[Bzoj1009][HNOI2008]GT考试(动态规划)

题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=1009

显而易见的动态规划加矩阵快速幂，不过转移方程不怎么好想，dp[i][j]表示长度为i的准考证号后j位与不吉利数字的前j位相同的方案数。则：

转移方程为$dp[i][j]=\sum_{k=0}^{m-1}dp[i-1][k]*g[k][j]$

答案为：$ans=\sum_{i=0}^{m}dp[n][i]$

g[i][j]表示长度为i的后缀变成长度为j的后缀的方案数。

而g数组可以用kmp预处理出来

附上洛谷40分不用矩阵优化的代码

   #include<bits/stdc++.h>
   using namespace std;
   typedef long long ll;
   typedef unsigned long long ull;
   const int maxn = 6e6 + ;
   ll Next[];
   ll g[][];
   ll dp[maxn][];
   char s[];
  void getN(int n) {
      Next[] = -;
      int i = , j = -;
      while (i < n) {
          if (j == - || s[i] == s[j])
              Next[++i] = ++j;
          else
              j = Next[j];
      }
  }
  int main() {
      ll n, m, mod;
      scanf("%lld%lld%lld", &n, &m, &mod);
      scanf("%s", s);
      getN(m);
      Next[] = ;
      for (int i = ; i < m; i++) {
          for (int j = ''; j <= ''; j++) {
              int t = i;
              while (t&& s[t] != j)
                  t = Next[t];
              if (s[t] == j)
                  t++;
              g[i][t]++;
          }
      }
      dp[][] = ;
      for (int i = ; i <= n; i++) {
          for (int j = ; j < m; j++) {
              for (int k = ; k < m; k++) {
                  dp[i][j] = (dp[i][j] + dp[i - ][k] * g[k][j]) % mod;
              }
          }
      }
      ll ans = ;
      for (int i = ; i < m; i++)
          ans = (ans + dp[n][i]) % mod;
      printf("%lld\n", ans);
  }

以及正解

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned long long ull;
 const int maxn = 6e6 + ;
 ll Next[];
 ll n, m, mod;
 ll dp[][];
 char s[];
 void getN(int n) {
     Next[] = -;
     int i = , j = -;
     while (i < n) {
         if (j == - || s[i] == s[j])
             Next[++i] = ++j;
         else
             j = Next[j];
     }
 }
 struct matrix {
     ll cnt[][];
     matrix() { memset(cnt, , sizeof(cnt)); }
     matrix operator *(const matrix a)const {
         matrix ans;
         for (int i = ; i <= m; i++) {
             for (int j = ; j <= m; j++) {
                 ans.cnt[i][j] = ;
                 for (int k = ; k <= m; k++)
                     ans.cnt[i][j] = (ans.cnt[i][j] + cnt[i][k] * a.cnt[k][j]) % mod;
             }
         }
         return ans;
     }
 };
 matrix powM(matrix a, int b) {
     matrix ans = matrix();
     for (int i = ; i <= m; i++)
         ans.cnt[i][i] = ;
     while (b) {
         if (b & )
             ans = ans * a;
         a = a * a;
         b /= ;
     }
     return ans;
 }
 int main() {
     matrix g, ans, dp = matrix();
     scanf("%lld%lld%lld", &n, &m, &mod);
     scanf("%s", s);
     getN(m);
     Next[] = ;
     memset(g.cnt, , sizeof(g.cnt));
     for (int i = ; i < m; i++) {
         for (int j = ''; j <= ''; j++) {
             int t = i;
             while (t&& s[t] != j)
                 t = Next[t];
             if (s[t] == j)
                 t++;
             g.cnt[i][t]++;
         }
     }
     dp.cnt[][] = ;
     ans = powM(g, n);
     ans = dp * ans;
     ll sum = ;
     for (int i = ; i < m; i++)
         sum = (sum + ans.cnt[][i]) % mod;
     printf("%lld\n", sum);
 }