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

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;
}