FFT加汉宁窗的N点插值函数C程序

我可以回答这个问题。以下是一个FFT加汉宁窗的N点插值函数的C程序示例：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define PI 3.14159265358979323846

void fft(double *x, double *y, int n)
{
    int i, j, k, m;
    double r, s, t, u, v, w, a, b, c, d;

    m = log(n) / log(2);

    for (i = 0; i < n; i++) {
        j = 0;
        for (k = 0; k < m; k++) {
            j <<= 1;
            if (i & (1 << k)) j |= 1;
        }
        if (j < i) {
            t = x[i];
            x[i] = x[j];
            x[j] = t;
            t = y[i];
            y[i] = y[j];
            y[j] = t;
        }
    }

    for (i = 1; i <= m; i++) {
        u = 1;
        v = 0;
        w = PI / (1 << (i - 1));
        a = cos(w);
        b = -sin(w);
        for (j = 0; j < (1 << (i - 1)); j++) {
            for (k = j; k < n; k += (1 << i)) {
                r = u * x[k + (1 << (i - 1))] - v * y[k + (1 << (i - 1))];
                s = v * x[k + (1 << (i - 1))] + u * y[k + (1 << (i - 1))];
                x[k + (1 << (i - 1))] = x[k] - r;
                y[k + (1 << (i - 1))] = y[k] - s;
                x[k] += r;
                y[k] += s;
            }
            c = u * a - v * b;
            d = v * a + u * b;
            u = c;
            v = d;
        }
    }
}

void hann(double *x, int n)
{
    int i;
    for (i = 0; i < n; i++) {
        x[i] *= 0.5 * (1 - cos(2 * PI * i / (n - 1)));
    }
}

void interpolate(double *x, double *y, int n, int m)
{
    int i, j, k;
    double *xx, *yy;

    xx = (double *) malloc(n * sizeof(double));
    yy = (double *) malloc(n * sizeof(double));

    for (i = 0; i < n; i++) {
        xx[i] = x[i];
        yy[i] = y[i];
    }

    hann(xx, n);

    fft(xx, yy, n);

    for (i = 0; i < m; i++) {
        x[i] = i * 1.0 / m;
        y[i] = 0;
        for (j = 0; j < n; j++) {
            k = (j + n / 2) % n;
            y[i] += xx[j] * cos(2 * PI * k * i / m) + yy[j] * sin(2 * PI * k * i / m);
        }
        y[i] /= n;
    }

    free(xx);
    free(yy);
}

int main()
{
    int n = 8;
    int m = 16;
    double x[] = {0, 1, 2, 3, 4, 5, 6, 7};
    double y[] = {0, 1, 0, -1, 0, 1, 0, -1};
    int i;

    interpolate(x, y, n, m);

    for (i = 0; i < m; i++) {
        printf("%f %f\n", x[i], y[i]);
    }

    return 0;
}

该程序实现了一个FFT加汉宁窗的N点插值函数，可以将一个长度为N的实数序列插值为一个长度为M的实数序列。程序中使用了FFT算法和汉宁窗函数，可以通过修改N和M的值来调整插值的精度。