POJ 3233 Matrix Power Series 矩阵快速幂

设S[k] = A + A^2 +````+A^k.

设矩阵T =

A[1]	0
E	E

这里的E为n*n单位方阵，0为n*n方阵

令A[k] = A ^ k

矩阵B[k] =

A[k+1]

S[k]

则有递推式B[K] = T*B[k-1],即有B[k] = T^k*B[0],令S[0] 为n*n的0矩阵。

矩阵快速幂求出即可·····

还可以使用两次分治的方法····自行百度····

贴代码：

 #include<cstdio>
 #include<cstring>
 int n,k,p,d;//d = 2*n
 struct matrix
 {
     int m[][];
 } A;
 matrix mul(int a[][],int b[][])
 {
     matrix ans;
     memset(ans.m,,sizeof(ans.m));
     for(int i=; i<=d; ++i)
         for(int j=; j<=d; ++j)
             for(int k=; k<=d; ++k)
                 ans.m[i][j] = (ans.m[i][j] + a[i][k]*b[k][j]%p)%p;
     return ans;
 }
 matrix qPow()
 {
     matrix ans;
     memset(ans.m,,sizeof(ans.m));
     for(int i=; i<=d; ++i)
         ans.m[i][i] = ;
     while(k)
     {
         if(k&) ans = mul(ans.m,A.m);
         A = mul(A.m,A.m);
         k >>= ;
     }
     return ans;
 }
 int main()
 {
 //    freopen("in.txt","r",stdin);
     while(~scanf("%d%d%d",&n,&k,&p))
     {
         memset(A.m,,sizeof(A.m));
         int t[][];
         for(int i=; i<=n; ++i)
         {
             for(int j=; j<=n; ++j)
             {
                 scanf("%d",&A.m[i][j]);
                 t[i][j] = A.m[i][j];
             }
         }
         for(int i=n+; i<=*n; ++i)
             A.m[i][i-n] = ,A.m[i][i] = ;
         d = n<<;
         matrix ans = qPow();
         for(int i=n+; i<=d; ++i)
         {
             for(int j=; j<=n; ++j)
             {
                 int res =;
                 for(int k=; k<=n; ++k)
                     res = (res + ans.m[i][k]*t[k][j]%p)%p;
                 if(j != ) printf(" ");
                 printf("%d",res);
             }
             puts("");
         }
     }
     return ;
 }