f(n)表示斐波那契數列的第n項

F(n) = a^f(n-1)*b^f(n)  這個推一下就知道了

然後f(n)可以用矩陣快速冪求

1 1                         f(n+1)    f(n)

1 0  的n次冪   =   f(n)         f(n-1)   就行了（至於為什麼有這個自己度娘把。。）

 但關鍵點在於f(n)可能很大 a^f(n-1)不好求

但是我們有費馬小定理    a^p-1 = 1(mod p) p是質數  1000000007是質數 所以a^1000000006=1 (mod 1000000007)

所以我們只需要求f(n-1)%1000000006的值就行了   然後答案就是a^(f(n-1)%1000000006)*b^((f(n)%1000000006))%1000000007

附一個Java的程式碼= =   最近在寫Java專案，一直開的eclipse  沒寫C++的  不過思路是一樣的。。。

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {

    private static InputReader in = new InputReader();

    private static final int MOD = 1000000007;
    private static final long[][] gen = { { 1, 1 }, { 1, 0 } };

    private static long pow(long n, long a) {
        if (a == 0)
            return 1;
        if (a == 1)
            return n;
        long ret = pow(n, a / 2);
        ret *= ret;
        ret %= MOD;
        if (a % 2 != 0)
            ret = ret * n;
        ret %= MOD;
        return ret;
    }

    private static long[][] quick(long[][] m, int a) {
        if (a == 0) {
            long[][] ret = new long[2][2];
            ret[0][0] = ret[1][1] = 1;
            ret[0][1] = ret[1][0] = 0;
            return ret;
        }
        if (a == 1)
            return gen;
        else {
            long[][] mm = quick(gen, a / 2);
            long[][] ret = new long[2][2];
            for (int i = 0; i < 2; i++)
                for (int j = 0; j < 2; j++) {
                    ret[i][j] = 0;
                    for (int k = 0; k < 2; k++) {
                        ret[i][j] += mm[i][k] * mm[k][j];
                        ret[i][j] %= MOD - 1;
                    }
                }
            if (a % 2 != 0) {
                long rett[][] = new long[2][2];
                for (int i = 0; i < 2; i++)
                    for (int j = 0; j < 2; j++) {
                        for (int k = 0; k < 2; k++) {
                            rett[i][j] += ret[i][k] * gen[k][j];
                            rett[i][j] %= MOD - 1;
                        }
                    }
                return rett;
            }
            return ret;
        }
    }

    public static void main(String[] args) {
        int a, b, n;
        try {
            while (true) {
                a = in.nextInt();
                b = in.nextInt();
                n = in.nextInt();
                if (n == 0) {
                    System.out.println(a);
                } else if (n == 1) {
                    System.out.println(b);
                } else {
                    long[][] ret = quick(gen, n - 1);
                    long ans = pow(b, ret[0][0]);
                    ans *= pow(a, ret[0][1]);
                    ans %= MOD;
                    System.out.println(ans);
                }
            }
        } catch (Exception e) {

        }
    }
}

class InputReader {
    BufferedReader br;
    StringTokenizer st;

    public InputReader(String s) {
        try {
            br = new BufferedReader(new FileReader(s));
        } catch (FileNotFoundException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }

    public InputReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    String nextToken() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    String nextLine() {
        try {
            return br.readLine();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return null;
    }

    int nextInt() {
        return Integer.parseInt(nextToken());
    }

    long nextLong() {
        return Long.parseLong(nextToken());
    }

    double nextDouble() {
        return Double.parseDouble(nextToken());
    }
}