2693: I 矩阵连乘_带i的连乘-优快云博客

本文探讨了矩阵运算的基本概念及其在快速幂算法中的应用，通过实例展示了如何使用矩阵来加速幂次运算过程，特别关注了快速幂算法在解决大规模幂次计算问题时的高效性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=101;
int mod=1000;
class Matrix
{
    int n,m;//矩阵的row and colum
    int num[maxn][maxn];//从1开始
    public:
        Matrix(){memset(num,0,sizeof(num));}
        Matrix(const Matrix &a) {n=a.n;m=a.m; memcpy(num,a.num,sizeof(a.num));}
        Matrix(int x,int y):n(x),m(y){}
        Matrix & operator =(const Matrix & a);
        friend Matrix operator + (const Matrix &a,const Matrix &b);
        friend Matrix operator - (const Matrix &a,const Matrix &b);
        friend Matrix operator * (const Matrix &a,const Matrix &b);
        friend Matrix operator % (const Matrix &a,int n);
        friend ostream & operator <<(ostream & cout,const Matrix & a);
        friend istream & operator >>(istream & cin,Matrix & a);
        void Modefy_n_m(int x,int y,int v);
        void out();//只在此题中有用，其他题目可删去
};
Matrix & Matrix:: operator =(const Matrix &a)
{
    for(int i=1;i<=a.n;i++) for(int j=1;j<=a.m;j++) num[i][j]=a.num[i][j];

    return *this;
}
istream & operator >>(istream & cin,Matrix & a)
{
    for(int i=1;i<=a.n;i++)
    {
        for(int j=1;j<=a.m;j++)
        {
            cin>>a.num[i][j];
        }
    }
    return cin;
}
ostream & operator <<(ostream & cout,const Matrix & a)
{
    for(int i=1;i<=a.n;i++)
    {
        for(int j=1;j<=a.m;j++)
        {
            cout<<a.num[i][j]<<"   ";
        }
        cout<<endl;
    }
    return cout;
}
Matrix operator % (const Matrix &a,int n)
{
    Matrix t;
    for(int i=1;i<=a.n;i++)
    {
        for(int j=1;j<=a.m;j++)
        {
            t.num[i][j]=a.num[i][j]%n;
        }
    }
    return t;
}
Matrix operator * (const Matrix &a,const Matrix &b)
{
    Matrix t(a.n,b.m);
    for(int i=1;i<=a.n;i++)
    {
        for(int j=1;j<=b.m;j++)
        {
            int cnt=0;
            for(int k=1;k<=a.m;k++)
            {
                cnt+=a.num[i][k]*b.num[k][j];
                cnt%=mod;//这是此题需要，别的题目可删去
            }
            t.num[i][j]=cnt;
        }
    }
    return t;
}
Matrix operator + (const Matrix &a,const Matrix &b)
{
    Matrix t;
    for(int i=1;i<=a.n;i++) for(int j=1;j<=a.m;j++) t.num[i][j]=a.num[i][j]+b.num[i][j];
    return t;
}
Matrix operator - (const Matrix &a,const Matrix &b)
{
    Matrix t;
    for(int i=1;i<=a.n;i++) for(int j=1;j<=a.m;j++) t.num[i][j]=a.num[i][j]-b.num[i][j];
    return t;
}
void Matrix:: Modefy_n_m(int x,int y,int v)
{
    num[x][y]=v;
}

void Matrix::out()
{
    printf("%d %d/n",num[1][m-1],num[1][m]);
}
Matrix a(5,5);
Matrix temp(5,5);
Matrix E(5,5);
void bin(int n)
{
    if(n==0)
    {
        a=E;
        return ;
    }
    if(n==1)
    {
        a=temp;
        return ;
    }
    if(n==2)
    {
        a=temp*temp;
        return ;
    }
    bin(n/2);
    a=a*a;
    if(n&1) a=a*temp;
}
int main()
{
    int n;
    int h[6]={0,2,1,0,3,3};
    int g[6][6]={{0,0,0,0,0,0},{0,2,1,0,3,3},{0,1,0,1,1,2},{0,0,0,1,0,1},{0,0,0,0,0,0},{0,0,0,0,0,0}};
    Matrix pre(1,5);
    for(int i=1;i<=5;i++) pre.Modefy_n_m(1,i,h[i]);
    for(int i=1;i<=5;i++) for(int j=1;j<=5;j++) temp.Modefy_n_m(i,j,g[i][j]);
    for(int i=1;i<=5;i++) for(int j=1;j<=5;j++) E.Modefy_n_m(i,j,i==j);
    while(scanf("%d%d",&n,&mod)==2)
    {
        bin(n-1);
        Matrix ans(1,5);ans=pre*a;
        ans.out();
    }
    return 0;
}