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[70][70];
} A;
matrix mul(int a[70][70],int b[70][70])
{
    matrix ans;
    memset(ans.m,0,sizeof(ans.m));
    for(int i=1; i<=d; ++i)
        for(int j=1; j<=d; ++j)
            for(int k=1; 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,0,sizeof(ans.m));
    for(int i=1; i<=d; ++i)
        ans.m[i][i] = 1;
    while(k)
    {
        if(k&1) ans = mul(ans.m,A.m);
        A = mul(A.m,A.m);
        k >>= 1;
    }
    return ans;
}
int main()
{
//    freopen("in.txt","r",stdin);
    while(~scanf("%d%d%d",&n,&k,&p))
    {
        memset(A.m,0,sizeof(A.m));
        int t[35][35];
        for(int i=1; i<=n; ++i)
        {
            for(int j=1; j<=n; ++j)
            {
                scanf("%d",&A.m[i][j]);
                t[i][j] = A.m[i][j];
            }
        }
        for(int i=n+1; i<=2*n; ++i)
            A.m[i][i-n] = 1,A.m[i][i] = 1;
        d = n<<1;
        matrix ans = qPow();
        for(int i=n+1; i<=d; ++i)
        {
            for(int j=1; j<=n; ++j)
            {
                int res =0;
                for(int k=1; k<=n; ++k)
                    res = (res + ans.m[i][k]*t[k][j]%p)%p;
                if(j != 1) printf(" ");
                printf("%d",res);
            }
            puts("");
        }
    }
    return 0;
}