Codeforces 1182E Product Oriented Recurrence 矩阵快速幂

Product Oriented Recurrence

先化简原式子

c ^ x * f[x]  = c ^ (x-1) * f[x-1] * c ^ (x-2) * f[x-2] * c ^ (x-3) * f[x-3]

及g[x] = c ^ x * f[x]

g[x] = g[x-1] * g[x-2] * g[x-3]

然后用矩阵快速幂计算g1, g2, g3的贡献， 计算出gn 之后 转回 fn

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

LL power(LL a, LL b) {
    LL ans = 1;
    while(b) {
        if(b & 1) ans = ans * a % mod;
        a = a * a % mod; b >>= 1;
    }
    return ans;
}

int MOD = (int)1e9 + 6;

struct Matrix {
    int a[3][3];
    Matrix() {
        memset(a, 0, sizeof(a));
    }
    void init() {
        for(int i = 0; i < 3; i++) {
            a[i][i] = 1;
        }
    }
    Matrix operator * (const Matrix &B) const {
        Matrix C;
        for(int i = 0; i < 3; i++) {
            for(int j = 0; j < 3; j++) {
                for(int k = 0; k < 3; k++) {
                    C.a[i][j] += 1LL * a[i][k] * B.a[k][j] % MOD;
                    if(C.a[i][j] >= MOD) C.a[i][j] -= MOD;
                }
            }
        }
        return C;
    }

    Matrix operator ^ (LL b) {
        Matrix C; C.init();
        Matrix A = (*this);
        while(b) {
            if(b & 1) C = C * A;
            A = A * A; b >>= 1;
        }
        return C;
    }
} M;

int mat[3][3] {
    {1, 1, 1},
    {1, 0, 0},
    {0, 1, 0}
};

LL n, f1, f2, f3, c;

int main() {
    for(int i = 0; i < 3; i++) {
        for(int j = 0; j < 3; j++) {
            M.a[i][j] = mat[i][j];
        }
    }

    scanf("%lld%lld%lld%lld%lld", &n, &f1, &f2, &f3, &c);

    Matrix ret = M ^ (n - 3);

    f1 = f1 * power(c, 1) % mod;
    f2 = f2 * power(c, 2) % mod;
    f3 = f3 * power(c, 3) % mod;

    LL ans = 1;

    LL cnt1 = ret.a[0][2];
    LL cnt2 = ret.a[0][1];
    LL cnt3 = ret.a[0][0];


    ans = ans * power(f1, cnt1) % mod;
    ans = ans * power(f2, cnt2) % mod;
    ans = ans * power(f3, cnt3) % mod;


    LL inv = power(power(c, n), mod - 2);

    ans = ans * inv % mod;

    printf("%lld
", ans);

    return 0;
}

/*
*/