如何使用快速傅立叶变换计算数组元素的乘积之和？

Question

I have some binary array. For example, let my array is:

int a[] = {1, 0, 0, 0, 1, 0, 1, 0, 1}

I want to calculate the values based on this formula:

How to calculate this function, using a fast Fourier transform? I have a large array and I have to calculate this function many times. So, I want to be able to calculate this function quick.

Answer 1

你正在计算基本上是在时域只是在频率domain.So乘法刚刚得到的FFT的卷积和卷积a与自身相乘，然后进行IFFT而在短返回时间domain.So ，您可以通过

b(2*i) = IFFT( FFT(a[0:2*i)]).FFT(a[0:2*i]) ) 

Answer 2

void main()
{
    int i,k,sum=0; 
    for(i=0;i<9;i++)
    {
        for(k=0;k<2*i;k++) 
        {
            sum + =(a[k]*a[2*i-k]; // calculating B
        }
    }
    cout<<sum;
}