POJ 2406 Power Strings【KMP判字符串循环节】

由KMP求出$fail[]$

设当前串长度为$len$

若$(len-fail[len])|len$

则循环节长度为$\frac{len}{len-fail[len]}$

否则无循环节。

#include <cmath> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define db double #define sg string #define ll long long #define rel(i,x,y) for(ll i=(x);i<(y);i++) #define rep(i,x,y) for(ll i=(x);i<=(y);i++) #define red(i,x,y) for(ll i=(x);i>=(y);i--) #define res(i,x) for(ll i=head[x];i;i=nxt[i]) using namespace std; const ll N=1e6+5; const ll Inf=1e18; const db Eps=1e-10; char ch[N]; ll len,fail[N]; void kmp() { len=strlen(ch+1); fail[1]=0;ll j=0; rep(i,2,len) { while(j>0&&ch[i]!=ch[j+1]) j=fail[j]; if(ch[i]==ch[j+1]) j++;fail[i]=j; } } int main() { while(~scanf("%s",ch+1)) { if(ch[1]=='.') return 0; kmp(); ll ret=1; if(len%(len-fail[len])==0) { printf("%lld\n",len/(len-fail[len])); } else printf("%lld\n",ret); } return 0; }