Fibonacci Sum

$$\begin{gathered} 斐波那契通項有a_n=\frac{1}{\sqrt{5}}\left((\frac{1+\sqrt{5}}{2}{)}^n?(\frac{1?\sqrt{5}}{2}{)}^n\right) \\ (\frac{1}{\sqrt{5}}{)}^k\sum _{i=0}^n{\left((\frac{1+\sqrt{5}}{2}{)}^{ic}?(\frac{1?\sqrt{5}}{2}{)}^{ic}\right)}^k \\ A=\frac{1+\sqrt{5}}{2},B=\frac{1?\sqrt{5}}{2} \\ (\frac{1}{\sqrt{5}}{)}^k\sum _{i=0}^n\sum _{j=0}^k(?1{)}^{k?j}{C}_k^j{A}^{icj}{B}^{ic(k?j)} \\ (\frac{1}{\sqrt{5}}{)}^k\sum _{j=0}^k(?1{)}^{k?j}{C}_k^j\left(\sum _{i=0}^n{A}^{cj\times i}{B}^{c(k?j)\times i}\right) \end{gathered}$$

#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; }

總結

以上是生活随笔為你收集整理的HDU 6755 Fibonacci Sum（二次剩余 + 二项式展开）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。