POJ - 1961 最小循环节

如果循环节存在那在前缀部分也肯定存在

如果循环节存在那至少是可以匹配的

而next是维护最大前缀的,意会意会

注意一定要先判整除,即使别的题目保证是存在循环的

特意画了一张灵魂草图帮助理解

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar('\n')
#define blank putchar(' ')
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int maxn = 1e6+11;
const int oo = 0x3f3f3f3f;
const double eps = 1e-7;
typedef long long ll;
ll read(){
    ll x=0,f=1;register char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int nxt[maxn];
char P[maxn];
void buildNext(){
	nxt[1]=0;
	int j=0,m=strlen(P+1);
	rep(i,2,m){
		while(j&&P[i]!=P[j+1])j=nxt[j];
		if(P[i]==P[j+1])j++;
		nxt[i]=j;
	}
}
int main(){
	int m,kase=0;
	while(~iin(m)){
		if(m==0)break;
		s1(P);
		buildNext();
		printf("Test case #%d\n",++kase);
		rep(i,2,m){
			if(i%(i-nxt[i])==0&&i/(i-nxt[i])>1){
				print(i); blank;
				println((i/(i-nxt[i])));
			}
		}
		putchar('\n');
	}
	return 0;
}