矩阵快速幂 学习笔记

(Updated: )
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举例1：Fibonacci

题目传送门

题意

$$f[1]=1,f[2]=1,f[3]=2,f[4]=3 \dots f[n]=f[n-1]+f[n-2]$$
那么输入$n$、$m$，求第n项Fibonacci的值$mod$ $m$，即$f[n]$ $mod$ $m$。
$$1\leq n \leq 2 \times 10^9$$
因为：$$f[i]=1 \times f[i-1]+1 \times f[i-2]$$$$f[i-1]=1\times f[i-1]+0 \times f[i-2]$$
所以，我们可以发现递推式可以转化为矩阵运算：
$$\left(\begin{array}{rcl}f[i]\\f[i-1]\end{array} \right) = \left(\begin{array}{rcl}1 \quad 1\\1\quad 0\end{array} \right) \times \left(\begin{array}{rcl}f[i-1]\\f[i-2]\end{array} \right)=\left(\begin{array}{rcl}1 \quad 1\\1\quad 0\end{array} \right)^2 \times \left(\begin{array}{rcl}f[i-2]\\f[i-3]\end{array} \right)$$
那么可得：
$$\left(\begin{array}{rcl}f[n]\\f[n-1]\end{array} \right)=\left(\begin{array}{rcl}1 \quad 1\\1\quad 0\end{array} \right)^{n-2} \times \left(\begin{array}{rcl}f[2]\\f[1]\end{array} \right)$$

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#include<algorithm>
#include<bitset>
#include<complex>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<iterator>
#include<limits>
#include<list>
#include<locale>
#include<map>
#include<memory>
#include<new>
#include<numeric>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<typeinfo>
#include<utility>
#include<valarray>
#include<vector>
#include<cctype>
#include<cerrno>
#include<cfloat>
#include<ciso646>
#include<climits>
#include<clocale>
#include<cmath>
#include<csetjmp>
#include<csignal>
#include<cstdarg>
#include<cstddef>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#define ll long long
using namespace std;
inline ll read(){
    ll res=0,f=1;char ch=getchar();
    while(ch<'0'ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') res=res*10+ch-'0',ch=getchar();
    return res*f;
}
inline void write(ll x){
    if(x<0) putchar('-'),x=-x;
    if(x<10) putchar(x+'0');
    else{
        write(x/10);
        putchar(x%10+'0');
    }
}
ll n,m;
struct node{
    ll f[3][3];
}a,f,s;
void build(node &x){
    for(ll i=0;i<=1;i++){
        for(ll j=0;j<=1;j++){
            if(i==j) x.f[i][j]=1;
            else x.f[i][j]=0;
        }
    }
}
void Mul(node &x,node &y,node &z){
    memset(z.f,0,sizeof(z.f));
    for(ll i=0;i<=1;i++){
        for(ll j=0;j<=1;j++){
            if(x.f[i][j]!=0){
                for(ll k=0;k<=1;k++){
                    z.f[i][k]+=x.f[i][j]*y.f[j][k];
                    z.f[i][k]%=m;
                }
            }
        }
    }
}
node Pow(ll b){
    node res;
    build(res);
    node tmp=f,t;
    while(b){
        if(b%2==1) Mul(res,tmp,t),res=t;
        Mul(tmp,tmp,t),tmp=t,b>>=1; 
    }
    return res;
}
int main(){
    n=read();m=read();
    if(n<=2){
        puts("1");
        return 0;
    }
    f.f[0][0]=1;f.f[1][0]=1;f.f[0][1]=1;f.f[1][1]=0;
    node q=Pow(n-2);
    write((q.f[0][0]+q.f[1][0]+m)%m);
    return 0;
}
//(f[n]  )=(1 1)^(n-2)*(f[2])
//(f[n-1])=(1 0)      *(f[1])

举例2：Fibonacci求和

题目传送门

题意

输入$n$、$m$，求出Fibonacci的前$n$项的和 $mod$ $m$的值，即：$f[n]\quad mod\quad m$
$$f[n]=f[n-1]+f[n-2]$$
$$f[n-1]=f[n-2]+f[n-3],f[n]=2\times f[n-2]+f[n-3]$$
$$f[n-2]=f[n-3]+f[n-4],f[n]=f[n-2]+2\times f[n-3]+f[n-4]$$
$$f[n-3]=f[n-4]+f[n-5],f[n]=f[n-2]+f[n-3]+2\times f[n-4]+f[n-5]$$
以此类推，可得：
$$f[n]=f[n-2]+f[n-3]+f[n-4]+f[n-5]+\dots f[2]+2 \times f[1]$$
那么：
$$f[n-2]+f[n-3]+f[n-4]+f[n-5]+\dots f[2]+f[1]=f[n]-f[1]$$
$$f[n-2]+f[n-3]+f[n-4]+f[n-5]+\dots f[2]+f[1]=f[n]-1$$
也就是：
$$f[n]+f[n-1]+f[n-2]+f[n-3]+\dots f[2]+f[1]=f[n+2]-1$$
所以只需要将上题代码改一改就好了：

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#include<algorithm>
#include<bitset>
#include<complex>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<iterator>
#include<limits>
#include<list>
#include<locale>
#include<map>
#include<memory>
#include<new>
#include<numeric>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<typeinfo>
#include<utility>
#include<valarray>
#include<vector>
#include<cctype>
#include<cerrno>
#include<cfloat>
#include<ciso646>
#include<climits>
#include<clocale>
#include<cmath>
#include<csetjmp>
#include<csignal>
#include<cstdarg>
#include<cstddef>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#define ll long long
using namespace std;
inline ll read(){
    ll res=0,f=1;char ch=getchar();
    while(ch<'0'ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') res=res*10+ch-'0',ch=getchar();
    return res*f;
}
inline void write(ll x){
    if(x<0) putchar('-'),x=-x;
    if(x<10) putchar(x+'0');
    else{
        write(x/10);
        putchar(x%10+'0');
    }
}
ll n,m;
struct node{
    ll f[3][3];
}a,f,s;
void build(node &x){
    for(ll i=0;i<=1;i++){
        for(ll j=0;j<=1;j++){
            if(i==j) x.f[i][j]=1;
            else x.f[i][j]=0;
        }
    }
}
void Mul(node &x,node &y,node &z){
    memset(z.f,0,sizeof(z.f));
    for(ll i=0;i<=1;i++){
        for(ll j=0;j<=1;j++){
            if(x.f[i][j]!=0){
                for(ll k=0;k<=1;k++){
                    z.f[i][k]+=x.f[i][j]*y.f[j][k];
                    z.f[i][k]%=m;
                }
            }
        }
    }
}
node Pow(ll b){
    node res;
    build(res);
    node tmp=f,t;
    while(b){
        if(b%2==1) Mul(res,tmp,t),res=t;
        Mul(tmp,tmp,t),tmp=t,b>>=1; 
    }
    return res;
}
int main(){
    n=read();m=read();
    if(n<=2){
        puts("1");
        return 0;
    }
    f.f[0][0]=1;f.f[1][0]=1;f.f[0][1]=1;f.f[1][1]=0;
    node q=Pow(n);
    write((q.f[0][0]+q.f[1][0]+m-1)%m);
    return 0;
}