处理双打时的模运算符用法

Question

I have to find the value of ( 1+sqrt(3) )^n where n < 10^9.As this number can be very large we have to print the ans%1000000007. I have written the following function for this.

double power(double x, int y)
{
    double temp;
    if( y == 0)
       return 1;
    temp = power(x, y/2);
    if (y%2 == 0)
       return temp*temp;
    else
    {
       if(y > 0)
           return x*temp*temp;
       else
           return (temp*temp)/x;
    }
}

Now, I unable to understand how to take care of the modulo condition.Can somebody please help.

Answer 1

You can't do that. You could use fmod , but since sqrt(3) cannot be exactly represented, you'd get bogus values for large exponents.

I'm rather confident that you actually need integer results ( (1 + sqrt(3))^n + (1 - sqrt(3))^n ), so you should use integer math, exponentiation by squaring with a modulo operation at each step. cf. this question

Answer 2

This approach is infeasible. As shown in this question , you would need hundreds of millions of decimal digits more precision than the double type supplies. The problem you are actually trying to solve is discussed here . Are you two in the same class?

Answer 3

模数需要整数类型，您可以将union用作双精度类型，并与整数结合使用模数（如果这是C）