矩阵快速幂学习入门

矩阵快速幂的原理是跟普通的快速幂一样，不过原来是数与数相乘，而矩阵快速幂是矩阵与矩阵相乘；

矩阵快速幂是运用于处理一些表达式进行多次递归会超时的问题；难点就是怎样构造矩阵，比如f[ n ] = f[n - 1] + f[ n - 2],

那么构造成的矩阵为：

1 1 

1 0;

模板：

typedef struct
{
    ll a[7][7];
    void Init()
    {
        memset(a,0,sizeof(0));
        for(int i = 1; i < 7; i ++)
        {
            a[i][i] = 1;
        }
    }
}Matrix;
Matrix p =
{
    1,2,1,4,6,4,1,
    1,0,0,0,0,0,0,
    0,0,1,4,6,4,1,
    0,0,0,1,3,3,1,
    0,0,0,0,1,2,1,
    0,0,0,0,0,1,1,
    0,0,0,0,0,0,1
};
Matrix Matrix_Mul(Matrix a,Matrix b)
{
    Matrix c;
    for(int i = 0; i < 7; i ++)
    {
        for(int j = 0; j < 7; j ++)
        {
            c.a[i][j] = 0;
            for(int k = 0; k < 7; k ++)
            {
                c.a[i][j] += (a.a[i][k] * b.a[k][j]) % MOD;
                c.a[i][j] %= MOD;
            }
        }
    }
    return c;
}
Matrix pow_Matrix(ll m)
{
    Matrix ans,b = p;
    memset(ans.a,0,sizeof(ans.a));
    for(int i = 0; i < 7; i ++)
    {
        ans.a[i][i] = 1;
    }
    while(m)
    {
        if(m & 1)
        {
            ans = Matrix_Mul(ans,b);
        }
        m >>= 1;
        b = Matrix_Mul(b,b);
    }
    return ans;
}

hdu2157

这是裸的矩阵快速幂，给出的图就是要进行相乘的矩阵，

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
typedef long long ll;
const int maxn = 100 + 10;
#define mod 1000

struct Matrix
{
    ll a[maxn][maxn];
}origin,res;
void Init(ll n)
{

    memset(res.a,0,sizeof(res.a));
    for(ll i = 0; i < n ; i ++)
    {
        res.a[i][i] = 1;
    }
}
Matrix Matrix_multi(Matrix aa,Matrix bb,ll n)//两矩阵相乘的算法
{
    Matrix ress;
    memset(ress.a,0,sizeof(ress));
    for(ll i = 0; i < n ; i ++)
    {
        for(ll  j = 0; j < n ; j ++)
        {
            for(ll k = 0; k < n ; k ++)
            {
                ress.a[i][j] += aa.a[i][k] * bb.a[k][j];
            }
        }
    }
    return ress;
}
void Matrix_calc(ll n,ll k)//矩阵快速幂
{
    Matrix origins = origin;
    while(k)
    {
        if(k & 1)
        {
            res = Matrix_multi(res,origins,n);
        }
        origins = Matrix_multi(origins,origins,n);
        k >>= 1;
    }
}

int main()
{
    int n,m;
    while( ~ scanf("%d%d",&n,&m) && (n || m) )
    {
        Init(n);
        memset(origin.a,0,sizeof(origin.a));
        for(ll i = 1; i <= m ; i ++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            origin.a[x][y] = 1;
        }
        int q;
        scanf("%d",&q);
        for(ll i = 1; i <= q; i ++)
        {
            Init(n);
            int x,y,k;
            scanf("%d%d%d",&x,&y,&k);
            Matrix_calc(n,k);
            cout << res.a[x][y] % mod <<endl;
        }
    }
    return 0;
}

hdu5950   Recursive sequence

题意：f[ n ] = f[n - 1] + 2 * f[n - 2] + n * 4;  n的范围< 2^31;

思路：数据范围很大，

首先一看这个题目，很容易想到构造矩阵：那么我们现在来分析一下怎么构造这个矩阵，那么 (n+1)^4 = n^4+4n^3+6n^2+4^n+1 所以光 (n+1)^4 这个矩阵就能构造出 5∗5 的一个矩阵来， 然后 f(n) = f(n−1)+2∗f(n−2) 这个是 2∗2 的矩阵，所以构造出来就应该是 7∗7 的转移矩阵 A ： 
{f(n),f(n−1),n^4,n^3,n^2,n,1}∗A={f(n+1),f(n),(n+1)^4,(n+1)^3,(n+1)^2,(n+1),1}
然后 f(n+1) = f(n)+2∗f(n−1)+(n+1)4， 可得矩阵 A： 

Matrix p =
{
    1,2,1,4,6,4,1,
    1,0,0,0,0,0,0,
    0,0,1,4,6,4,1,
    0,0,0,1,3,3,1,
    0,0,0,0,1,2,1,
    0,0,0,0,0,1,1,
    0,0,0,0,0,0,1
};

然后就是直接代码了：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MOD 2147493647
const int maxn = 200+ 20;
typedef struct
{
    ll a[7][7];
    void Init()
    {
        memset(a,0,sizeof(0));
        for(int i = 1; i < 7; i ++)
        {
            a[i][i] = 1;
        }
    }
}Matrix;
Matrix p =
{
    1,2,1,4,6,4,1,
    1,0,0,0,0,0,0,
    0,0,1,4,6,4,1,
    0,0,0,1,3,3,1,
    0,0,0,0,1,2,1,
    0,0,0,0,0,1,1,
    0,0,0,0,0,0,1
};
Matrix Matrix_Mul(Matrix a,Matrix b)
{
    Matrix c;
    for(int i = 0; i < 7; i ++)
    {
        for(int j = 0; j < 7; j ++)
        {
            c.a[i][j] = 0;
            for(int k = 0; k < 7; k ++)
            {
                c.a[i][j] += (a.a[i][k] * b.a[k][j]) % MOD;
                c.a[i][j] %= MOD;
            }
        }
    }
    return c;
}
Matrix pow_Matrix(ll m)
{
    Matrix ans,b = p;
    memset(ans.a,0,sizeof(ans.a));
    for(int i = 0; i < 7; i ++)
    {
        ans.a[i][i] = 1;
    }
    while(m)
    {
        if(m & 1)
        {
            ans = Matrix_Mul(ans,b);
        }
        m >>= 1;
        b = Matrix_Mul(b,b);
    }
    return ans;
}
int main()
{
    int Tcase;
    scanf("%d",&Tcase);
    for(int ii = 1; ii <= Tcase; ii ++)
    {
        ll n,a,b;
        scanf("%I64d%I64d%I64d",&n,&a,&b);
        if(n == 1)
        {
            cout << a << endl;
            continue;
        }
        if(n == 2)
        {
            cout << b << endl;
            continue;
        }
        Matrix ans = pow_Matrix(n - 2);
        ll anss = 0;
        anss = (anss +  ans.a[0][0] * b % MOD) % MOD;
        anss = (anss +  ans.a[0][1] * a % MOD) % MOD;
        anss = (anss +  ans.a[0][2] * 16 % MOD) % MOD;
        anss = (anss +  ans.a[0][3] * 8 % MOD) % MOD;
        anss = (anss +  ans.a[0][4] * 4 % MOD) % MOD;
        anss = (anss +  ans.a[0][5] * 2 % MOD) % MOD;
        anss = (anss +  ans.a[0][6] * 1 % MOD) % MOD;
        cout << anss << endl;
    }
    return 0;
}