hdu-4549 M斐波那契数列 && nyoj - 1000

运用费马小定理&&矩阵快速幂  求出 a , b 的个数

运用快速幂求解 a^num1 * b ^ num2 % MOD

#include
#include
typedef __int64 LL;
#define MOD 1000000007
#define mod 1000000006
struct matrix//矩阵
{
    LL Matrix[2][2];
};
matrix s =
{
    1,1,
    1,0,
};
matrix m =
{
    1,0,
    0,1,
};
matrix matrix_multiplication(matrix a,matrix b)//矩阵乘法
{
    matrix c;
    for(int i = 0;i < 2;i ++)
    {
        for(int j =0;j < 2;j++)
        {
            c.Matrix[i][j] = 0;
            for(int k = 0;k < 2;k++)
                c.Matrix[i][j] = (c.Matrix[i][j] + a.Matrix[i][k]*b.Matrix[k][j]) % mod;
        }
    }
    return c;
}
matrix Fast_power1(LL n)//矩阵快速幂
{
    matrix ret = m,p = s;
    while(n)
    {
        if(n&1) ret = matrix_multiplication(ret,p);
        p = matrix_multiplication(p,p);
        n >>= 1;
    }

    return ret;
}
LL Fast_power2(LL x,LL num)//x^num%MOD的快速幂
{
    LL res = 1;
    while(num)
    {
        if(num&1) res = (res * x) % MOD;
        x = (x*x)%MOD;
        num>>=1;
    }
    return res;
}
int main()
{
    LL a,b,n,x,y;
    while(~scanf("%I64d%I64d%I64d",&a,&b,&n))
    {
        matrix k;
        k = Fast_power1(n);
        LL x = k.Matrix[1][1],y = k.Matrix[1][0];
        LL z = (Fast_power2(a,x) * Fast_power2(b,y))%MOD;
        printf("%I64d\n",z);
    }
}