Practice II String

I wanted to do the number theory ... But all the other dalao are making the leap.

So......

Chapter I KMP

KMP, the most critical one is not this semi-violent single-mode match.

But this NXT array often has some strange problems, especially the Loop section can be obtained directly from t-nxt[t] ... Oh, God.

In short, remember that the NXT is the longest public prefix in the suffix of the tail pointer is OK

T1 poj3461 Oulipo

Time Cost:10min

The pure board really doesn't have anything to say.

Is that every time you empty the NXT, OK

Code:

1#include <cstdio>2#include <cstring>3 using namespacestd;4 Const intN =1000005;5 #defineRep (i,a,n) for (int i = a;i <= n;i++)6 #defineMS (A, b) memset (a,b,sizeof a)7 8 CharS[n],t[n];9 ints,t;Ten intNxt[n]; One intans; A voidCLR () {MS (NXT,0), ans =0;} - voidgnxt () { -     intTMP =0; theRep (I,2, T) { -      while(TMP && t[tmp+1] = T[i]) tmp =nxt[tmp]; -     if(t[tmp+1] = = T[i]) Nxt[i] = + +tmp; -     } + } - voidKMP () { +     intTMP =0; ARep (I,1, S) { at      while(TMP && t[tmp+1] = S[i]) tmp =nxt[tmp]; -     if(t[tmp+1] ==S[i]) { -tmp++; -         if(tmp = = T) ans++; -     } -     } in } -  to  + intMain () { -     intQ; thescanf"%d",&q); *      while(q--) { $ CLR ();Panax Notoginsengscanf"%s", t+1), scanf ("%s", s+1); -S = strlen (s+1), T = strlen (t+1); the gnxt (); + KMP (); Aprintf"%d\n", ans); the     } +}

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T2 POJ 2406 Power strings

Time Cost:55min

Have you ever met before? The last violent roll of rough

Today, I understand it by playing the table .

As the NXT is defined, the string itself is the longest common prefix of its own

In the last time we asked for NXT, we considered the part that was nxt[t] discarded (recorded as Len)

If this segment is divisible by T, consider the second Len long string

It is equal to the first Len long string.

The same can be pushed to T/len times

At the same time NXT take maximum goo t-nxt[t] take the smallest

So if T is an integral multiple of t-nxt[t] then t-nxt[t] is the smallest cyclic section

Code:

1#include <cstdio>2#include <cstring>3 using namespacestd;4 Const intN =1000005;5 #defineRep (i,a,n) for (int i = a;i <= n;i++)6 #defineMS (A, b) memset (a,b,sizeof a)7 8 CharS[n],t[n];9 ints,t;Ten intNxt[n]; One intans; A voidCLR () {MS (NXT,0), ans =0;} - voidgnxt () { -     intTMP =0; theRep (I,2, T) { -      while(TMP && t[tmp+1] = T[i]) tmp =nxt[tmp]; -     if(t[tmp+1] = = T[i]) Nxt[i] = + +tmp; -     } + } - voidKMP () { +     intTMP =0; ARep (I,1, S) { at      while(TMP && t[tmp+1] = S[i]) tmp =nxt[tmp]; -     if(t[tmp+1] ==S[i]) { -tmp++; -         if(tmp = = T) ans++; -     } -     } in } -  to  + intMain () { -      while(1) { the CLR (); *scanf"%s", t+1); $T = strlen (t+1);Panax Notoginseng     if(T = =1&& t[1] =='.')return 0; - gnxt (); the     if(T% (t-nxt[t]) = =0) printf ("%d\n", t/(tnxt[t])); +     ElsePuts"1"); A     } the}

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T3 cf562d Om Nom and necklace

Time Cost:45min

It's easier to think of a previous problem.

Cross can't do we see AB as the cycle section

Because K is given, it is possible to find a cyclic section length range by I and K

Ok if this length is an integer multiple of the shortest loop section obtained by NXT

The time to achieve this is to save Len and then determine if the legal interval exists so that you can automatically subtract the remainder of a

However, it is important to note that a or B is empty and is divided into K or k+1 segments (without suffixes) to be

(The actual A is empty already processed)

Code:

#include <cstdio>#include<cstring>using namespacestd;Const intN =1000005;#defineRep (i,a,n) for (int i = a;i <= n;i++)#defineMS (A, b) memset (a,b,sizeof a)CharS[n],t[n];ints,t;intNxt[n];intans;voidCLR () {MS (NXT,0), ans =0;}voidgnxt () {intTMP =0; Rep (I,2, T) {     while(TMP && t[tmp+1] = T[i]) tmp =Nxt[tmp]; if(t[tmp+1] = = T[i]) Nxt[i] = + +tmp; }}voidKMP () {intTMP =0; Rep (I,1, S) {     while(TMP && t[tmp+1] = S[i]) tmp =Nxt[tmp]; if(t[tmp+1] ==S[i]) {tmp++; if(tmp = = T) ans++; }    }}intn,k;intMain () {scanf ("%d%d",&n,&k); scanf ("%s", t+1); T=N;    GNXT (); Rep (I,1, T) {    intLen = i-Nxt[i]; if(i% (k +1) ==0&& (I/(k +1))% Len = =0) Putchar ('1');//lenb==0    Else {        //Circular subsequence Range        intUPP = I/k,down = I/(k +1) +1; UPP= UPP/Len; down= (down + len-1) /Len; if(Upp >= down) putchar ('1'); ElsePutchar ('0'); }} puts (""); return 0;}

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To be Continued ...

Practice II String