bzoj 1009 DP 矩阵优化

原来的DP：

dp[i][j]表示长度为i的合法串，并且它的长度为j的后缀是给定串的长度为j的前缀。

转移：

i==0

dp[0][0] = 1 dp[0][1~m-1] = 0

i>=1

dp[i][0] = dp[i-1][0]*10-dp[i-1][m-1]

dp[i][1] = dp[i-1][0]-(a[m]==a[1])*dp[i-1][m-1]

dp[i][2] = dp[i-1][1]-(a[m-1~m]==a[1~2])*dp[i-1][m-1]

dp[i][3] = dp[i-1][2]-(a[m-2~m]==a[1~3])*dp[i-1][m-1]

...

dp[i][m-1] = dp[i-1][m-2]-(a[2~m]==a[1~m-1])*dp[i-1][m-1]

（a[i~j]==a[p~q]表示原串的[i,j]区间与[p,q]区间是否相等）

然后就是一般的矩阵求线性递推的东西了。

 /**************************************************************
     Problem: 1009
     User: idy002
     Language: C++
     Result: Accepted
     Time:48 ms
     Memory:804 kb
 ****************************************************************/

 #include <cstdio>

 int n, m, mod;
 int a[];
 bool mat[];

 struct Matrix {
     int a[][];
     void make_zero() {
         for( int i=; i<m; i++ )
             for( int j=; j<m; j++ )
                 a[i][j] = ;
     }
     void make_unit() {
         for( int i=; i<m; i++ ) {
             for( int j=; j<m; j++ )
                 a[i][j] = ;
             a[i][i] = ;
         }
     }
     Matrix operator*( const Matrix & B ) const {
         const Matrix & A = *this;
         Matrix C;
         C.make_zero();
         for( int i=; i<m; i++ )
             for( int j=; j<m; j++ )
                 for( int k=; k<m; k++ ) {
                     C.a[i][j] += A.a[i][k]*B.a[k][j];
                     C.a[i][j]%=mod;
                 }
         return C;
     }
 };

 void prep() {
     for( int i=; i<m-; i++ ) {
         mat[i] = true;
         for( int j=; j<=i; j++ )
             if( a[j]!=a[m-i-+j] ) {
                 mat[i] = false;
                 break;
             }
     }
 }
 Matrix mpow( Matrix a, int b ) {
     Matrix rt;
     rt.make_unit();
     for( ; b; b>>=,a=a*a )
         if( b& ) rt=rt*a;
     return rt;
 }

 void work() {
     Matrix a;
     a.make_zero();
     a.a[][] = ;
     a.a[][m-] = -;
     for( int i=; i<m; i++ ) {
         a.a[i][i-] = ;
         a.a[i][m-] = -mat[i-];
     }
     a = mpow(a,n);
     int ans = (a.a[][]%mod+mod)%mod;
     printf( "%d\n", ans );
 }

 int main() {
     scanf( "%d%d%d", &n, &m, &mod );
     char ch;
     while( scanf( "%[\n ]", &ch ) );
     for( int i=; i<m; i++ ) {
         scanf( "%c", &ch );
         a[i] = ch-'';
     }
     prep();
     work();
 }