矩阵快速幂入门

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NEFU459 A Simple Math Problem

这题的矩阵。

挂一张比较棒的图。

更新后的写法

#include 
using namespace std;
const int MAX = 10;
typedef long long ll;
typedef struct{
    ll m[MAX][MAX];
}Matrix;

Matrix P;//构造出的矩阵

ll k,mod,a[MAX];
Matrix matrixmul(Matrix a,Matrix b) //矩阵乘法
{
    int i,j,k;
    Matrix c;
    for (i = 0 ; i < MAX; i++)
        for (j = 0; j < MAX;j++)
        {
            c.m[i][j] = 0;
            for (k = 0; k < MAX; k++)
                c.m[i][j] =(c.m[i][j]+(a.m[i][k]*b.m[k][j]))%mod;
            c.m[i][j] %=mod;
        }
    return c;
}

Matrix quickpow(Matrix m , ll n)
{
    Matrix b;//单位矩阵在这构造也可以
    for(int i=0;i<MAX;i++)
        for(int j=0;j<MAX;j++)
        {
            if(i==j)b.m[i][j]=1;
            else b.m[i][j]=0;
        }
    while (n >= 1)
    {
        if (n & 1)
                b = matrixmul(b,m);
        n = n >> 1;
        m = matrixmul(m,m);
    }
    return b;
}
int main()
{
    while(cin>>k>>mod){
        for(int i=0;i<MAX;i++){
            cin>>a[i];
            P.m[0][i]=a[i];
            if(i>=1) P.m[i][i-1]=1;
        }
        if(k<10){printf("%lld\n",a[k]%mod);continue;}
        else{
            Matrix tmp=quickpow(P,k-9);
            ll ans=0;
            for(int i=0;i<MAX;i++){
                ans=(ans+tmp.m[0][i]*(9-i)%mod)%mod;
            }
            printf("%lld\n",ans%mod);
        }
    }
    return 0;
}

NEFU461 fibs的位数

广义斐波那契数列。
$$an=u\ast a_{n-1}+v\ast a_{n-2}$$
$$a_0=a,a_1=b$$
$$x=\frac{u+\sqrt{u^2+4v}}{2},y=\frac{u-\sqrt{u^2+4v}}{2}$$
则$$a_n=\frac{(b-ay)x^n}{x-y}$$
对$a_n$取对数加一即可：
$$len=nlo{g}_{10}(x)+lo{g}_{10}(b-ay)-lo{g}_{10}(x-y)+1$$

#include 
using namespace std;
typedef long long ll;
ll n,a,b,u,v;
int main()
{
    while(scanf("%lld%lld%lld%lld%lld",&n,&a,&b,&u,&v)!=EOF)
    {
        double c=sqrt(u*u+4*v);
        double x=(u+c)/2.0;
        double y=(u-c)/2.0;
        double len=n*log10(x)+log10(b-a*y)-log10(c);
        printf("%lld\n",(ll)len+1);
    }
    return 0;
}

NEFU462 fibs的组合

一开始看这题一点思路都没有，那就先打表！

#include 
const int maxn=100;
using namespace std;

inline int f(int x) {
	if (x == 1 || x == 2) return 1;
	else return f(x - 1) + f(x - 2);
};
int c(int n,int k){
    if (n==k||k==0)
	{
		return 1;
	}
	else
	{
		return c(n-1,k)+c(n-1,k-1);
	}
}
int ans;

int main()
{
    for(int i=1;i<=10;i++){
        ans=0;
        printf("f(%d)=%d   ",i,f(i));
        for(int j=1;j<=i;j++){
            ans+=(f(j)*c(i,j));
        }
        printf("s(%d)=%d\n",i,ans);
    }
    // f(1)=1;f(2)=1;f(3)=2;f(4)=3
    // s(1)=1;s(2)=3;
    return 0;
}

观察到$S(n)=f(2n)$
那就好做多了！找循环节即可！

#include 

using namespace std;
inline int f(int x) {
	if (x == 1 || x == 2) return 1;
	else return f(x - 1) + f(x - 2);
};
int main()
{
    for(int i=1;i<=200;i++){
        if(f(2*i)%8==0)
            cout<<"yes"<<endl;
        else
            cout<<"no"<<endl;
    }
    return 0;
}

显而易见， 代码孕育而生！

#include 
using namespace std;
typedef long long ll;
ll n,x;
int main()
{
    while(cin>>n){
        x=n%3;
        if(x==1||x==2){
            cout<<"no\n";
        }
        else{
            cout<<"yes\n";
        }
    }
    return 0;
}

NEFU463 Fibs之和

可以推出规律为：
$$S(n)=f(n+2)-f(2)$$
所以求$a$到$b$的和,就是
$$\begin{gathered} S(b)-S(a-1) \\ =f(b+2)-f(a+1) \end{gathered}$$
代码孕育而生。

#include 
using namespace std;
const int MAX = 2;
const int mod=1000000000;
typedef long long ll;
typedef struct{
    ll m[MAX][MAX];
}Matrix;

Matrix P = {1,1,
            1,0//推出来的矩阵
};
Matrix I = {1,0,//单位阵
            0,1
};

Matrix matrixmul(Matrix a,Matrix b) //矩阵乘法
{
    int i,j,k;
    Matrix c;
    for (i = 0 ; i < MAX; i++)
        for (j = 0; j < MAX;j++)
        {
            c.m[i][j] = 0;
            for (k = 0; k < MAX; k++)
                c.m[i][j] =(c.m[i][j]+(a.m[i][k]*b.m[k][j]))%mod;
            c.m[i][j] %=mod;
        }
    return c;
}

Matrix quickpow(ll n)
{
    Matrix m = P, b = I;
    while (n >= 1)
    {
        if (n & 1)
                b = matrixmul(b,m);
        n = n >> 1;
        m = matrixmul(m,m);
    }
    return b;
}
ll a,b;
int main()
{
    while(cin>>a>>b){
        if(a==0&&b==0)
            break;
        Matrix tmp1,tmp2;
        tmp1=quickpow(a+1);
        tmp2=quickpow(b+2);
        int ans=((tmp2.m[1][0]-tmp1.m[1][0])%mod+(tmp2.m[1][1]-tmp1.m[1][1])%mod)%mod;
        cout<<(ans+mod)%mod<<endl;
        }
    return 0;
}

NEFU457 Not Fibonacci

如果结果是负数可以这么处理(ans+mod)%mod.
$f(n)=p\ast f(n-1)+q\ast f(n-2)$
$f(0)=a,f(1)=b$

记录一下怎么求出来的$n-1$
$${\left\{\begin{array}{ccc}1& p& q\\ 0& p& q\\ 0& 1& 0\end{array}\right\}}^1\ast \left\{\begin{array}{c}s(1)\\ f(1)\\ f(0)\end{array}\right\}=\left\{\begin{array}{c}s(2)\\ f(2)\\ f(1)\end{array}\right\}$$

• 首先就是推出上面的这个式子，等号左边只有一列的这个矩阵，一定要出现最小的值，比如$f(0)$。
• 然后，推出来的矩阵的1次方，就对应一下等号右边的1，比如$f(1)$。
• 然后就可以推出下面的式子了！

$${\left\{\begin{array}{ccc}1& p& q\\ 0& p& q\\ 0& 1& 0\end{array}\right\}}^n\ast \left\{\begin{array}{c}s(1)\\ f(1)\\ f(0)\end{array}\right\}=\left\{\begin{array}{c}s(n+1)\\ f(n+1)\\ f(n)\end{array}\right\}$$
要求的是$s(n)$，把$n$转换成$n-1$就行了。
$${\left\{\begin{array}{ccc}1& p& q\\ 0& p& q\\ 0& 1& 0\end{array}\right\}}^{n-1}\ast \left\{\begin{array}{c}a+b\\ b\\ a\end{array}\right\}=\left\{\begin{array}{c}s(n)\\ f(n)\\ f(n-1)\end{array}\right\}$$

#include 
using namespace std;
const int MAX = 3;
const int mod=1e7;
typedef long long ll;
typedef struct{
    ll m[MAX][MAX];
}Matrix;

Matrix P;//构造出的矩阵

Matrix matrixmul(Matrix a,Matrix b) //矩阵乘法
{
    int i,j,k;
    Matrix c;
    for (i = 0 ; i < MAX; i++)
        for (j = 0; j < MAX;j++)
        {
            c.m[i][j] = 0;
            for (k = 0; k < MAX; k++)
                c.m[i][j] =(c.m[i][j]+(a.m[i][k]*b.m[k][j]))%mod;
            c.m[i][j] %=mod;
        }
    return c;
}

Matrix quickpow(Matrix m , ll n)
{
    Matrix b;//单位矩阵在这构造也可以
    for(int i=0;i<MAX;i++)
        for(int j=0;j<MAX;j++)
        {
            if(i==j)b.m[i][j]=1;
            else b.m[i][j]=0;
        }
    while (n >= 1)
    {
        if (n & 1)
                b = matrixmul(b,m);
        n = n >> 1;
        m = matrixmul(m,m);
    }
    return b;
}
ll t,a,b,p,q,s,e,ans1,ans2;
int main()
{
    cin>>t;
    while(t--){
        cin>>a>>b>>p>>q>>s>>e;
        P={1,p,q,0,p,q,0,1,0};
        if(e==0) ans1=a;
        else{
            Matrix tmp=quickpow(P,e-1);
            ans1=(tmp.m[0][0]*(a+b)+tmp.m[0][1]*b+tmp.m[0][2]*a)%mod;
        }
        if(s==0) ans2=0;//为什么这儿是0，不是a呢？因为我们要求的是S(s-1)的值，s如果等于0的话，那么就没有意义了。
        else if(s==1) ans2=a;
        else{
            Matrix tmp=quickpow(P,s-2);
            ans2=(tmp.m[0][0]*(a+b)+tmp.m[0][1]*b+tmp.m[0][2]*a)%mod;
        }
        printf("%lld\n",((ans1-ans2)%mod+mod)%mod);
    }

    return 0;
}

NEFU458 Another kind of Fibonacci

$${\left\{\begin{array}{cccc}1& 1& 0& 0\\ 0& x^2& y^2& 2xy\\ 0& 1& 0& 0\\ 0& x& 0& y\end{array}\right\}}^n\ast \left\{\begin{array}{c}s(0)\\ f^2(1)\\ f^2(0)\\ f(1)f(0)\end{array}\right\}=\left\{\begin{array}{c}s(n)\\ f^2(n+1)\\ f^2(n)\\ f(n+1)f(n)\end{array}\right\}$$

#include 
using namespace std;
const int MAX = 4;
const int mod=10007;
typedef long long ll;
typedef struct{
    ll m[MAX][MAX];
}Matrix;

Matrix P;//构造出的矩阵

Matrix matrixmul(Matrix a,Matrix b) //矩阵乘法
{
    int i,j,k;
    Matrix c;
    for (i = 0 ; i < MAX; i++)
        for (j = 0; j < MAX;j++)
        {
            c.m[i][j] = 0;
            for (k = 0; k < MAX; k++)
                c.m[i][j] =(c.m[i][j]+(a.m[i][k]*b.m[k][j]))%mod;
            c.m[i][j] %=mod;
        }
    return c;
}

Matrix quickpow(Matrix m , ll n)
{
    Matrix b;//单位矩阵在这构造也可以
    for(int i=0;i<MAX;i++)
        for(int j=0;j<MAX;j++)
        {
            if(i==j)b.m[i][j]=1;
            else b.m[i][j]=0;
        }
    while (n >= 1)
    {
        if (n & 1)
                b = matrixmul(b,m);
        n = n >> 1;
        m = matrixmul(m,m);
    }
    return b;
}
ll n,x,y,ans;
int main()
{
    while(cin>>n>>x>>y){
        ans=0;
        ll x_2=(x*x)%mod;
        ll y_2=(y*y)%mod;
        ll xy=(2*x*y)%mod;
        P={1,1,0,0,
           0,x_2,y_2,xy,
           0,1,0,0,
           0,x,0,y};
        Matrix tmp=quickpow(P,n);
        for(int i=0;i<4;i++){
            ans=(ans+tmp.m[0][i])%mod;
        }
        printf("%lld\n",ans%mod);
    }
    return 0;
}

完结。