题目链接
题解
三倍经验！
我们要快速求区间最小循环节
我们知道循环节有如下性质：
①当(L)为循环节长度，那么(s[l...r - L] = s[l + L...r])且(L | (r - l + 1))
②如果(L)为循环节，那么(L x)也为循环节
一个比较暴力的思想是枚举(len = r - l + 1)的因子，用(hash)去判是否相同，这样做是(O(qsqrt{n}))的，过于暴力
由性质②我们知道，最后的答案(L)一定是(len)删除若干个质因子的结果，所以我们只需枚举质因子即可
由于质因子个数是(O(logn))的，所以预处理一下即可(O(logn))，枚举质因子
复杂度(O(qlogn))
要注意，(P3647)卡自然溢出
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define ULL unsigned long long int
using namespace std;
const int maxn = 500005,maxm = 100005,INF = 1000000000,P = 2001611;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
ULL pw[maxn],h[maxn];
LL pw2[maxn],h2[maxn];
char s[maxn];
int n,q,fac[maxn],p[maxn],pi,isn[maxn];
void init(){
for (int i = 2; i <= n; i++){
if (!isn[i]) p[++pi] = i,fac[i] = i;
for (int j = 1; j <= pi && i * p[j] <= n; j++){
isn[i * p[j]] = true; fac[i * p[j]] = p[j];
if (i % p[j] == 0) break;
}
}
}
inline bool check(int l,int r,int ll,int rr){
return h[r] - h[l - 1] * pw[r - l + 1] == h[rr] - h[ll - 1] * pw[rr - ll + 1]
&& ((h2[r] - h2[l - 1] * pw2[r - l + 1] % P) % P + P) % P == ((h2[rr] - h2[ll - 1] * pw2[rr - ll + 1] % P) % P + P) % P;
}
int main(){
n = read();
pw[0] = 1; for (int i = 1; i <= n; i++) pw[i] = pw[i - 1] * 107;
pw2[0] = 1; for (int i = 1; i <= n; i++) pw2[i] = pw2[i - 1] * 107 % P;
scanf("%s",s + 1);
init();
for (int i = 1; i <= n; i++){
h[i] = h[i - 1] * 107 + s[i];
h2[i] = (h2[i - 1] * 107 % P + s[i]) % P;
}
q = read();
int l,r,len,ans;
while (q--){
l = read(); r = read(); len = r - l + 1;
ans = len;
for (int x = len; x > 1; x /= fac[x]){
int L = ans / fac[x];
if (check(l,r - L,l + L,r))
ans = L;
}
printf("%d
",ans);
}
return 0;
}