FFT 高精度乘法模板

#define L(x) (1 << (x))
const double PI = acos(-1.0);
const int N = 1e7 + 10;
double ax[N], ay[N], bx[N], by[N];
char sa[N / 2], sb[N / 2];
int sum[N];
int x1[N], x2[N];
int revv(int x, int bits) {
    int ret = 0;
    for (int i = 0; i < bits; i++) {
        ret <<= 1;
        ret |= x & 1;
        x >>= 1;
    }
    return ret;
}
void fft(double * a, double * b, int n, bool rev) {
    int bits = 0;
    while (1 << bits < n) ++bits;
    for (int i = 0; i < n; i++) {
        int j = revv(i, bits);
        if (i < j)
            swap(a[i], a[j]), swap(b[i], b[j]);
    }
    for (int len = 2; len <= n; len <<= 1) {
        int half = len >> 1;
        double wmx = cos(2 * PI / len), wmy = sin(2 * PI / len);
        if (rev) wmy = -wmy;
        for (int i = 0; i < n; i += len) {
            double wx = 1, wy = 0;
            for (int j = 0; j < half; j++) {
                double cx = a[i + j], cy = b[i + j];
                double dx = a[i + j + half], dy = b[i + j + half];
                double ex = dx * wx - dy * wy, ey = dx * wy + dy * wx;
                a[i + j] = cx + ex, b[i + j] = cy + ey;
                a[i + j + half] = cx - ex, b[i + j + half] = cy - ey;
                double wnx = wx * wmx - wy * wmy, wny = wx * wmy + wy * wmx;
                wx = wnx, wy = wny;
            }
        }
    }
    if (rev) {
        for (int i = 0; i < n; i++)
            a[i] /= n, b[i] /= n;
    }
}
int sol(int a[], int na, int b[], int nb, int ans[]) {
    int len = max(na, nb), ln;
    for (ln = 0; L(ln) < len; ++ln);
    len = L(++ln);
    for (int i = 0; i < len ; ++i) {
        if (i >= na) ax[i] = 0, ay[i] = 0;
        else ax[i] = a[i], ay[i] = 0;
    }
    fft(ax, ay, len, 0);
    for (int i = 0; i < len; ++i) {
        if (i >= nb) bx[i] = 0, by[i] = 0;
        else bx[i] = b[i], by[i] = 0;
    }
    fft(bx, by, len, 0);
    for (int i = 0; i < len; ++i) {
        double cx = ax[i] * bx[i] - ay[i] * by[i];
        double cy = ax[i] * by[i] + ay[i] * bx[i];
        ax[i] = cx, ay[i] = cy;
    }
    fft(ax, ay, len, 1);
    for (int i = 0; i < len; ++i)
        ans[i] = (int)(ax[i] + 0.5);
    return len;
}
string mul(string sa, string sb) {
    int l1, l2, l;
    int i;
    string ans;
    memset(sum, 0, sizeof(sum));
    l1 = sa.size();
    l2 = sb.size();
    for (i = 0; i < l1; i++)
        x1[i] = sa[l1 - i - 1] - '0';
    for (i = 0; i < l2; i++)
        x2[i] = sb[l2 - i - 1] - '0';
    l = sol(x1, l1, x2, l2, sum);
    for (i = 0; i < l || sum[i] >= 10; i++) {//进位
        sum[i + 1] += sum[i] / 10;
        sum[i] %= 10;
    }
    l = i;
    while (sum[l] <= 0 && l > 0)    l--; // 检索最高位
    for (i = l; i >= 0; i--)
        ans += sum[i] + '0'; // 倒序输出
    return ans;
}