BZOJ 1009 [HNOI2008]GT考试 (KMP + 矩阵快速幂)

时间：2024-08-19 12:36:26

1009: [HNOI2008]GT考试

Time Limit: 1 Sec  Memory Limit: 162 MB
Submit: 4266  Solved: 2616
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Description

　　阿申准备报名参加GT考试，准考证号为N位数X1X2....Xn(0<=Xi<=9),他不希望准考证号上出现不吉利的数字。
他的不吉利数学A1A2...Am(0<=Ai<=9)有M位，不出现是指X1X2...Xn中没有恰好一段等于A1A2...Am. A1和X1可以为
0

Input

　　第一行输入N,M,K.接下来一行输入M位的数。 N<=10^9,M<=20,K<=1000

Output

　　阿申想知道不出现不吉利数字的号码有多少种，输出模K取余的结果.

Sample Input

4 3 100
111

Sample Output

81

HINT

析：先用KMP把不能出现的串先匹配出来，然后再对每种状态进行计算，dp[i][j] 表示已经匹配到 i 并转移到 j 有多少种，但是时间复杂度太高，所以要用矩阵快速幂来进行优化。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define pu push_up

#define pd push_down

#define cl clear()

#define all 1,n,1

#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-3;

const int maxn = 20 + 10;

const int maxm = 3e5 + 10;

const int mod = 100003;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, -1, 0, 1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

char s[maxn];

int f[maxn], K;

struct Matrix{

  int a[maxn][maxn];

  int n;

  Matrix(){ ms(a, 0);  }

  void toOne(){ FOR(i, 0, n)  a[i][i] = 1; }

  Matrix operator * (const Matrix &rhs){

    Matrix res;

    res.n = n;

    FOR(i, 0, n)  FOR(j, 0, n)

      for(int k = 0; k < n; ++k)

        res.a[i][j] = (res.a[i][j] + a[i][k] * rhs.a[k][j]) % K;

    return res;

  }

};

Matrix fast_pow(Matrix a, int n){

  Matrix res; res.n = a.n; res.toOne();

  while(n){

    if(n & 1)  res = res * a;

    n >>= 1;

    a = a * a;

  }

  return res;

}

int main(){

  scanf("%d %d %d", &n, &m, &K);

  scanf("%s", s);

  f[0] = f[1] = 0;

  for(int i = 1; i < m; ++i){

    int j = f[i];

    while(j && s[j] != s[i])  j = f[j];

    f[i+1] = s[j] == s[i] ? j+1: 0;

  }

  Matrix x;  x.n = m;

  for(int i = 0; i < m; ++i){

    for(int t = 0; t < 10; ++t){

      int j = i;

      while(j && s[j] - '0' != t)  j = f[j];

      if(s[j] - '0' == t)  ++j;

      ++x.a[i][j];

    }

  }

  Matrix ans = fast_pow(x, n);

  int res = 0;

  for(int i = 0; i < m; ++i)  res = (res + ans.a[0][i]) % K;

  printf("%d\n", res);

  return 0;

}

　　

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