Codechef Eugene and big number（矩阵快速幂）

题目链接 Eugene and big number

题目转化为

$f(n) = m * f(n - 1) + a$

$f(n + 1) = m * f(n) + a$

两式相减得

$f(n + 1) = (m + 1) * f(n) - m * f(n - 1)$

求$f(n)$

其中$m$为$10^{k}$ ($k$为$a$的位数)

那么利用矩阵快速幂加速就可以了。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)    for (int i(a); i <= (b); ++i)

typedef long long LL;

struct Matrix{ LL arr[5][5];}  init, unit, Unit;

int T;
LL a, n, mod;

inline LL Pow(LL a, LL b, LL Mod){
       LL ret(1); for (; b; b >>= 1, (a *= a) %= Mod) if (b & 1) (ret *= a) %= Mod; return ret;
}

Matrix Mul(Matrix a, Matrix b){
    Matrix c;
    rep(i, 1, 2) rep(j, 1, 2){
        c.arr[i][j] = 0;
        rep(k, 1, 2) (c.arr[i][j] += (a.arr[i][k] * b.arr[k][j] % mod)) %= mod;
    }
    return c;
}

Matrix MatrixPow(Matrix a, LL k){
    Matrix ret(Unit); for (; k; k >>= 1, a = Mul(a, a)) if (k & 1) ret = Mul(ret, a); return ret;
}    


inline LL calc(LL n){
    LL ret(0); for (; n; n /= 10) ++ret;
    return ret;
}

int main(){

    scanf("%d", &T);
    while (T--){
        Unit.arr[1][1] = Unit.arr[2][2] = 1;
        scanf("%lld%lld%lld", &a, &n, &mod);
        if (a == 0LL){
            puts("0");
            continue;
        }
        LL exp = calc(a), m = Pow(10, exp, mod);
        LL f1 = a % mod, f2 = ((f1 * m) % mod + a) % mod;
        LL f3 = ((f2 * m) % mod + a) % mod;
        if (n == 1LL){
            printf("%lld\n", f1 % mod);
            continue;
        }

        if (n == 2LL){
            printf("%lld\n", f2 % mod);
            continue;
        }

        if (n == 3LL){
            printf("%lld\n", f3 % mod);
            continue;
        }

        LL num = (2 * mod - m) % mod;

        init.arr[1][1] = f3, init.arr[1][2] = init.arr[2][1] = f2, init.arr[2][2] = f1;
        unit.arr[1][1] = m + 1, unit.arr[1][2] = num, unit.arr[2][1] = 1, unit.arr[2][2] = 0;
        Matrix mul = MatrixPow(unit, n - 3);
        Matrix ret = Mul(mul, init);
        printf("%lld\n", ret.arr[1][1]);
    }



    return 0;
}