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：/article/2368247.html

结论就是计算：

#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
,A
,B
;

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