LightOJ 1268 Unlucky Strings (KMP+矩阵快速幂)

题意：给出一个字符集和一个字符串和正整数n，问由给定字符集组成的所有长度为n的串中不以给定字符串为连续子串的有多少个？

析：n 实在是太大了，如果小的话，就可以用动态规划做了，所以只能用矩阵快速幂来做了，dp[i][j] 表示匹配完 i 到匹配 j 个有多少种方案，利用矩阵的性质，就可以快速求出长度为 n 的个数，对于匹配的转移，正好可以用KMP的失配函数来转移。

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>
#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 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 = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int maxn = 50 + 10;
const int mod = 1e9 + 7;
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], t[maxn];
int f[maxn];
 
struct Matrix{
  unsigned int a[maxn][maxn];
  int n;
  Matrix(int nn) : n(nn){
    memset(a, 0, sizeof a);
  }
 
  void toOne(){
    for(int i = 0; i < n; ++i)
      a[i][i] = 1;
  }
 
  friend Matrix operator * (const Matrix &lhs, const Matrix &rhs){
    Matrix res(lhs.n);
    for(int i = 0; i < lhs.n; ++i)
      for(int j = 0; j < rhs.n; ++j)
        for(int k = 0; k < lhs.n; ++k)
          res.a[i][j] += lhs.a[i][k] * rhs.a[k][j];
    return res;
  }
};
 
Matrix fast_pow(Matrix a, int n){
  Matrix res(a.n);
  res.toOne();
  while(n){
    if(n&1) res = res * a;
    a = a * a;
    n >>= 1;
  }
  return res;
}
 
void getFail(char *s, int n){
  f[0] = f[1] = 0;
  for(int i = 1; i < n; ++i){
    int j = f[i];
    while(j && s[j] != s[i])  j = f[j];
    f[i+1] = s[i] == s[j] ? j+1 : 0;
  }
}
 
int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d", &n);
    scanf("%s %s", t, s);
    int lens = strlen(s);
    getFail(s, lens);
    Matrix ans(lens);
    for(int i = 0; i < lens; ++i)
      for(int k = 0; t[k]; ++k){
        int j = i;
        while(j && s[j] != t[k])  j = f[j];
        if(s[j] == t[k])  ++j;
        ++ans.a[i][j];
      }
    ans = fast_pow(ans, n);
    unsigned int res = 0;
    for(int i = 0; i < lens; ++i)  res += ans.a[0][i];
    printf("Case %d: %u\n", kase, res);
  }
  return 0;
}