poj2406 (KMP next array)

The main idea: give a string to ask it up by how many of the same string composition

If the abababab is made up of 4 AB

Analysis:

The application of the next array in KMP to the minimum cyclic section

For example

Ababab next[6] = 4; That

Ababab

Ababab

The 1~4 bit is the same as the 2~6 bit

So the first two

is equal to 3, 4 bits

3, 4 bits equals 5, 6 bits

......

So if we can divide it, we'll loop to the end.

If not divisible

The last remaining few are not in the loop.

For example

1212121

1212121

The last remaining 1 cannot be equal to the cyclic section

Code:

1#include <iostream>2#include <cstdio>3#include <cstring>4 using namespacestd;5 6 Const intMAXN =1000005;7 8 intNEXT[MAXN];9 Ten void Get(Char*s) { One     intL =strlen (s); A     intj =0, k =-1; -next[0] = -1; -      while(J <l) { the         if(k = =-1|| S[J] = =S[k]) { -NEXT[++J] = + +K; -}Else { -K =Next[k]; +         } -     } + } A CharS[MAXN]; at  - intMain () { -      while(gets (s)) { -         if(strcmp (s),".") ==0) { -              Break; -         } in         Get(s); -         intAns =1; to         intL =strlen (s); +         if(l% (l-next[l]) = =0) { -ans = l/(L-next[l]); the         } *printf"%d\n", ans); $     }Panax Notoginseng}

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poj2406 (KMP next array)