P3193 [HNOI2008] KMP + 矩阵快速幂

题解

传送门 P3193 [HNOI2008]GT考试

题解

以准考证号后缀与不吉利数字的最长匹配长度为状态，$KMP$ 预处理处状态转移矩阵，进行矩阵快速幂即可。总时间复杂度为 $O(M^3\log N)$。

#include <bits/stdc++.h>
using namespace std;
#define rep(i, l, r) for (int i = l, _ = r; i < _; ++i)
const int MAXN = 22;
int N, M, K;
string ban;
struct matrix
{
    
    
    int n, a[MAXN][MAXN];
    matrix() {
    
     n = 0, memset(a, 0, sizeof(a)); }
    void operator*=(const matrix &o)
    {
    
    
        int b[MAXN][MAXN];
        memset(b, 0, sizeof(b));
        rep(i, 0, n) rep(k, 0, n) rep(j, 0, n) b[i][j] = (b[i][j] + a[i][k] * o.a[k][j]) % K;
        memcpy(a, b, sizeof(a));
    }
} A, res;

int nxt[MAXN];

void kmp(string a, int n, int tran[][MAXN])
{
    
    
    int i = 0, j = -1;
    nxt[i] = j;
    while (i < n)
    {
    
    
        while (j >= 0 && a[i] != a[j])
            j = nxt[j];
        ++i, ++j;
        nxt[i] = j;
    }
    rep(k, 0, n) rep(num, 0, 10)
    {
    
    
        i = j = k;
        char c = num + '0';
        while (j >= 0 && a[j] != c)
            j = nxt[j];
        ++j;
        if (j == n)
            continue;
        ++tran[i][j];
    }
}

int main()
{
    
    
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> N >> M >> K;
    cin >> ban;
    A.n = res.n = M;
    kmp(ban, M, A.a);
    rep(i, 0, res.n) res.a[i][i] = 1;
    while (N)
    {
    
    
        if (N & 1)
            res *= A;
        A *= A, N >>= 1;
    }
    int sum = 0;
    rep(i, 0, res.n) sum = (sum + res.a[0][i]) % K;
    cout << sum << '\n';
    return 0;
}