求快速傅立叶算法程序，最好是c/c++的

[Jack3]

回复者有分，分不够再加

[niejx]

VOID FFT(complex<double> * TD, complex<double> * FD, int r)
{
// 付立叶变换点数
LONG count;

// 循环变量
int i,j,k;

// 中间变量
int bfsize,p;

// 角度
double angle;

complex<double> *W,*X1,*X2,*X;

// 计算付立叶变换点数
count = 1 << r;

// 分配运算所需存储器
W = new complex<double>[count / 2];
X1 = new complex<double>[count];
X2 = new complex<double>[count];

// 计算加权系数
for(i = 0; i < count / 2; i++)
{
angle = -i * PI * 2 / count;
W[i] = complex<double> (cos(angle), sin(angle));
}

// 将时域点写入X1
memcpy(X1, TD, sizeof(complex<double>) * count);

// 采用蝶形算法进行快速付立叶变换
for(k = 0; k < r; k++)
{
for(j = 0; j < 1 << k; j++)
{
bfsize = 1 << (r-k);
for(i = 0; i < bfsize / 2; i++)
{
p = j * bfsize;
X2[i + p] = X1[i + p] + X1[i + p + bfsize / 2];
X2[i + p + bfsize / 2] = (X1[i + p] - X1[i + p + bfsize / 2]) * W[i * (1<<k)];
}
}
X = X1;
X1 = X2;
X2 = X;
}

// 重新排序
for(j = 0; j < count; j++)
{
p = 0;
for(i = 0; i < r; i++)
{
if (j&(1<<i))
{
p+=1<<(r-i-1);
}
}
FD[j]=X1[p];
}

// 释放内存
delete W;
delete X1;
delete X2;
}