Fibonacci Sum
斐波那契通项有an=15((1+52)n−(1−52)n)(15)k∑i=0n((1+52)ic−(1−52)ic)kA=1+52,B=1−52(15)k∑i=0n∑j=0k(−1)k−jCkjAicjBic(k−j)(15)k∑j=0k(−1)k−jCkj(∑i=0nAcj×iBc(k−j)×i)斐波那契通项有a_n = \frac{1}{\sqrt 5}\left((\frac{1 + \sqrt 5}{2}) ^ n - (\frac{1 - \sqrt 5}{2}) ^ n\right)\\ (\frac{1}{\sqrt 5}) ^ k \sum_{i = 0} ^ {n} \left((\frac{1 + \sqrt 5}{2}) ^{ic} - (\frac{1 - \sqrt 5}{2}) ^{ic} \right) ^ k\\ A = \frac{1 + \sqrt 5}{2}, B = \frac{1 - \sqrt 5}{2}\\ (\frac{1}{\sqrt 5}) ^ k\sum_{i = 0} ^{n} \sum_{j = 0} ^{k}(-1) ^{k - j} C_k ^ j A ^{icj} B^{ic(k - j)}\\ (\frac{1}{\sqrt 5}) ^ k\sum_{j = 0} ^{k}(-1) ^{k - j} C_{k} ^{j} \left(\sum_{i = 0} ^{n} A^{cj \times i} B ^{c(k - j) \times i} \right)\\ 斐波那契通项有an=51((21+5)n−(21−5)n)(51)ki=0∑n((21+5)ic−(21−5)ic)kA=21+5,B=21−5(51)ki=0∑nj=0∑k(−1)k−jCkjAicjBic(k−j)(51)kj=0∑k(−1)k−jCkj(i=0∑nAcj×iBc(k−j)×i)
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int mod = 1e9 + 9, Mod = 1e9 + 8;ll n, c, k;const ll sqrt5 = 383008016, A = 691504013, B = 308495997;const int N = 1e6 + 10;ll quick_pow(ll a, ll n) {ll ans = 1;while (n) {if (n & 1) {ans = ans * a % mod;}a = a * a % mod;n >>= 1;}return ans;
}int fac[N], inv[N];int C(int n, int m) {if(n < 0 || m < 0 || m > n) return 0;if(m == 0 || m == n) return 1;return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod;
}void init() {fac[0] = 1;for(int i = 1; i < N; i++)fac[i] = 1ll * fac[i - 1] * i % mod;inv[N - 1] = quick_pow(fac[N - 1], mod - 2);for(int i = N - 2; i >= 0; i--)inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);const ll inv_sqrt5 = quick_pow(sqrt5, mod - 2);init();int T;scanf("%d", &T);while (T--) {scanf("%lld %lld %lld", &n, &c, &k);ll A_C = quick_pow(A, c % Mod), B_C = quick_pow(B, c % Mod);ll inv_B_C = quick_pow(B_C, mod - 2), base = quick_pow(B_C, k);ll multi = A_C * inv_B_C % mod, ans = 0;for (int i = 0; i <= k; i++){ll temp;if (base == 1){temp = n % mod;}else {temp = base * (quick_pow(base, n % Mod) + Mod) % mod * quick_pow(base - 1, mod - 2) % mod;}ll temp2 = C(k, i) * temp % mod;if ((k - i) & 1){ans = (ans - temp2 + mod) % mod;}else{ans = (ans + temp2) % mod;}base = base * multi % mod;}ans = ans * quick_pow(inv_sqrt5, k) % mod;printf("%lld\n", ans);}return 0;
}