[NOI 2014]动物园

思路:

比较单纯的求一下next数据和失配的次数。

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

const int MaxN = 1000005;
const int M = 1000000007;

char s[MaxN];
int dep[MaxN], fail[MaxN];

int main()
{
	int nT;
	scanf("%d", &nT);
	while (nT--)
	{
		scanf("%s", s + 1);

		int u = 0, v = 0;
		int res = dep[1] = 1;
		for (int i = 2; s[i]; ++i)
		{
			while (u && s[i] != s[u + 1])
				u = fail[u];
			fail[i] = (u += s[i] == s[u + 1]);
			dep[i] = dep[u] + 1;

			while (v && s[i] != s[v + 1])
				v = fail[v];
			v += s[i] == s[v + 1];
			while (v > i >> 1)
				v = fail[v];
			res = (ll)res * (dep[v] + 1) % M;
		}

		printf("%d\n", res);
	}
	return 0;
}