2020杭电多校第一场 1005.Fibonacci Sum

Posted 2020-12-18 2462478392lee

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

题意：求一个式子，其中F是斐波那契数列

 

 

 思路：因为斐波那契数列的通式为

 

 

 所以式子可以化简为

 

 又根据2mod 1e9+9的逆元为500000005，所以x^2≡5(mod 1e9+9)解得x为383008016

所以a≡(1+sqrt(5))/2≡(1+383008016)*500000005≡691504013(mod 1e9+9)

同理b≡308495997(mod 1e9+9)。

详细的链接在这：https://blog.csdn.net/acdreamers/article/details/23039571

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=1e9+9;
const ll MAXN=1e5;
ll factor[MAXN+10];
ll invFactor[MAXN+10];
ll invn[MAXN+10];
const ll A=691504013;
const ll invA=691504012;
const ll B=308495997;
const ll D=276601605;
inline ll quick_pow(ll a, ll b)
{
    ll ans=1, base=a;
    while(b!=0)
    {
        if (b&1)
            ans=(ll) ans*base%mod;
        base=(ll) base*base%mod;
        b>>=1;
    }
    return ans;
}
inline ll MOD(ll a, ll b)
{
    a+=b;
    if (a>=mod)
        a-=mod;
    return a;
}
inline void init()
{
    factor[0]=invFactor[0]=invn[0]=factor[1]=invFactor[1]=invn[1]=1;
    for(int i=2; i<=MAXN; i++)
    {
        factor[i]=factor[i-1]*i%mod;
        invn[i]=(ll) (mod-mod/i)*invn[mod%i]%mod;
        invFactor[i]=invFactor[i-1]*invn[i]%mod;
    }
}
inline ll getC(ll m, ll n)
{
    if (n<0 || m<0 || m>n)
        return 0;
    ll ans=factor[n];
    ans=(ll) ans*invFactor[m]%mod;
    ans=(ll) ans*invFactor[n-m]%mod;
    return ans;
}
int main()
{
    init();
    int t;
    scanf("%d", &t);
    while(t--)
    {
        ll n, c, k;
        scanf("%lld%lld%lld", &n, &c, &k);
        ll ans=0;
        ll a1=quick_pow(quick_pow(A, k), c%(mod-1));
        ll q=quick_pow((ll) invA*B%mod, c%(mod-1));
        ll n1=n%mod;
        ll n2=n%(mod-1);
        ll a1power=quick_pow(a1, n2);
        ll qpower=quick_pow(q, n2);
        for(int i=0; i<=k; i++)
        {
            ll sum=getC(i, k);
            if (i&1)
            {
                sum=mod-sum;
            }
            if (a1==1)
            {
                ans=(ans+(ll) sum*n1%mod)%mod;
            }
            else
            {
                sum=(ll) sum*((ll) a1*(a1power-1+mod)%mod)%mod;
                sum=(ll) sum*quick_pow(a1-1, mod-2)%mod;
                ans=MOD(ans, sum);
            }
            a1=(ll)a1*q%mod;
            a1power=(ll)a1power*qpower%mod;
        }
        printf("%lld
",(ll)ans*quick_pow(D,k)%mod);
    }
}