POJ -3233，HDU-5015(矩阵快速幂)

题目链接： POJ-3233

思路

利用矩阵快速幂推出转移矩阵。注意什么时候取模，同时用printf一定要注意格式的问题，与类型要匹配。
这次的转移矩阵是由一些子矩阵构成，不再是一些普通的常数了。
$$\left\{\begin{array}{cc}E& E\\ O& A\end{array}\right\}\times \left\{\begin{array}{c}{S}_{k-1}\\ A_k\end{array}\right\}=\left\{\begin{array}{c}{S}_k\\ A_{k+1}\end{array}\right\}$$

#include
#include 
#include 
#include 
using namespace std;
typedef long long ll;
struct Matrix {
   int m[100][100];
};
int mod;
Matrix mul(Matrix A,Matrix B,int n)
{
    Matrix tmp;
    int i,j,k;
    memset(tmp.m,0,sizeof(tmp.m));
    for(i = 1; i <= n; i++)
    for(j = 1; j <= n ; j++)
        for(k = 1; k <= n ;k++)
    {
        tmp.m[i][j] = (tmp.m[i][j] + A.m[i][k] * B.m[k][j]) % mod;
        tmp.m[i][j] %= mod;
    }
    return tmp;
}
Matrix pow(Matrix A,int N,int n)
{
    int i;
    Matrix res;
    memset(res.m,0,sizeof(res.m));
    for(i = 1; i <= n ;i++)
        res.m[i][i] = 1;
    while(N)
    {
        if(N & 1)
            res = mul(res,A,n);
        A = mul(A,A,n);
        N >>= 1;
    }
    return res;
}
int main()
{
    int n,k;
    int i,j;
    while(scanf("%d %d %d",&n,&k,&mod)!=EOF)
    {
    Matrix T,A,D;
    memset(T.m,0,sizeof(T.m));
    memset(D.m,0,sizeof(D.m));
    for(i = 1; i <= n ;i++)
        for(j = 1; j <= n ;j++)
    {
        scanf("%d",&A.m[i][j]);
    }
    for(i = 1; i <= n ; i++)
        T.m[i][i] = 1;
    for(i = 1; i <=  n ;i++)
        for(j = n+1; j <= 2*n; j++)
    {
        T.m[i][j] = T.m[i+n][j] = A.m[i][j-n];
    }
    for(i = 1; i <= n ;i++)
    for(j = 1; j <= n ;j++)
    {
        D.m[i][j] = D.m[i+n][j] = A.m[i][j];
    }
    n <<= 1;
    T = pow(T,k - 1,n);
    T = mul(T,D,n);
    n >>= 1;
    for(i = 1; i <= n ;i++)
    for(j = 1; j <= n ;j++)
    printf("%d%c",T.m[i][j],j==n?'\n':' ');
    }
}

题目链接： HDU -5015

题意

告诉你第一行和第一列分别是什么数，然后告诉你递推公式，让你求a[n][m]是多少。

思路

我们发现m特别大，因此考虑快速幂，主要是递推矩阵不好想，这里是以列转移。
$$n+2\left\{\begin{array}{ccccccc}10& \cdots & \cdots & \cdots & \cdots & \cdots & 1\\ 10& 1& \cdots & \dots & \cdots & \cdots & 1\\ 10& 1& 1& \cdots & \cdots & \cdots & 1\\ 10& 1& 1& 1\cdots & \cdots & \cdots & 1\\ 0& 0& \cdots & & & & 1\end{array}\right\}\times \left\{\begin{array}{c}{a}_{0,m-1}\\ a_{1,m-1}\\ \cdots \\ a_{n,m-1}\\ 3\end{array}\right\}=\left\{\begin{array}{c}{a}_{0,m}\\ a_{1,m}\\ \cdots \\ a_{n,m}\\ 3\end{array}\right\}$$

#include
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int mod = 10000007;
struct Matrix{
   ll m[20][20];
};
int  n;
Matrix mul(Matrix A,Matrix B)
{
    Matrix tmp;
    int i,j,k;
    memset(tmp.m,0,sizeof(tmp.m));
    for(i = 0; i <= n + 1 ;i++)
    for(j = 0; j <= n + 1 ;j++)
    for(k = 0; k <= n + 1 ;k++)
    {
        tmp.m[i][j] = (tmp.m[i][j] + A.m[i][k] * B.m[k][j]) % mod;
        tmp.m[i][j] %= mod;
    }
    return tmp;
}
Matrix pow(Matrix A,int N)
{
    Matrix res;
    int i;
    memset(res.m,0,sizeof(res.m));
    for(i = 0; i <= n + 1; i++)
        res.m[i][i] = 1;
    while(N)
    {
        if(N & 1)
            res = mul(res,A);
        A = mul(A,A);
        N >>= 1;
    }
    return res;
}
int main()
{
    ll sum;
    int m,d[100][100];
    while(cin >> n >> m)
    {
    d[0][0] = 23;
    d[n + 1][0] = 3;
    for(int i = 1; i <= n ;i++)
        cin >> d[i][0];
    Matrix A,B;
    memset(A.m,0,sizeof(A.m));
    memset(B.m,0,sizeof(B.m));
    for(int i = 0; i <= n+1; i ++)
        A.m[i][0] = d[i][0];
    for(int i = 0 ; i <= n ;i++ )
    {
        for(int j = 0 ; j <= i  ;j++ )
        {
            B.m[i][j] = 1;
        }
    }
    for(int i = 0; i <= n; i++)
    {
        B.m[i][0] = 10;
        B.m[i][n + 1] = 1;
    }
    B.m[n + 1][n + 1] = 1;
    B = pow(B , m);
    A = mul(B , A);
    cout << A.m[n][0] << endl;
    }
}