[HNOI 2008]GT考试

Description

题库链接

问你长度为 (n) 的可含前导零的数字串中，不含长度为 (m) 的子串 (X) 有多少个，取模。

(1leq nleq 10^9,1leq mleq 20)

Solution

一个显然的 (DP) ，就是令 (f_{i,j}) 表示已经生成出的串长为 (i) 位，后 (j) 位与 (X) 串的前 (j) 位匹配的方案数。

那么 (f_{i,j}=sum_{x=1}^m f_{i-1,x}cdot[x 可以转移到 j]) 。

由于 (n) 过大，我们可以用矩阵加速。在构造矩阵时可以用 (KMP) 中的 (next) 数组的思想来简化计算过程，省去了暴力的计算。

Code

//It is made by Awson on 2018.3.14
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('
'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(LL x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(LL x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, m, k, nxt[25];
char ch[25];
struct mat {
    int a[25][25];
    mat() {memset(a, 0, sizeof(a)); }
    mat operator * (const mat &b) const {
    mat ans;
    for (int i = 0; i <= m; i++)
        for (int j = 0; j <= m; j++)
        for (int p = 0; p <= m; p++)
            ans.a[i][j] = (ans.a[i][j]+a[i][p]*b.a[p][j]%k)%k;
    return ans;
    }
}S, T;

void build() {
    for (int i = 2; i <= m; i++) {
    int j = nxt[i-1];
    while (j && ch[j+1] != ch[i]) j = nxt[j];
    if (ch[j+1] == ch[i]) nxt[i] = j+1;
    }
    for (int i = 1; i <= m; i++) {
    for (char j = '0'; j <= '9'; j++) {
        if (j == ch[i]) continue;
        int p = nxt[i-1];
        while (p && j != ch[p+1]) p = nxt[p];
        if (j == ch[p+1]) ++p; ++T.a[i-1][p];
    }
    ++T.a[i-1][i];
    }
}
mat quick_pow(mat a, int b) {
    mat ans = a; --b;
    while (b) {
    if (b&1) ans = ans*a;
    b >>= 1, a = a*a;
    }
    return ans;
}
void work() {
    read(n), read(m), read(k); scanf("%s", ch+1); build();
    S.a[0][0] = 1; S = S*quick_pow(T, n); int ans = 0;
    for (int i = 0; i < m; i++) ans += S.a[0][i]; writeln(ans%k);
}
int main() {
    work(); return 0;
}