写一段CCS5.5的代码，内容是实现基2的FFT

CCS5.5是一个编程环境，无法直接实现算法。但是我可以给您提供一个基2的FFT的C++代码，您可以在CCS5.5中进行编译和运行。

#include <bits/stdc++.h>
using namespace std;

typedef complex<double> Complex;

const double PI = acos(-1);

void fft(vector<Complex>& a, int n, int inv) {
    for (int i = 1, j = 0; i < n - 1; i++) {
        for (int s = n; j ^= s >>= 1, ~j & s;);
        if (i < j) swap(a[i], a[j]);
    }
    for (int i = 1; i < n; i <<= 1) {
        double x = inv ? PI / i : -PI / i;
        Complex w = { cos(x), sin(x) };
        for (int j = 0; j < n; j += i << 1) {
            Complex wk = { 1, 0 };
            for (int k = 0; k < i; k++, wk *= w) {
                Complex x = a[j + k], y = wk * a[j + k + i];
                a[j + k] = x + y, a[j + k + i] = x - y;
            }
        }
    }
    if (inv) for (int i = 0; i < n; i++) a[i] /= n;
}

vector<int> multiply(vector<int>& A, vector<int>& B) {
    vector<Complex> a(A.begin(), A.end()), b(B.begin(), B.end());
    int n = 1;
    while (n < A.size() + B.size()) n <<= 1;
    a.resize(n), b.resize(n);
    fft(a, n, 0), fft(b, n, 0);
    for (int i = 0; i < n; i++) a[i] *= b[i];
    fft(a, n, 1);
    vector<int> res(n);
    for (int i = 0; i < n; i++) res[i] = round(a[i].real());
    return res;
}

int main() {
    vector<int> A = { 1, 2, 3 }, B = { 4, 5, 6 };
    vector<int> C = multiply(A, B);
    for (int i : C) cout << i << " ";
    cout << endl;
    return 0;
}

这段代码实现了基于2的FFT算法，可以计算多项式的乘积。您可以在CCS5.5中运行它并进行测试。