标签：scan   air   int   set   const   ORC   欧拉   就是   stack   

传送门

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\(c^{x_1},f_1^{x_2},f_2^{x_3},f_3^{x_4}\)相乘得到，因此我们可以分别用矩阵快速幂求出\(x_1,x_2,x_3,x_4\)，最后用快速幂求得答案。
对\(f_1,f_2,f_3\)：
\[ \begin{aligned} (x_n&&x_{n-1}&&x_{n-2})=(x_{n-1}&&x_{n-2}&&x_{n-3}) \left[ \begin{matrix} 1 & 1 & 0\1 & 0 & 1\1 & 0 & 0\\end{matrix} \right] \end{aligned} \]
对\(c\):
\[ \begin{aligned} (x_n&&x_{n-1}&&x_{n-2}&&n&&1)=(x_{n-1}&&x_{n-2} && x_{n-3} && n-1 && 1) \left[ \begin{matrix} 1 & 1 & 0 & 0 & 0\1 & 0 & 1 & 0 & 0\1 & 0 & 0 & 0 & 0\2 & 0 & 0 & 1 & 0\2 & 0 & 0 & 1 & 1\\end{matrix} \right] \end{aligned} \]
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;
}

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

标签：scan   air   int   set   const   ORC   欧拉   就是   stack   

原文地址：https://www.cnblogs.com/Dillonh/p/11007112.html