BZOJ2795/2890/3647 [Poi2012]A Horrible Poem 【字符串hash】

题目链接

BZOJ2795
BZOJ2890
BZOJ3647

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\(O(q\sqrt{n})\)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\n",ans);
	}
	return 0;
}