BZOJ 2326 HNOI 2011 数学作业 矩阵乘法

题目大意

求一个这样的数：“12345678910111213……”对m取模的值。

思路

观察这个数字的特点，每次向后面添加一个数。也就是原来的数乘10^k之后在加上一个数。而且处理每个数量级的时候是相似的。所以就可以用矩阵乘法来加速。我构造的矩阵是这样的。
$\begin{bmatrix} 当前数字 & 累加数字 & 1\end{bmatrix} \times \begin{bmatrix} 数量级 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix} =\begin{bmatrix} 新的数字 & 累加数字+1 &1\end{bmatrix}$

CODE

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MO p
using namespace std;

long long k,p;

struct Matrix{
    long long num[5][5];
    int w,h;

    Matrix(int _,int __):w(_),h(__) {
        memset(num,0,sizeof(num));
    }
    Matrix() {
        memset(num,0,sizeof(num));
    }
    Matrix operator *(const Matrix &a)const {
        Matrix re(w,a.h);
        for(int i = 1; i <= w; ++i)
            for(int j = 1; j <= a.h; ++j)
                for(int k = 1; k <= h; ++k)
                    re.num[i][j] = (re.num[i][j] + num[i][k] * a.num[k][j] % MO) % MO;
        return re;
    }
};

inline Matrix QuickPower(Matrix a,long long k)
{
    Matrix re(3,3);
    re.num[1][1] = re.num[2][2] = re.num[3][3] = 1;
    while(k) {
        if(k&1) re = re * a;
        a = a * a;
        k >>= 1;
    }
    return re;
}

int main()
{
    cin >> k >> p;
    long long now = 10,ans_num = 0;
    while(k >= now - now / 10 && now < 1e18) {
        k -= now - now / 10;
        Matrix right(3,3);
        right.num[1][1] = now % MO;
        right.num[2][1] = right.num[2][2] = right.num[3][2] = right.num[3][3] = 1;
        right = QuickPower(right,now - now / 10);
        Matrix left(1,3);
        left.num[1][1] = ans_num;
        left.num[1][2] = (now / 10) % MO;
        left.num[1][3] = 1;
        ans_num = (left * right).num[1][1];
        now *= 10;
    }
    Matrix right(3,3);
    right.num[1][1] = now % MO;
    right.num[2][1] = right.num[2][2] = right.num[3][2] = right.num[3][3] = 1;
    right = QuickPower(right,k);
    Matrix left(1,3);
    left.num[1][1] = ans_num;
    left.num[1][2] = (now / 10) % MO;
    left.num[1][3] = 1;
    cout << (left * right).num[1][1] << endl;
    return 0;
}