题目

\[\begin{aligned}
&f_n=c^{2*n-6}f_{n-1}f_{n-2}f_{n-3}&\\
\end{aligned}
\]

思路

我们通过迭代发现\(f_n\)其实就是由\(c^{t_1},f_1^{t_2},f_2^{t_3},f_3^{t_4}\)相乘得到，因此我们可以分别用矩阵快速幂求出\(t_1,t_2,t_3,t_4\)，最后用快速幂求得答案。

对于\(n<=3\)的我们直接输出即可，\(n>3\)的我们先将\(n\)减去\(3\)，然后进行求解。

对\(f_1,f_2,f_3\)的指数，我们可以推出\(x_n=x_{n-1}+x_{n-2}+x_{n-3}\)：

\[\begin{aligned}
(x_n&&x_{n-1}&&x_{n-2})=(x_{n-1}&&x_{n-2}&&x_{n-3})
\left[
\begin{matrix}
1 & 1 & 0\\
1 & 0 & 1\\
1 & 0 & 0\\
\end{matrix}
\right]
\end{aligned}
\]

对\(c\)的指数，我们可以推出\(x_n=x_{n-1}+x_{n-2}+x_{n-3}+2n=x_{n-1}+x_{n-2}+x_{n-3}+2(n-1)+2\):

\[\begin{aligned}
(x_n&&x_{n-1}&&x_{n-2}&&n&&1)=(x_{n-1}&&x_{n-2}&&x_{n-3}&&n-1&&1)
\left[
\begin{matrix}
1 & 1 & 0 & 0 & 0\\
1 & 0 & 1 & 0 & 0\\
1 & 0 & 0 & 0 & 0\\
2 & 0 & 0 & 1 & 0\\
2 & 0 & 0 & 1 & 1\\
\end{matrix}
\right]
\end{aligned}
\]

注意，由于我们处理出来的\(x_1,x_2,x_3,x_4\)都是指数部分，这里如果膜\(1e9+7\)的话是不对的，我们还需要对其进行欧拉降幂。

代码实现

#include <set>

#include <map>

#include <deque>

#include <queue>

#include <stack>

#include <cmath>

#include <ctime>

#include <bitset>

#include <cstdio>

#include <string>

#include <vector>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

typedef long long LL;

typedef pair<LL, LL> pLL;

typedef pair<LL, int> pLi;

typedef pair<int, LL> pil;;

typedef pair<int, int> pii;

typedef unsigned long long uLL;

#define lson rt<<1

#define rson rt<<1|1

#define lowbit(x) x&(-x)

#define name2str(name) (#name)

#define bug printf("*********\n")

#define debug(x) cout<<#x"=["<<x<<"]" <<endl

#define FIN freopen("D://code//in.txt","r",stdin)

#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;

const int mod = 1000000007;

const int maxn = 2e5 + 7;

const double pi = acos(-1);

const int inf = 0x3f3f3f3f;

const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int f[10], a[10][10];

void mulself(int a[10][10]) {

    int c[10][10];

    memset(c, 0, sizeof(c));

    for(int i = 0; i < 3; i++) {

        for(int j = 0; j < 3; j++) {

            for(int k = 0; k < 3; k++) {

                c[i][j] = (c[i][j] + (long long)a[i][k] * a[k][j] % (mod - 1)) % (mod - 1);

            }

        }

    }

    memcpy(a, c, sizeof(c));

}

void mul(int f[10], int a[10][10]) {

    int c[10];

    memset(c, 0, sizeof(c));

    for(int i = 0; i < 3; i++) {

        for(int j = 0; j < 3; j++) {

            c[i] = (c[i] + (long long)f[j] * a[j][i] % (mod - 1)) % (mod - 1);

        }

    }

    memcpy(f, c, sizeof(c));

}

void mulself1(int a[10][10]) {

    int c[10][10];

    memset(c, 0, sizeof(c));

    for(int i = 0; i < 5; i++) {

        for(int j = 0; j < 5; j++) {

            for(int k = 0; k < 5; k++) {

                c[i][j] = (c[i][j] + (long long)a[i][k] * a[k][j] % (mod - 1)) % (mod - 1);

            }

        }

    }

    memcpy(a, c, sizeof(c));

}

void mul1(int f[10], int a[10][10]) {

    int c[10];

    memset(c, 0, sizeof(c));

    for(int i = 0; i < 5; i++) {

        for(int j = 0; j < 5; j++) {

            c[i] = (c[i] + (long long)f[j] * a[j][i] % (mod - 1)) % (mod - 1);

        }

    }

    memcpy(f, c, sizeof(c));

}

int qpow(int x, int n) {

    int res = 1;

    while(n) {

        if(n & 1) res = 1LL * res * x % mod;

        x = 1LL * x * x % mod;

        n >>= 1;

    }

    return res;

}

LL n;

int f1, f2, f3, c;

int main(){

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

    if(n == 1) return printf("%d\n", f1) * 0;

    if(n == 2) return printf("%d\n", f2) * 0;

    if(n == 3) return printf("%d\n", f3) * 0;

    n -= 3;

    LL ans = 1;

    f[0] = 1, f[1] = 0, f[2] = 0;

    a[0][0] = 1, a[0][1] = 1, a[0][2] = 0;

    a[1][0] = 1, a[1][1] = 0, a[1][2] = 1;

    a[2][0] = 1, a[2][1] = 0, a[2][2] = 0;

    LL x = n;

    while(x) {

        if(x & 1) mul(f, a);

        mulself(a);

        x >>= 1;

    }

    ans = ans * qpow(f3, f[0]) % mod;

    f[0] = 0, f[1] = 1, f[2] = 0;

    a[0][0] = 1, a[0][1] = 1, a[0][2] = 0;

    a[1][0] = 1, a[1][1] = 0, a[1][2] = 1;

    a[2][0] = 1, a[2][1] = 0, a[2][2] = 0;

    x = n;

    while(x) {

        if(x & 1) mul(f, a);

        mulself(a);

        x >>= 1;

    }

    ans = ans * qpow(f2, f[0]) % mod;

    f[0] = 0, f[1] = 0, f[2] = 1;

    a[0][0] = 1, a[0][1] = 1, a[0][2] = 0;

    a[1][0] = 1, a[1][1] = 0, a[1][2] = 1;

    a[2][0] = 1, a[2][1] = 0, a[2][2] = 0;

    x = n;

    while(x) {

        if(x & 1) mul(f, a);

        mulself(a);

        x >>= 1;

    }

    ans = ans * qpow(f1, f[0]) % mod;

    if(n == 1) f[0] = 2;

    if(n == 2) f[0] = 6;

    if(n == 3) f[0] = 14;

    if(n > 3) {

        n -= 3;

        f[0] = 14, f[1] = 6, f[2] = 2, f[3] = 3, f[4] = 1;

        memset(a, 0, sizeof(a));

        a[0][0] = a[0][1] = 1;

        a[1][0] = a[1][2] = 1;

        a[2][0] = 1;

        a[3][0] = 2, a[3][3] = 1;

        a[4][0] = 2, a[4][3] = a[4][4] = 1;

        while(n) {

            if(n & 1) mul1(f, a);

            mulself1(a);

            n >>= 1;

        }

    }

    ans = ans * qpow(c, f[0]) % mod;

    printf("%lld\n", ans);

    return 0;

}

巴特西

Product Oriented Recurrence(Codeforces Round #566 (Div. 2)E+矩阵快速幂+欧拉降幂)

题目

思路

代码实现

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