Codeforces Round #566 (Div. 2)E. Product Oriented Recurrence【矩阵】

最新推荐文章于 2025-03-09 23:17:34 发布

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最新推荐文章于 2025-03-09 23:17:34 发布

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本文链接：https://blog.youkuaiyun.com/qq_42671946/article/details/91488692

本文介绍了一种使用矩阵快速幂求解特定递归序列第n项的高效算法，适用于形如fn=c2n-6*fn-1*fn-2*fn-3的递归公式。通过构造矩阵并利用快速幂算法，可以显著减少计算复杂度，从而在大数运算中快速求得结果。

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E. Product Oriented Recurrence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Let fx=c2x−6⋅fx−1⋅fx−2⋅fx−3fx=c2x−6⋅fx−1⋅fx−2⋅fx−3 for x≥4x≥4.

You have given integers nn, f1f1, f2f2, f3f3, and cc. Find fnmod(109+7)fnmod(109+7).

Input

The only line contains five integers nn, f1f1, f2f2, f3f3, and cc (4≤n≤10184≤n≤1018, 1≤f11≤f1, f2f2, f3f3, c≤109c≤109).

Output

Print fnmod(109+7)fnmod(109+7).

Examples

input

Copy

5 1 2 5 3

output

Copy

input

Copy

17 97 41 37 11

output

Copy

317451037

Note

In the first example, f4=90f4=90, f5=72900f5=72900.

In the second example, f17≈2.28×1029587f17≈2.28×1029587.

分析：

我们可以将通项写成如下形式：

$f[x]=c^{C[x]}*f[1]^{F1[x]}*f[2]^{F2[x]}*f[3]^{F3[x]}$

其中：

$F1[x]=F1[x-1]+F1[x-2]+F1[x-3]$

$F2[x]=F2[x-1]+F2[x-2]+F2[x-3]$

$F3[x]=F3[x-1]+F3[x-2]+F3[x-3]$

$C[x]=C[x-1]+C[x-2]+C[x-3]+2*x-6$

构造3维矩阵

$\left[ \begin{matrix} F[i+1] \\ F[i] \\ F[i-1] \end{matrix} \right] \tag{3} =\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right] \tag{3} \left[ \begin{matrix} F[i] \\ F[i-1] \\ F[i-2] \end{matrix} \right] \tag{3}$

得到F1，F2，F3。

将C[x]转化一下

$C[x]=C[x-1]+C[x-2]+C[x-3]+G[x]$

$G[x]=2x-6=G[x-1]+2$

构造矩阵

$\left[ \begin{matrix}C[i+1]\\ C[i]\\ C[i-1]\\G[i+2]\\ 2 \end{matrix} \right] \tag{3} =\left[ \begin{matrix} 1 & 1 & 1 & 1& 0\\ 1 & 0& 0& 0& 0\\ 0& 1 & 0& 0& 0\\ 0& 0& 0& 1& 1\\ 0& 0& 0& 0& 1 \end{matrix} \right] \tag{3} \left[ \begin{matrix}C[i]\\ C[i-1]\\ C[i-2]\\G[i+1]\\ 2 \end{matrix} \right] \tag{3}$

得到C。

然后欧拉降幂后直接计算即可。

#include "bits/stdc++.h"

using namespace std;
typedef long long LL;
const int maxn = 1e6 + 5;
const int MOD = 1e9 + 6;
const int mod = 1e9 + 7;
using namespace std;

long long qk(long long a, long long n) {
    long long ans = 1;
    while (n) {
        if (n & 1)ans = ans * a % mod;
        n >>= 1;
        a = a * a % mod;
    }
    return ans;
}

struct Matrix {
    int n;
    LL a[5][5];

    Matrix(int n) : n(n) {
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)a[i][j] = 0;
    }

};

Matrix operator*(Matrix A, Matrix B) {
    int n = A.n;
    Matrix res(n);
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            for (int k = 0; k < n; k++) {
                res.a[i][j] += (LL) A.a[i][k] * B.a[k][j] % MOD;
                res.a[i][j] %= MOD;
            }
    return res;
}

Matrix operator^(Matrix A, long long b) {
    int n = A.n;
    Matrix res(n), base = A;
    for (int i = 0; i < n; i++)res.a[i][i] = 1;
    while (b) {
        if (b & 1)res = res * base;
        base = base * base;
        b >>= 1;
    }
    return res;
}

Matrix a(3), b(5);

int main() {
    long long n, f1, f2, f3, c;
    cin >> n >> f1 >> f2 >> f3 >> c;
    a.a[0][0] = 1;
    a.a[0][1] = 1;
    a.a[0][2] = 1;
    a.a[1][0] = 1;
    a.a[2][1] = 1;
    b.a[0][0] = 1;
    b.a[0][1] = 1;
    b.a[0][2] = 1;
    b.a[0][3] = 1;
    b.a[1][0] = 1;
    b.a[2][1] = 1;
    b.a[3][3] = 1;
    b.a[3][4] = 1;
    b.a[4][4] = 1;
    a = a ^ (n - 3);
    b = b ^ (n - 3);
    long long ans = 1;
    long long F1 = (a.a[0][2] ) % MOD;
    long long F2 = ( a.a[0][1] ) % MOD;
    long long F3 = (a.a[0][0] ) % MOD;
    ans = qk(f1, F1) * qk(f2, F2) % mod * qk(f3, F3) % mod;
    long long C = (b.a[0][3]*2 + b.a[0][4] * 2) % MOD;
    ans = ans * qk(c, C) % mod;
    cout << ans << endl;

}