2019 牛客多校第五场 B - generator 1

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题目链接：2019 牛客多校第五场 B

如果使用高精度每次除以2的操作是 \(O(10^6)\) 的，这里我们需要每次倍增规模让指数规模变小的操作是 \(O(1)\) 的，可以使用以10为单位倍增：

\[ A^b =\left\{  \begin{aligned}  &  [A^{\frac{b}{10}}] ^ {10} , & b \bmod 10 = 0 \\ &  [A^{\frac{b}{10}}] ^ {10} \times A^{b\bmod 10} ,  & b \bmod 10 \neq 0   \end{aligned} \right . \]

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#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAX_N = 1000000 + 5;
LL x[2], a, b, mod;
char n[MAX_N];
struct Matrix {
    LL v[2][2];
    Matrix(LL lu = 0, LL ru = 0, LL ld = 0, LL rd = 0) {
        v[0][0] = lu; v[0][1] = ru; v[1][0] = ld; v[1][1] = rd;
    }
    inline friend Matrix operator * (const Matrix &l, const Matrix &r) {
        Matrix ret;
        ret.v[0][0] = ((l.v[0][0] * r.v[0][0]) % mod + (l.v[0][1] * r.v[1][0]) % mod) % mod;
        ret.v[0][1] = ((l.v[0][0] * r.v[0][1]) % mod + (l.v[0][1] * r.v[1][1]) % mod) % mod;
        ret.v[1][0] = ((l.v[1][0] * r.v[0][0]) % mod + (l.v[1][1] * r.v[1][0]) % mod) % mod;
        ret.v[1][1] = ((l.v[1][0] * r.v[0][1]) % mod + (l.v[1][1] * r.v[1][1]) % mod) % mod;
        return ret;
    }
};
inline Matrix getPowStepTwo(Matrix o, LL y) {
    Matrix ret = Matrix(1, 0, 0, 1);
    while (y) {
        if (y & 1) ret = ret * o;
        o = o * o;
        y >>= 1;
    }
    return ret;
}
Matrix getPowStepTen(Matrix o, char *s) {
    int len = strlen(s);
    Matrix tmp;
    Matrix ret = Matrix(1, 0, 0, 1);
    for (int i = len - 1; i >= 0; --i) {
        int x = s[i] - '0';
        if (x) ret = ret * getPowStepTwo(o, x);
        // o = getPowStepTwo(o, 10);
        tmp = o * o;
        tmp = tmp * tmp;
        tmp = tmp * tmp;
        o = o * o * tmp;
    }
    return ret;
}
int main() {
    scanf("%lld%lld%lld%lld", x, x + 1, &a, &b);
    scanf("%s%lld", n, &mod);
    Matrix ori = Matrix(a, b, 1, 0);
    Matrix ans = getPowStepTen(ori, n);
    LL out = ((ans.v[1][0] * x[1]) % mod + (ans.v[1][1] * x[0]) % mod) % mod;
    printf("%lld\n", out);
    return 0;
}