【bzoj2326】[HNOI2011]数学作业

题目链接

Description

题解

[fnn1]=[fn−1n−11]×⎡⎣⎢10k11011001⎤⎦⎥

对于 0…9 ， 10…99 ， 100…999 ，…， 10t…n 分段计算。

#include
using namespace std;

#define rep(i, l, r) for(int i = (l); i <= (r); i++)
typedef long long ll;

ll n, m;

struct Mat{
    ll dat[5][5];
    Mat(){memset(dat, 0, sizeof(dat));}
    friend Mat operator * (const Mat a, const Mat b){
        Mat mul;
        rep(i, 1, 3) rep(j, 1, 3) rep(k, 1, 3)
            mul.dat[i][j] = (mul.dat[i][j] + a.dat[i][k] * b.dat[k][j]) % m;
        return mul;
    }
    friend Mat operator ^ (const Mat p, ll n){
        Mat q = p, ret;
        rep(i, 1, 3) ret.dat[i][i] = 1;
        while(n){
            if(n & 1) ret = ret * q;
            q = q * q;
            n >>= 1;
        }
        return ret;
    }
}a;

void init(){
    scanf("%lld%lld", &n, &m);
}

void cal(ll t, ll fin){
    Mat b;
    b.dat[1][1] = t % m;
    b.dat[2][1] = b.dat[2][2] = b.dat[3][1] = b.dat[3][2] = b.dat[3][3] = 1;
    b = b ^ (fin - t / 10 + 1);
    a = a * b;
}

void work(){
    ll t = 10;
    a.dat[1][3] = 1;
    while(t <= n){
        cal(t, t-1);
        t *= 10;
    }
    cal(t, n);
    printf("%lld\n", a.dat[1][1]);
}

int main(){
    init();
    work();
    return 0;
}