2014 Super Training #7 F Power of Fibonacci --数学+逆元+快速幂

原题：ZOJ 3774  http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3774

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这题比较复杂，看这篇比较详细：http://blog.csdn.net/acdreamers/article/details/23039571

结论就是计算：

充分利用了快速幂及求逆元。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <cstdlib>

#include <algorithm>

#define Mod 1000000009

#define ll long long

using namespace std;

#define N 100007



ll fac[N],A[N],B[N];



void init()

{

    int i;

    fac[0] = 1;

    for(i=1;i<N;i++)

        fac[i] = fac[i-1]*i%Mod;

    A[0] = B[0] = 1;

    for(i=1;i<N;i++)

    {

        A[i] = A[i-1]*691504013 % Mod;

        B[i] = B[i-1]*308495997 % Mod;

    }

}



ll fastm(ll n,ll k,ll MOD)

{

    ll res = 1LL;

    n %= MOD;

    while(k)

    {

        if(k&1LL)

            res = (res*n)%MOD;

        k >>= 1;

        n = n*n%MOD;

    }

    return res;

}



ll Inv(ll n,ll MOD)

{

    return fastm(n,MOD-2,MOD);

}



int main()

{

    int cs;

    ll n,k;

    init();

    ll ans,r;

    scanf("%d",&cs);

    while(cs--)

    {

        scanf("%lld%lld",&n,&k);

        ans = 0;

        for(r=0;r<=k;r++)

        {

            ll t = A[k-r]*B[r] % Mod;

            ll x = fac[k];            // k!

            ll y = fac[k-r]*fac[r] % Mod;   // (k-r)!*(r)!

            ll c = x*Inv(y,Mod) % Mod;      // c = C(k,r) = x/y = x*Inv(y)

            ll tmp = t*(fastm(t,n,Mod)-1)%Mod*Inv(t-1,Mod)%Mod; //t(t^n-1)/(t-1) = t(t^n-1)*Inv(t-1)

            if(t == 1)

                tmp = n%Mod;

            tmp = tmp*c%Mod;

            if(r&1LL)   // (-1)^r

                ans -= tmp;

            else

                ans += tmp;

            ans %= Mod;

        }

        //ans = ans*(1/sqrt(5))^k

        ll m = Inv(383008016,Mod)%Mod;

        ans = ans*fastm(m,k,Mod)%Mod;

        ans = (ans%Mod+Mod)%Mod;

        printf("%lld\n",ans);

    }

    return 0;

}

View Code