To simplify the original formula
c ^ x * f[x] = c ^ (x-1) * f[x-1] * c ^ (x-2) * f[x-2] * c ^ (x-3) * f[x-3]
And g [x] = c ^ x * f [x]
g[x] = g[x-1] * g[x-2] * g[x-3]
Then calculate the power matrix fast g1, g2, g3 contribution, calculated back after the fn gn
//#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 ]; years = years * Power (f1, CNT1)% mod; years = years * power (f2, cnt2)% mod; years = years * Power (f3, CNT 3)% mod; Inv LL = power (power (c, n) mod - 2 ); years = years * inv% mod; printf ( " % lld \ n " , year); return 0 ; } / * * /