阿申准备报名参加 GT 考试，准考证号为 nn位数 X1X2⋯Xn(0≤Xi≤9)X_1X_2\cdots X_n(0\le X_i\le 9)，他不希望准考证号上出现不吉利的数字。

他的不吉利数字 A1A2⋯Am(0≤Ai≤9)A_1A_2\cdots A_m(0\le A_i\le 9)有 mm位，不出现是指 X1X2⋯XnX_1X_2\cdots X_n中没有恰好一段等于A1A2⋯AmA_1A_2\cdots A_m

给出AA，求XX可能的方案数

n≤109,m≤20,K≤1000n\le10^9, m\le20, K\le 1000

LOJ 10224

Solution

设f[i][j]f[i][j]表示长度为ii的串，最后jj个可以匹配模板串前jj位的方案数，答案为∑i=0m−1f[n][i]\sum_{i=0}^{m-1}f[n][i]

转移ff的话可以考虑设g[i][j]g[i][j]表示模板串从前缀ii失配后转移到前缀jj的方案数，这个可以通过枚举下一个字符，然后用kmp的nextnext直接暴力算

那么f[i+1][k]=∑j=1mf[i][j]∗g[j][k]f[i + 1][k] = \sum_{j=1}^{m}f[i][j]* g[j][k]，矩阵快速幂优化即可

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 20 + 5;

int N, M, Mod;
int Next[Maxn];
int A[Maxn];

inline void Add (int &a, int b) { if ((a += b) >= Mod) a -= Mod; }

struct Matrix
{
	int A[Maxn][Maxn];
	inline int* operator [] (const int &x) { return A[x]; }

	inline void init ()
	{
		for (int i = 0; i < M; ++i)
			for (int j = 0; j < M; ++j) 
			{
				A[i][j] = 0;
				if (i == j) A[i][j] = 1;
			}
	}

	inline Matrix operator * (const Matrix &rhs) const
	{
		Matrix ans;
		for (int i = 0; i < M; ++i)
			for (int j = 0; j < M; ++j)
			{
				ans[i][j] = 0;
				for (int k = 0;  k < M; ++k)
					Add (ans[i][j], (LL)A[i][k] * rhs.A[k][j] % Mod);
			}
		return ans;
	}

} trans;

inline Matrix Pow (Matrix a, int b)
{
	Matrix ans; ans.init();
	for (int i = b; i; i >>= 1, a = a * a) if (i & 1) ans = ans * a;
	return ans;
}

inline void Init ()
{
	int j = 0;
	for (int i = 2; i <= M; ++i)
	{
		while (j && A[j + 1] != A[i]) j = Next[j];
		if (A[j + 1] == A[i]) ++j;
		Next[i] = j;
	}

	for (int i = 0; i < M; ++i)
	{
		for (int j = 0; j <= 9; ++j)
		{
			int k = i;
			while (k && A[k + 1] != j) k = Next[k];
			if (A[k + 1] == j) ++k;
			++trans[i][k];
		}
	}
}

inline void Solve ()
{
	Init ();
	Matrix res = Pow (trans, N);

	int ans = 0;
	for (int i = 0; i < M; ++i) Add (ans, res[0][i]);
	cout << ans << endl;
}

inline void Input ()
{
	N = read<int>(), M = read<int>(), Mod = read<int>();
	char S[Maxn];
	scanf("%s", S + 1);
	for (int i = 1; i <= M; ++i) A[i] = S[i] - '0';
	A[0] = 2003;
	A[M + 1] = 216;
}

int main()
{

#ifndef ONLINE_JUDGE
	freopen("A.in", "r", stdin);
	freopen("A.out", "w", stdout);
#endif

	Input ();
	Solve ();

	return 0;
}