洛谷P3193 [HNOI2008]GT考试(dp 矩阵乘法)

题意

题目链接

Sol

设(f[i][j])表示枚举到位置串的第i位，当前与未知串的第j位匹配，那么我们只要保证在转移的时候永远不会匹配即可

预处理出已知串的每个位置加上某个字符后能转移到的位置，矩阵快速幂优化一下

复杂度(O(M^3 log n))

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 22;
int N, M, mod,  s[MAXN], trans[MAXN][10], p[MAXN], g[MAXN], base[MAXN];
char ss[MAXN];
template<typename A, typename B> inline void add2(A &x, B y) {
	if(x + y < 0) x = x + y + mod;
	else x = x + y >= mod ? x + y - mod : x + y;
}
int Lim;
struct Ma {
	int m[MAXN][MAXN];
	Ma() {
		memset(m, 0, sizeof(m));
	}
	void init() {
		for(int i = 0; i <= Lim; i++) m[i][i] = 1;
	}
	Ma operator * (const Ma &rhs) const {
		Ma ans;
		for(int i = 0; i <= Lim; i++)
			for(int j = 0; j <= Lim; j++) {
				__int128 tmp = 0;
				for(int k = 0; k <= Lim; k++) tmp += 1ll * m[i][k] * rhs.m[k][j] % mod;
				ans.m[i][j] = tmp % mod;
			}
		return ans;
	}
}f;
void GetNxt() {
	int j = 0;
	for(int i = 0; i <= M; i++) {
		if(i > 1) {
			while(j && s[i] != s[j + 1]) j = p[j];
			if(s[i] == s[j + 1]) j++;
			p[i] = j;
		}
		for(int t = 0; t <= 9; t++) {
			int k = i;
			while(k && t != s[k + 1]) k = p[k];
			if(t == s[k + 1]) k++;
			trans[i][t] = k;
		}
	}
}
Ma MPow(Ma a, int p) {
 	Ma base; base.init();
 	while(p) {
 		if(p & 1) base = base * a;
		 a = a * a; p >>= 1;	
	}
	return base;
}
int main() {
	cin >> N >> M >> mod; Lim = M + 1;
	scanf("%s", ss + 1);
	for(int i = 1; i <= M; i++) s[i] = ss[i] - '0';
	for(int i = 0; i <= 9; i++) g[i == s[1]]++;
	GetNxt();
	for(int j = 0; j <= M; j++) 
		for(int k = 0; k <= 9; k++) 
			if(trans[j][k] != M)
				f.m[trans[j][k]][j]++;
	Ma tmp = MPow(f, N - 1);
	for(int i = 0; i <= Lim; i++) 
		for(int j = 0; j <= Lim; j++)
			add2(base[i], 1ll * tmp.m[i][j] * g[j] % mod);
	int ans = 0;
	for(int i = 0; i <= M - 1; i++) add2(ans, base[i]);
	cout << ans;
	return 0;
}
/*
4 3 100
121
*/