Fibonacci Sum （HDU - 6755，斐波那契数列幂之和 + 亿点小优化）

一.题目链接

\quad Fibonacci Sum

二.题目大意

\quad T组数据，每次给出三个整数 N,C,K，求 ${\sum }_{i=0}^{N}fi{b}_{i\times C}^K(mod10^9+9)$.
\quad $1\le T\le 10^2,1\le N,C\le 10^{18},1\le K\le 10^5$.

三.分析

\quad 根据斐波那契数列通项公式 $fi{b}_n=\frac{1}{\sqrt{5}}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^{i\times C}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{i\times C}\right]$，可得下式

\quad $\begin{array}{rl}\sum _{i=0}^{N}fi{b}_{i\times C}^K& =\sum _{i=0}^N{\left\{\frac{1}{\sqrt{5}}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^{i\times C}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{i\times C}\right]\right\}}^K\\ & ={\left(\frac{1}{\sqrt{5}}\right)}^K\sum _{i=0}^N{\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^{i\times C}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{i\times C}\right]}^K\end{array}$

\quad 令 $p=10^9+9,a=\frac{1+\sqrt{5}}{2}(modp),b=\frac{1-\sqrt{5}}{2}(modp),c=\frac{1}{\sqrt{5}}(modp)$

\quad 使用小学知识二次剩余和费马小定理可得 $a=308495997,b=691504013,c=723398404$.

\quad 于是 $\begin{array}{r}\sum _{i=0}^{N}fi{b}_{i\times C}^K=c^K\sum _{i=0}^N{\left(a^{i\times C}-b^{i\times C}\right)}^K\end{array}$.

\quad 根据幼儿园学的二项式定理展开可得 $\begin{array}{r}\sum _{i=0}^{N}fi{b}_{i\times C}^K=c^K\sum _{i=0}^N\sum _{j=0}^K\left(\begin{array}{c}K\\ j\end{array}\right)(-1{)}^j{\left(b^{iC}\right)}^j{\left(a^{iC}\right)}^{K-j}\end{array}$

\quad 整理一下

\quad $\begin{array}{rl}\sum _{i=0}^{N}fi{b}_{i\times C}^K& =c^K\sum _{i=0}^N\sum _{j=0}^K\left(\begin{array}{c}K\\ j\end{array}\right)(-1{)}^j{\left(b^{iC}\right)}^j{\left(a^{iC}\right)}^{K-j}\\ & =c^K\sum _{j=0}^K\left(\begin{array}{c}K\\ j\end{array}\right)(-1{)}^j\sum _{i=0}^N{\left[b^{jC}{a}^{(K-j)C}\right]}^i\end{array}$

\quad 令 $t_j={\left[b^{jC}{a}^{(K-j)C}\right]}^i$，根据等比数列求和公式得

\quad $\begin{array}{rl}\sum _{i=0}^{N}fi{b}_{i\times C}^K& =\{\begin{array}{c}{c}^K{\sum }_{j=0}^K\left(\begin{array}{c}K\\ j\end{array}\right)(-1{)}^j(N+1),t_j=1\\ c^K{\sum }_{j=0}^K\left(\begin{array}{c}K\\ j\end{array}\right)(-1{)}^j\frac{t_j^{N+1}-1}{t_j-1},t_j\ne 1\end{array}\end{array}$

\quad 至此，这道题的大体做法就口胡完了.

\quad 这道题比较卡时间，因此我们可以预处理一些东西，同时维护某些量，总的时间复杂度为 $O(10^{5}log10^9+TKlog10^9)$.

四.代码实现

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int M = (int)1e5;
const int inf = 0x3f3f3f3f;
const ll mod = (ll)1e9 + 9;
const double eps = 1e-6;
const ll a = 308495997, b = 691504013, c = 723398404;

ll fac[M + 5], invfac[M + 5], ck[M + 5];
ll ack[M + 5], bck[M + 5];

ll quick(ll a, ll b, ll p = mod)
{
    
    
    ll s = 1;
    while(b)
    {
    
    
        if(b & 1) s = s * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return s;
}

ll inv(ll n, ll p = mod)
{
    
    
    return quick(n, p - 2, p);
}

ll C(int n, int m)
{
    
    
    return fac[n] * invfac[m] % mod * invfac[n - m] % mod;
}

void work()
{
    
    
    ll N, C, K; scanf("%lld %lld %lld", &N, &C, &K);
    ll ac = quick(a, C), bc = quick(b, C);
    ack[0] = 1, bck[0] = 1;
    for(int i = 1; i <= K; ++i)
    {
    
    
        ack[i] = ack[i - 1] * ac % mod;
        bck[i] = bck[i - 1] * bc % mod;
    }
    ll acn = quick(quick(a, C), N + 1), bcn = quick(quick(b, C), N + 1);
    ll invacn = inv(acn), invbcn = inv(bcn);
    ll nowacn = quick(acn, K), nowbcn = 1;
    ll s = 0, m = (N + 1) % mod, w, t;
    for(int i = 0; i <= K; ++i)
    {
    
    
        w = ::C(K, i); if(i & 1) w = -w % mod;
        t = ack[K - i] * bck[i] % mod;
        if(t == 1)  w = w * m % mod;
        else        w = w * (nowacn * nowbcn % mod - 1) % mod * inv(t - 1) % mod;
        s = (s + w) % mod;
        nowacn = nowacn * invacn % mod, nowbcn = nowbcn * bcn % mod;
    }
    s = s * ck[K] % mod;
    s = (s % mod + mod) % mod;
    printf("%lld\n", s);
}

int main()
{
    
    
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    fac[0] = invfac[0] = ck[0] = 1;
    for(int i = 1; i <= M; ++i)
    {
    
    
        fac[i] = fac[i - 1] * i % mod;
        invfac[i] = inv(fac[i]);
        ck[i] = ck[i - 1] * c % mod;
    }
    int T; scanf("%d", &T);
    while(T--) work();
    return 0;
}