2020 Multi-University Training Contest 1 1005 ( HDU 6755 ) Fibonacci Sum

Posted 2020-12-18 zeronera

(Fibonacci)数列的通项公式：

[F_n=frac{1}{sqrt5}igg[(frac{1+sqrt5}{2})^n-(frac{1-sqrt5}{2})^nigg] ]

令：

[a=frac{1+sqrt5}{2} ]

[b=frac{1-sqrt5}{2} ]

则：

[F_n=frac{1}{sqrt5}ig(a^n-b^nig) ]

[F_n^k=igg(frac{1}{sqrt5}igg)^kig(a^n-b^nig)^k ]

[F_n^k=igg(frac{1}{sqrt5}igg)^kdisplaystylesum_{r=0}^{k}ig(-1ig)^rC_k^rig(a^{k-r}b^{r}ig)^n ]

根据题意：

[F_c^k=igg(frac{1}{sqrt5}igg)^kdisplaystylesum_{r=0}^{k}ig(-1ig)^rC_k^rig(a^{k-r}b^{r}ig)^c ]

[F_{2c}^k=igg(frac{1}{sqrt5}igg)^kdisplaystylesum_{r=0}^{k}ig(-1ig)^rC_k^rig(a^{k-r}b^{r}ig)^{2c} ]

[... ]

[F_{nc}^k=igg(frac{1}{sqrt5}igg)^kdisplaystylesum_{r=0}^{k}ig(-1ig)^rC_k^rig(a^{k-r}b^{r}ig)^{nc} ]

令：

[t_r=(a^{k-r}b^r)^c ]

则：

[F_c^k+F_{2c}^k+...+F_{nc}^k=igg(frac{1}{sqrt5}igg)^kdisplaystylesum_{r=0}^{k}ig(-1ig)^rC_k^rfrac{t_r(t_r^n-1)}{t_r-1} ]

最后处理根式，在本题中(5)是模(1e9+9)的二次剩余，所以有(x^2equiv5(mod)?(1000000009))，解得(x=383008016)或(616991993)，任取一解代替(sqrt5)
若取(x=383008016)，则有：

[a=frac{1+sqrt5}{2}equiv(1+sqrt5)*inv(2)equiv(1+383008016)*500000005equiv691504013(mod1000000009) ]

[b=frac{1-sqrt5}{2}equiv(1-sqrt5)*inv(2)equiv(1-383008016)*500000005equiv308495997(mod1000000009) ]

特判(t_r=1)的情况

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1000000009;
const int INVS5 = 276601605; // 1/sqrt(5)
const int A = 691504013;
const int B = 308495997;
const int maxn = 100010;
int base, last;

int qpow(int x, int y)
{
    int ans = 1;
    while (y)
    {
        if (y & 1)
            ans = (ll)x * ans % MOD;
        x = (ll)x * x % MOD;
        y >>= 1;
    }
    return ans;
}

int fac[maxn], invf[maxn], a[maxn], b[maxn];

void init(int n)
{
    fac[0] = a[0] = b[0] = 1;

    for (int i = 1; i <= n; i++)
    {
        fac[i] = (ll)fac[i - 1] * i % MOD;
        a[i] = (ll)a[i - 1] * A % MOD;
        b[i] = (ll)b[i - 1] * B % MOD;
    }
    invf[n] = qpow(fac[n], MOD - 2);
    for (int i = n - 1; i >= 0; i--)
        invf[i] = (ll)invf[i + 1] * (i + 1) % MOD;
}

inline int C(int n, int m){ // n >= m >= 0
    return n < m || m < 0 ? 0 : (ll)fac[n] * invf[m] % MOD * invf[n - m] % MOD;
}
int solve(ll n, ll c, int k)
{
    int ans = 0;
    for (int i = 0; i <= k; i++)
    {
        int tr=last,q;
        last = (ll)last * base % MOD;
        if (tr == 1)
            q = n % MOD;
        else
            q = (ll)tr * (qpow(tr, n % (MOD - 1)) - 1) % MOD * qpow(tr - 1, MOD - 2) % MOD;
        int res = (ll)C(k, i) * q % MOD;
        if (i & 1)
            ans -= res;
        else
            ans += res;
        ans %= MOD;
    }
    ans = (ll)ans * qpow(INVS5, k) % MOD;
    ans = (ans + MOD) % MOD;
    return ans;
}

int t, k;
ll n, c;
int main()
{
    init(100000);
    scanf("%d", &t);
    while (t--)
    {
        scanf("%lld%lld%d", &n, &c, &k);
        base = (ll)qpow(B, c % (MOD - 1)) * qpow(qpow(A, c % (MOD - 1)), MOD - 2) % MOD;
        last = qpow(a[k], c % (MOD - 1));
        printf("%d
", solve(n, c, k));
    }
    return 0;
}

(refer to:) https://blog.csdn.net/acdreamers/article/details/23039571