给出一个字符串,求每个前缀的最小循环节长度,并输出整个字符串的最小循环节。字符串长度为3*10^6

找循环节这种问题还是要用KMP
对于长度为i的字符串 i%(i-f[i])==0 此时,它的最小循环节长度就是i-f[i],f[i]是字符串的第i位的失配函数(注意是从1开始算)

#include<bits/stdc++.h>
#define eps 1e-9
#define FOR(i,j,k) for(int i=j;i<=k;i++)
#define MAXN 1005
#define MAXM 40005
#define INF 0x3fffffff
#define PB push_back
#define MP make_pair
#define X first
#define Y second
#define lc (k<<1)
#define rc ((k<<1)1)
using namespace std;
typedef long long LL;
int i,j,k,n,m,x,y,T,ans,big,cas,num,len;
bool flag;
char s[];
int f[]; void getFail(char *s,int *f)
{
int m=strlen(s),j;
f[]=;f[]=;
for (int i=;i<m;i++)
{
j=f[i];
while (j&&s[i]!=s[j]) j=f[j];
f[i+]=(s[i]==s[j]?j+:);
}
} int main()
{
scanf("%s",s);
getFail(s,f);
len=strlen(s);
for (i=;i<=len;i++)
{
if (i%(i-f[i])==)
{
printf("%d ",i-f[i]);
}else
printf("%d ",i);
}
printf("\t\n");
printf("%s\n",s+f[len]);
return ;
}

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