!Q.empty ()) { * $ intU=q.front ();//coutPanax Notoginseng Q.pop (); - for(intI=0; i -; i++){ the if(Ac[u].nxt[i]) { +Ac[ac[u].nxt[i]].fail=Ac[ac[u].fail].nxt[i]; Aac[ac[u].nxt[i]].e|=ac[ac[ac[u].fail].nxt[i]].e; the Q.push

=0;i -; i++)if(PN) Pnf=p,q[++r]=pn,pn->th=R; + while(lR) { -p=q[++l]; the for(intI=0;i -; i++){ * if(PN) { $Q[++r]=pn,pn->th=R;Panax Notoginseng for(Pnf=p->fail; pnf!=root !pnf->next[i]; pnf=pnf->fail); - if(Pnf->next[i]) pnf=pnf->Next[i]; thePn->w|=pnf->W; + } A } the } + } - $ intQ_pow (intXinty) { $ intans=1; - for(; y; x=x*x%p, y=y>>1)if(y1) ans=ans*x%P; - returnans; the } - voiddp () {Wuyif[0][0

Always do not understand how to do ah, think for a long time $twt$Finally understand why find the first $fail$ to meet the criteria to calculate, because avoid repetition, this answer,,,Then $root$ below to connect to 26 nodes, here 26 letters are not in the dictionary is replaced with $f[i][0]$, this also thought for a long time, $==$ I still go Home farm.#include Finally made it out, because looking at someone else's template, which also,,,"Bzoj 1030" "Jsoi 2007"

Very old topic, a long time ago to learn AC automata when the A onceIt's a review today.We can turn the problem into a string that does not appear in a given string.We set up AC automata, set DP[I][J] to show that I have walked the current on the J-nodeThe word node is not transferred during DP#include 　　Bzoj 1030 Text Generator

"Topic link" http://www.lydsy.com/JudgeOnline/problem.php?id=1030"The main topic"Find the number of strings that contain any given stringExercisesWe find the quantity that does not contain any given string, and use the complete set to reduce it,For a given string set up AC automata, with 1 nodes as the root, 0 nodes to 1 even full character set transfer as a super-source,So 0->match can match all strings that don't contain a given string,DP[I][J] Indicates the number of strings matched to the I

];inlinevoidReadintk) { intf=1; k=0;CharC=GetChar (); while(c'0'|| C>'9') c=='-' (f=-1), c=GetChar (); while(c'9' c>='0') k=k*Ten+c-'0', c=GetChar (); K*=F; } InlinevoidInsert () {intLen=strlen (s+1), now=0; for(intI=1, ch;i) { if(!tree[now].nxt[ch=s[i]-'A']) Tree[now].nxt[ch]=++Tott; now=Tree[now].nxt[ch]; } Cnt[now]=1;} InlinevoidGetfail () {intFront=1, rear=0; tree[0].fail=-1; for(intI=0, too;i -; i++) if((too=tree[0].nxt[i]) h[++rear]=too; while(frontrear) { intn