hdu 1757 A Simple Math Problem 矩阵快速幂

Posted 2020-09-29 Pacify

快速幂套模板呀！！！！

矩阵快速幂模板

#include<bits/stdc++.h>
using namespace std;

const int N = 12;
int n,Mod;

struct Mat
{
    int mat[N][N];
};

Mat operator *(Mat a, Mat b)
{
    Mat c;
    memset(c.mat, 0, sizeof(c.mat));
    for(int k = 0; k < n; ++k)
        for(int i = 0; i < n; ++i)
            if(a.mat[i][k])
                for(int j = 0; j < n; ++j)
                    if(b.mat[k][j])
                        c.mat[i][j] = (c.mat[i][j] +a.mat[i][k] * b.mat[k][j])%Mod;
    return c;
}

Mat operator ^(Mat a, int k)
{
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    for(int i = 0; i < n; ++i)
        c.mat[i][i]=1;
    for(; k; k >>= 1)
    {
        if(k&1) c = c*a;
        a = a*a;
    }
    return c;
}

int main()
{
//    freopen("Input.txt","r",stdin);
    int k;
    int aa[12];
    n=10;
    while(~scanf("%d%d",&k,&Mod))
    {
        for(int i=0;i<10;i++)
            scanf("%d",aa+i);
        if(k<10)
        {
            printf("%d\\n",k);
            continue;
        }
        Mat a;
        memset(a.mat,0,sizeof(a.mat));
        for(int i=0;i<10;i++)
            a.mat[i][0]=aa[i];
        for(int i=1;i<10;i++)
            a.mat[i-1][i]=1;
        a=a^(k-9);
        int ans=0;
        for(int i=0;i<10;i++)
            ans=(ans+a.mat[i][0]*(9-i))%Mod;
        printf("%d\\n",ans);
    }
    return 0;
}