矩阵乘法和快速幂模板

#include <string.h>
const int mod=1e9+7;
struct MM
{
    int r,c;
    int mx[105][105];
    MM(int rr=0,int cc=0){r=rr,c=cc;}
    friend MM operator *(MM ta,MM tb)
    {
        MM tc(ta.r,tb.c);
        for(int i=1;i<=tc.r;i++)
            for(int j=1;j<=tc.c;j++)
            {
                tc.mx[i][j]=0;
                for(int k=1;k<=tb.r;k++)
                    tc.mx[i][j]=(tc.mx[i][j]+ta.mx[i][k]*tb.mx[k][j])%mod;
            }
        return tc;
    }
    friend MM operator ^(MM ta,int num)
    {
        MM ret(ta.r,ta.c);                          //r==c
        memset(ret.mx,0,sizeof ret.mx);
        for(int i=1;i<=ta.r;i++) ret.mx[i][i]=1;    //单位矩阵
        while(num)
        {
            if(num&1)
                ret=ta*ret;
            num>>=1;
            ta=ta*ta;
        }
        return ret;
    }
};
原文地址:https://www.cnblogs.com/weeping/p/5943561.html