HDU 2243 Entrance Examination Road boundless--word complex AC automata + matrix fast Power

Links: http://acm.hdu.edu.cn/showproblem.php?pid=2243

Test instructions: given n (1<= N < 6) a root with a length not exceeding 5, the number of words with a length not exceeding L (l <231) containing at least one root word; MoD 264.

Base: The POJ 2778DNA sequence is the number of legitimate strings with a given length without a pattern string, and the string length is equal to the upper bound of int;

1.mod 264 Direct use of unsigned long long natural overflow can be, said some implicit. It is also easy to think of the use of built-in type directly, otherwise high-precision MoD 2^64 to kneel;

2. The subject is an upgraded version of the POJ 2778, inside the range, not a specific length, and still contain at least one, then directly from the negative solution, that is, the total number-does not contain the number of roots;

3. The operation of the Matrix, the realization of continuous discrete exponential summation; Add a column to the matrix, all 1, the added row is 1 except the lower right corner, and the others are 0;

$\begin{pmatrix}
A\rightarrow \begin{pmatrix}
A & 1\\
0 & 1
\end{pmatrix}
\end{pmatrix}$

$ans = \sum \limits_{i=0}^{\rm size}mat\left [0 \right]\left [i \right]$

The same solution: $\sum \limits_{i=0}^{\rm size}26^i$ only need to establish two-dimensional matrix

\begin{vmatrix}
& 1\\
0 & 1
\end{vmatrix} immediately after the direct matrix to the power;

15MS 1632k
#include <iostream>#include<cstdio>#include<cstring>#include<string.h>#include<algorithm>#include<vector>#include<cmath>#include<stdlib.h>#include<time.h>#include<stack>#include<Set>#include<map>#include<queue>using namespacestd;#defineRep0 (I,L,R) for (int i = (l); i < (R); i++)#defineREP1 (I,L,R) for (int i = (l); I <= (r); i++)#defineRep_0 (i,r,l) 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 << 1#defineRson m+1, R, RT << 1|1#defineSqr (a) (a) * (a)typedef pair<int,int>PII;#defineA First#defineB Second#defineMK Make_pairtypedef __int64 LL;TYPEDEF unsignedLong Longull;template<typename t>voidRead1 (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;} Template<typename t>voidRead2 (T &a,t &b) {Read1 (a); Read1 (b);} Template<typename t>voidRead3 (T &a,t &b,t &c) {Read1 (a); Read1 (b); Read1 (c);} Template<typename t>void  out(T a) {if(a>9) out(ATen); Putchar (A%Ten+'0');}intT,kase =0, I,j,k,n,m,top;#defineMoD 100000structmatrix{ull d[Panax Notoginseng][Panax Notoginseng]; intSz//can't also make ull, otherwise int = Ull will go wrongMatrix (intR) {MS0 (d); sz =R;} voidinit () { for(inti =0; I < sz;i++) D[i][i]=1; } Matrixoperator*(Constmatrix& a)Const{Matrix ans (A.SZ);  for(inti =0; I < sz;i++)         for(intj =0; J < sz;j++){             for(intK =0; k < sz;k++) Ans.d[i][j]+ = d[i][k]*A.d[k][j]; }        returnans; }    voidDebug () {Rep0 (I,0, SZ) {Rep0 (J,0, SZ) cout<<d[i][j]<<" "; cout<<Endl; }    }};    Matrix Pow (Matrix a,ll m) {matrix ans (A.SZ);    Ans.init ();  while(m) {if(m&1) ans = ans*A; A= A *A; M>>=1; }    returnans;}Const intSigma_size = -;Const intMAXN =Panax Notoginseng;structaho_corasick{intCh[maxn][sigma_size]; intVAL[MAXN],F[MAXN],LAST[MAXN],CNT[MAXN]; intsz; Map<string,int>MS; Aho_corasick () {}voidInit () {sz =1; val[0] =0; MS0 (ch[0]);    MS0 (CNT); Ms.clear ();} intIdxCharc) {returnC-'a';} voidInsert (Char*s,intv) {        intU =0, n =strlen (s);  for(inti =0; I < n;i++){            intc =idx (s[i]); if(!Ch[u][c])                {MS0 (Ch[sz]); VAL[SZ]=0; CH[U][C]= sz++; } u=Ch[u][c]; } Val[u]=v; }    voidGetfail () {Queue<int>Q; f[0] =0;  for(intc =0; c < sigma_size;c++) {//Initialize Queue            intU = ch[0][c]; if(u) {f[u] =0; Q.push (U); Last[u] =0;} }         while(!Q.empty ()) {            intR =Q.front (); Q.pop ();  for(intc =0; c < sigma_size;c++){                intU =Ch[r][c]; if(!u) {Ch[r][c] = Ch[f[r]][c];Continue;}//Implementing Compressionq.push (U); intv =F[r];  while(v &&!ch[v][c]) v =F[v]; F[u]=Ch[v][c]; Last[u]= Val[f[u]]?F[u]:last[f[u]]; }}} Matrix Find () {Matrix ans (sz+1);  for(inti =0; I < sz;i++){            if(Val[i] | | last[i])Continue;  for(intj =0; J < sigma_size;j++){                intv =Ch[i][j]; if(Val[v] | | last[v])Continue; ANS.D[I][V]++; } Ans.d[i][sz]=1;//add a column to save the previous Sigma and;} Ans.d[sz][sz]=1;//Lower right corner        returnans; }}ac;Charp[ the];intMain () {ull n,l; //freopen ("Data.txt", "R", stdin);     while(SCANF ("%i64d%i64d", &n,&l) = =2) {ac.init (); REP1 (i,1, N) {scanf ("%s", p); Ac.        Insert (P,i);        } ac.getfail (); Matrix ans=AC.        Find (); Ans=Pow (ans,l); Ull CNT=0; Rep0 (i,0, ANS.SZ) cnt + = ans.d[0][i]; Matrix A (2);//Seek the sum a.d[0][0] = -, a.d[0][1] = a.d[1][1] =1; A=Pow (a,l); printf ("%i64u\n", a.d[0][0]+a.d[0][1]-CNT); }    return 0;}

HDU 2243 Entrance Examination Road boundless--word complex AC automata + matrix fast Power