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 100007ll 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)^rans -= tmp;elseans += tmp;ans %= Mod;}//ans = ans*(1/sqrt(5))^kll 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

转载于:https://www.cnblogs.com/whatbeg/p/3836551.html

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