[SCU 4503] Tooeasy Or Toodifficult (Manacher+xor-trie)

SCU-4503

Board question, according to test instructions description is divided into three steps
1. For each location-centered palindrome string, and the maximum palindrome string
2. For each palindrome string the length of the XOR prefix and
3. Ask for two different or prefix and XOR maximum (Xor-trie)

Then use the quick power to calculate the JD, and then the size of the FJD

Take a look at the trie. Insert a 0 first

#pragma COMMENT (linker, "/stack:102400000,102400000")#include <cstdio>#include <iostream>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <map>#include <set>#include <queue>using namespace STD;typedefpair<int,int> Pii;typedef Long LongLL;typedef unsigned Long LongULL;typedef DoubleDBL;typedef Long DoubleLdbl;#define MST (A, B) memset (A,b,sizeof (a) )#define CLR (a) MST (a,0)#define SQR (a) (a*a)Const intMaxl=1e5+Ten;structmanacher{Char*STR;int*mana,len,res; Manacher (Char*,int); ~manacher () {Delete[]str;Delete[]mana;}intQueryint);};structtrie{structNode {CharChrintval,nxt[2]; } *trie;intSiz Trie (intSize) {trie=NewNode[size];memset(Trie,0,sizeof(node) *size); siz=0;} ~trie () {Delete[]trie;}intChnum (Charx) {returnX-' 0 ';}voidADDSTR (Char*);intQueryChar*STR);};CharINPT[MAXL];intMANA[MAXL];Charstr01[ -];intSpr_pow (int, LL,int);Char* NTS (int,Char*);intMain () {intTscanf("%d", &t); for(intck=1; ck<=t; ck++) {scanf('%s ', INPT);intlen=strlen(INPT); Manacher Mana (Inpt,len);intJd= (LL) Spr_pow (Mana.res, (LL) len*len*len,mana.res/3*5+1) +mana.res*4/5; Trie Trie (3e5);intPxor=0, Fjd=pxor; TRIE.ADDSTR (NTS (0, Str01));#ifdef LOCAL         for(intI=1; i<2*len+2; i++)printf("%c", Mana.str[i]);puts(""); for(intI=1; i<2*len+1; i++)printf("%d", Mana.mana[i]);puts("");#endif         for(intI=0; i<len; i++) {pxor^=mana.query (i);//printf ("I_len:%d\n", Mana.query (i));NTS (PXOR,STR01);            Fjd=max (FJD, Trie.query (Str01));        Trie.addstr (STR01); }if(JD&GT;FJD)puts("LIUJC");Else puts("Luoxinchen"); }return 0;}intSpr_pow (intNum,ll mi,intMOD) {intres=1; while(MI) {if(mi&1) res= (LL) res*num)%mod;        Num= (LL) num*num)%mod; mi>>=1; }returnRes;} Manacher::manacher (Char*TSTR,intTlen): Len (tlen) {str=New Char[2*len+Ten]; Mana=New int[2*len+Ten]; res=0; str[0]='! '; for(intI=0; i<len; i++) {str[2*i+1]=' # '; str[2*i+2]=tstr[i]; } str[2*len+1]=' # '; str[2*len+2]=0; mana[1]=1;intp=1, rm=1; for(intI=2; i<2*len+1; i++) {if(rm>i) Mana[i] = min (rm-i+1, mana[2*p-i]);Elsemana[i]=1; while(Str[i-mana[i]] = = Str[i+mana[i]]) mana[i]++;if(i+mana[i]-1&GT;RM) {rm=i+mana[i]-1;        P=i; }if(Res<mana[i]) res=mana[i]; } res--;}intManacher::query (intNP) {returnmana[2*np+2]-1;}voidTRIE::ADDSTR (Char*STR) {intlen=strlen(str), np=0; for(intI=0; i<len; i++) {intCh=chnum (Str[i]);if(Trie[np].nxt[ch]) np=trie[np].nxt[ch];Else{trie[np].nxt[ch]=++siz;            Np=siz;        Trie[np].chr=str[i]; }    }}intTrie::query (Char*STR) {intres=0, np=0; for(intI=0; i< to; i++) {intnow=str[i]-' 0 ';if(trie[np].nxt[now^1]) {np=trie[np].nxt[now^1]; res^=1<< ( --I.); }Else if(Trie[np].nxt[now])        {Np=trie[np].nxt[now]; }    }returnRes;}Char* NTS (intNumCharStr[]) { for(intI= -; i>=0; i--) {str[i]= (num&1)+' 0 '; num>>=1;} str[ to]=0;returnSTR;}

[SCU 4503] Tooeasy Or Toodifficult (Manacher+xor-trie)