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：

 1 #include<cstdio>
 2 #include<cstring>
 3 int n,k,p,d;//d = 2*n
 4 struct matrix
 5 {
 6     int m[70][70];
 7 } A;
 8 matrix mul(int a[70][70],int b[70][70])
 9 {
10     matrix ans;
11     memset(ans.m,0,sizeof(ans.m));
12     for(int i=1; i<=d; ++i)
13         for(int j=1; j<=d; ++j)
14             for(int k=1; k<=d; ++k)
15                 ans.m[i][j] = (ans.m[i][j] + a[i][k]*b[k][j]%p)%p;
16     return ans;
17 }
18 matrix qPow()
19 {
20     matrix ans;
21     memset(ans.m,0,sizeof(ans.m));
22     for(int i=1; i<=d; ++i)
23         ans.m[i][i] = 1;
24     while(k)
25     {
26         if(k&1) ans = mul(ans.m,A.m);
27         A = mul(A.m,A.m);
28         k >>= 1;
29     }
30     return ans;
31 }
32 int main()
33 {
34 //    freopen("in.txt","r",stdin);
35     while(~scanf("%d%d%d",&n,&k,&p))
36     {
37         memset(A.m,0,sizeof(A.m));
38         int t[35][35];
39         for(int i=1; i<=n; ++i)
40         {
41             for(int j=1; j<=n; ++j)
42             {
43                 scanf("%d",&A.m[i][j]);
44                 t[i][j] = A.m[i][j];
45             }
46         }
47         for(int i=n+1; i<=2*n; ++i)
48             A.m[i][i-n] = 1,A.m[i][i] = 1;
49         d = n<<1;
50         matrix ans = qPow();
51         for(int i=n+1; i<=d; ++i)
52         {
53             for(int j=1; j<=n; ++j)
54             {
55                 int res =0;
56                 for(int k=1; k<=n; ++k)
57                     res = (res + ans.m[i][k]*t[k][j]%p)%p;
58                 if(j != 1) printf(" ");
59                 printf("%d",res);
60             }
61             puts("");
62         }
63     }
64     return 0;
65 }

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