51nod 1236 序列求和 V3
Fib(n)表示斐波那契数列的第n项,Fib(n) = Fib(n-1) + Fib(n-2)。Fib(0) = 0, Fib(1) = 1。
(1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...)
F(n, k) = Fib(n)^k(Fib(n)的k次幂)。S(n, k) = F(1, k) + F(2, k) + ...... F(n, k)。例如:S(4, 2) = 1^2 + 1^2 + 2^2 + 3^2 = 15。给出n和k,求S(n, k) Mod 1000000009的结果。输入
第1行:一个数T,表示后面用作输入测试的数的数量。(1 <= T <= 10) 第2 - T + 1行:每行2个数,N, K中间用空格分割。(1 <= N <= 10^18, 1 <= K <= 100000)输出
输出共T行,对应S(n, k) Mod 1000000009的结果。
$$
f(i)=\frac{1}{\sqrt 5}\left[\left(\frac{1+\sqrt 5}{2}\right)^i-\left(\frac{1-\sqrt 5}{2}\right)^i\right]
$$
设 $a=\frac{1+\sqrt 5}{2},b=\frac{1-\sqrt 5}{2}$
$$
\begin{aligned}
&\sum_{i=1}^{n}\left\{\frac{1}{\sqrt 5}\left[\left(\frac{1+\sqrt 5}{2}\right)^i-\left(\frac{1-\sqrt 5}{2}\right)^i\right]\right\}^k \\
=&\left(\frac{1}{\sqrt 5}\right)^k\sum_{i=1}^{n} \sum_{j=0}^{k} {k \choose j} a^{ij}b^{i(k-j)}(-1)^{k-j} \\
=&\left(\frac{1}{\sqrt 5}\right)^k \sum_{j=0}^{k} {k \choose j} (-1)^{k-j} \sum_{i=1}^{n}\left(a^{j}b^{k-j}\right)^i \\
\end{aligned}
$$
设 $q=a^jb^{k-j}$,则可以化为:
$$
\left(\frac{1}{\sqrt 5}\right)^k \sum_{j=0}^{k} {k \choose j} (-1)^{k-j} \frac{q^{n+1}-q}{q-1}
$$
同时有:$\sqrt 5 \equiv 616991993 \pmod {10^9+9}$
然后似乎就做完了啊
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 5 const int mod = 1e9 + 9; 6 7 const int s5 = 616991993; 8 9 ll pw(ll a, ll b) { 10 ll res = 1; 11 for( ; b ; b >>= 1, a = a * a % mod) 12 if(b & 1) 13 res = res * a % mod; 14 return res; 15 } 16 17 const int N = 1e5 + 10; 18 19 ll fac[N], invfac[N]; 20 21 ll C(int n, int m) { 22 return n < m ? 0 : fac[n] * invfac[m] % mod * invfac[n - m] % mod; 23 } 24 25 const ll a = (1 + s5) * pw(2, mod - 2) % mod; 26 const ll b = (1 - s5) * pw(2, mod - 2) % mod; 27 28 void sol() { 29 ll n, k; cin >> n >> k; 30 ll ans = 0; 31 for(int j = 0 ; j <= k ; ++ j) { 32 ll q = pw(a, j) * pw(b, k - j) % mod; 33 if(q == 1) { 34 ans += C(k, j) * pw(-1, k - j) % mod * (n % mod) % mod; 35 } else { 36 ans += C(k, j) * pw(-1, k - j) % mod * ((pw(q, n + 1) - q) % mod * pw(q - 1, mod - 2) % mod) % mod; 37 } 38 ans %= mod; 39 } 40 ans = ans * pw(pw(s5, mod - 2), k) % mod; 41 cout << (ans % mod + mod) % mod << endl; 42 } 43 44 int main() { 45 ios :: sync_with_stdio(0); 46 const int T = 1e5; 47 fac[0] = invfac[0] = 1; 48 for(int i = 1 ; i <= T ; ++ i) fac[i] = fac[i - 1] * i % mod; 49 invfac[T] = pw(fac[T], mod - 2); 50 for(int i = T - 1 ; i >= 1 ; -- i) invfac[i] = invfac[i + 1] * (i + 1) % mod; 51 int fafa; cin >> fafa; 52 while(fafa --) sol(); 53 }