题目
[egin{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})：
[egin{aligned}
(x_n&&x_{n-1}&&x_{n-2})=(x_{n-1}&&x_{n-2}&&x_{n-3})
left[
egin{matrix}
1 & 1 & 0\
1 & 0 & 1\
1 & 0 & 0\
end{matrix}
ight]
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):
[egin{aligned}
(x_n&&x_{n-1}&&x_{n-2}&&n&&1)=(x_{n-1}&&x_{n-2}&&x_{n-3}&&n-1&&1)
left[
egin{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}
ight]
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("*********
")
#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
", f1) * 0;
if(n == 2) return printf("%d
", f2) * 0;
if(n == 3) return printf("%d
", 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
", ans);
return 0;
}