2020杭电多校 Fibonacci Sum 数论

链接

http://acm.hdu.edu.cn/showproblem.php?pid=6755

题意

求和
其中

题解

根据公式

首先要知道通项公式且能将其转化为mod意义下整数（解二次同余方程）。
得出：
A = 691504013（（1+根号5）/2）, B = 308495997（（1-根号5）/2）;
sqrt5 = 383008016（根号5）, invsqrt5 = 276601605（根号5分之一）
将每一项展开，观察到纵向为一个n+1项的等比数列，等比数列求和可以有优化掉些快速幂，

#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define ll long long
const ll mod = 1e9+9;
const int maxn = 1e5+5;
// 逆元
ll A = 691504013, B = 308495997;
ll sqrt5 = 383008016, invsqrt5 = 276601605;

ll fac[maxn], f[maxn], inv[maxn];
//快速幂模板
ll qpow(ll a, ll n)
{
    ll res = a%mod, sum = 1;
    while(n)
    {
        if(n&1) sum = (sum*res)%mod;
        res = (res*res)%mod;
        n >>= 1;
    }
    return sum;
}
//数据预处理
void init()
{
    fac[1] = 1;
    f[1] = 1;
    inv[1] = 1;
    for(ll i = 2; i <= 100000; i++)
    {
        fac[i] = fac[i-1]*i%mod;//计算阶乘
        f[i] = (mod - mod/i)*f[mod%i]%mod;
        inv[i] = inv[i-1]*f[i]%mod;//计算阶乘逆元
    }
}
//计算组合数 C（m,n）=m!/n!(m-n)!
ll C(ll m, ll n)
{
    if(m < n) return 0;
    if(n == 0 || m == n) return 1;
    return fac[m]*inv[n]%mod*inv[m-n]%mod;
}
ll sac[maxn], sbc[maxn];
int main()
{
    init();
    int t;
    scanf("%d", &t);
    while(t--)
    {
        ll n, c, k;
        scanf("%lld%lld%lld", &n,&c,&k);
        // n = 1e18, c = 1e18, k = 1e5;
        ll res = qpow(invsqrt5, k);
        ll ans = 0, flag = -1;
        ll x, y;
        ll ac = qpow(A, c), bc = qpow(B, c);
        sac[0] = 1, sbc[0] = 1;
        sac[1] = ac, sbc[1] = bc;
        for(int i=1; i<=k; i++)
        {
            sac[i] = sac[i-1]*ac%mod;
            sbc[i] = sbc[i-1]*bc%mod;
        }
        ll acn = qpow(qpow(A, c), n);
        ll bcn = qpow(qpow(B, c), n);
        ll invacn = qpow(acn, mod-2);
        ll invbcn = qpow(bcn, mod-2);
        ll now_acn = qpow(acn, k), now_bcn = 1;
        for(ll i = 0; i <= k; i++)
        {
            flag *= -1;
            ll q = sac[k-i] * sbc[i] % mod;
            if(q == 1)
            {
                q = (n%mod)*C(k, i) % mod;
                ans = (ans + flag*q%mod + mod) % mod;
            }
            else
            {
                x = (C(k, i)*sac[k-i]%mod) * sbc[i] % mod;
                x = (1-(now_acn * now_bcn % mod) + mod) % mod * x % mod;
                y = (1 - (sac[k-i] * sbc[i])%mod ) % mod;
                y = qpow(y, mod-2);
                ans = (ans + flag*x*y%mod + mod)%mod;
            }
            now_acn = (now_acn * invacn) % mod;
            now_bcn = (now_bcn * bcn) % mod;
        }
        printf("%lld\n",ans*res%mod);
    }
    return 0;
}