Think:
1知識點：費馬小定理降冪+矩陣快速冪+快速冪
（１）：費馬小定理降冪：
定理：若gcd(A, M) == 1，則A^x = A^(x%Eular(M))(mod M)
注：Eular(M)為M的尤拉函式值
2題意：
已知：
F[0] = a
F[1] = b
F[n] = F[n-1] * F[n-2] ( n > 1 )
輸入a, b, n，詢問F[n] = ?
3解題方法：
（１）：
F[0] = a
F[1] = b
F[2] = b*a
F[3] = b^(2)*a
F[4] = b^(3)*a^(2)
F[5] = b^(5)*a^(3)
.
.
.
F[n] = b^(fib(n-1))*a^(fib(n-2))
（２）：由於快速冪指數太高，因此需要通過費馬小定理降冪

以下為Accepted程式碼

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int mod = 1000000007;

struct Matrix{
    LL v[4][4];
};

Matrix multiply(const Matrix &a, const Matrix &b, int Matrix_len);
Matrix matrix_pow(Matrix &x, int
 
 k, int Matrix_len);
LL num_pow(LL num, LL k);

int main(){
    LL a, b;
    int n;
    while(~scanf("%lld %lld %d", &a, &b, &n)){
        if(n == 0){
            printf("%lld\n", a%mod);
            continue;
        }
        if(n == 1){
            printf("%lld\n", b%mod);
            continue
 
;
        }
        Matrix x;
        x.v[0][0] = 1, x.v[0][1] = 1;
        x.v[1][0] = 1, x.v[1][1] = 0;
        Matrix y = matrix_pow(x, n-2, 2);

        LL t1 = (y.v[0][0] + y.v[0][1])%(mod-1);
        LL t2 = (y.v[1][0] + y.v[1][1])%(mod-1);

        LL ans1 = num_pow(b, t1);
        LL ans2 = num_pow(a, t2);
        LL ans = ans1*ans2%mod;
        printf("%lld\n", ans);
    }
    return 0;
}
Matrix multiply(const Matrix &a, const Matrix &b, int Matrix_len){
    Matrix tmp;
    memset(tmp.v, 0, sizeof(tmp.v));
    for(int i = 0; i < Matrix_len; i++){
        for(int j = 0; j < Matrix_len; j++){
            for(int k = 0; k < Matrix_len; k++){
                tmp.v[i][j] += (a.v[i][k]*b.v[k][j]);
                tmp.v[i][j] %= (mod-1);
            }
        }
    }
    return tmp;
}
Matrix matrix_pow(Matrix &x, int k, int Matrix_len){
    Matrix tmp;
    memset(tmp.v, 0, sizeof(tmp.v));
    for(int i = 0; i < Matrix_len; i++)
        tmp.v[i][i] = 1;
    while(k){
        if(k & 1)
            tmp = multiply(tmp, x, Matrix_len);
        x = multiply(x, x, Matrix_len);
        k >>= 1;
    }
    return tmp;
}
LL num_pow(LL num, LL k){
    LL ans = 1;
    while(k){
        if(k & 1)
            ans = ans*num%mod;
        num = num*num%mod;
        k >>= 1;
    }
    return ans;
}