題意： M斐波那契數列F[n]是一種整數數列，它的定義如下：

F[0] = a F[1] = b F[n] = F[n-1] * F[n-2] ( n > 1 )

現在給出a, b, n，你能求出F[n]的值嗎？ 分析： f[0]=a; f[1]=b; f[2]=a * b; f[3]=a * b^2; f[4]=a^2 * b^3; f[5]=a^3 * b^5; … f[n]=a^fib(n-1) * b^fib(n); 得到這個式子後，可以用杜教bm求出fib數列的第n項和第n-1項，然後尤拉降冪一下快速冪求一下就完事了； 尤拉降冪： https://blog.csdn.net/hjsss3/article/details/81562792

程式碼：

#include <cstdio> 
#include<iostream> 
#include <cstring> 
#include <cmath> 
#include <algorithm> 
#include<vector>
#include<assert.h>
 
using namespace std;
#define rep(i,a,n) for (long long i=a;i<n;i++)
#define per(i,a,n) for (long long i=n-1;i>=a;i--)
 

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((long long)(x).size())
typedef vector<long long> VI;
typedef long long ll;
typedef pair<long long,long long> PII;
ll mod=1e9+7;
ll powmod(ll a,ll b) {ll res=1
 
;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
// head
 
long long _,n;
namespace linear_seq
{
    const long long N=10010;
    ll res[N],base[N],_c[N],_md[N];
 
    vector<long long> Md;
    void mul(ll *a,ll *b,long long k)
    {
        rep(i,0,k+k) _c[i]=0;
        rep(i,0,k) if (a[i]) rep(j,0,k)
            _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (long long i=k+k-1;i>=k;i--) if (_c[i])
            rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        rep(i,0,k) a[i]=_c[i];
    }
    long long solve(ll n,VI a,VI b)
    { // a 係數 b 初值 b[n+1]=a[0]*b[n]+...
//        printf("%d\n",SZ(b));
        ll ans=0,pnt=0;
        long long k=SZ(a);
        assert(SZ(a)==SZ(b));
        rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
        Md.clear();
        rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
        rep(i,0,k) res[i]=base[i]=0;
        res[0]=1;
        while ((1ll<<pnt)<=n) pnt++;
        for (long long p=pnt;p>=0;p--)
        {
            mul(res,res,k);
            if ((n>>p)&1)
            {
                for (long long i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
                rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
    VI BM(VI s)
    {
        VI C(1,1),B(1,1);
        long long L=0,m=1,b=1;
        rep(n,0,SZ(s))
        {
            ll d=0;
            rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
            if (d==0) ++m;
            else if (2*L<=n)
            {
                VI T=C;
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                L=n+1-L; B=T; b=d; m=1;
            }
            else
            {
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                ++m;
            }
        }
        return C;
    }
    long long gao(VI a,ll n)
    {
        VI c=BM(a);
        c.erase(c.begin());
        rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
    }
};
 ll phi(ll n)
{
     ll i,rea=n;
     for(i=2;i*i<=n;i++)
     {
         if(n%i==0)
         {
             rea=rea-rea/i;
             while(n%i==0)
                 n/=i;
          }
     }
     if(n>1)
         rea=rea-rea/n;
     return rea;
}
ll ksm(ll a,ll b,ll p){
	ll ret=1;
	while(b){
		if(b&1){
			ret=ret*a%p;
		}
		b>>=1;
		a=a*a%p;
	}
	return ret;
}
ll a,b;
VI f;
int fr[1010];
int main()
{
	ll x,y,n,m,ans,i;
	while(scanf("%lld%lld%lld",&a,&b,&n)!=EOF){
		if(n==0){
			cout<<a<<endl;
			continue;
		}
		if(n==1){
			cout<<b<<endl;
			continue;
		}
		mod=phi(1e9+7);
		f.clear();
		fr[1]=1;
		fr[2]=1;
		f.push_back(fr[1]);
		f.push_back(fr[2]);
		for(int i=3;i<=11;i++){
			fr[i]=fr[i-1]+fr[i-2]%mod;
			f.push_back(fr[i]);
		}
		ll p1=linear_seq::gao(f,n-2);
		ll p2=linear_seq::gao(f,n-1);
		ll k1=ksm(a,p1+mod,1e9+7);
		ll k2=ksm(b,p2+mod,1e9+7);
		ll pp=1e9+7;
		ll ans=k1*k2%(pp);
		printf("%I64d\n",ans);
	}
}