NTT模板

//一个n项和一个m项求卷积
typedef long long LL;
const int p = 998244353, G = 3, Gi = 332748118;//这里的Gi是G的除法逆元
const int N = 5000007;
const double PI = acos(-1);
int n, m;
int res, ans[N];
int len = 1;//
int L;//二进制的位数
int RR[N];
LL a[N], b[N];
inline int qread()
{ 
    int x = 0, y = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')
            y = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * y;
}
LL ksm(LL a, LL b, LL p)
{
    LL ans = 1;
    while (b) {
        if (b & 1) ans = ans * a % p;
        b >>= 1;
        a = a * a % p;
    }
    return ans;
}
LL inv(LL x) { return ksm(x, p - 2, p); }
void NTT(LL* A, int type)
{
    for (int i = 0; i < len; ++i)
        if (i < RR[i])
            swap(A[i], A[RR[i]]);
    for (int mid = 1; mid < len; mid <<= 1) {
        LL wn = ksm(G, (p - 1) / (mid * 2), p);
        if (type == -1) wn = ksm(wn, p - 2, p);
        //如果超时了上面if这句话删掉，在下面的if(type == -1)里加上下面这个循环
        /*for (int i = 1; i < len / 2; i ++)
        swap(A[i], A[len - i]); */
        for (int j = mid << 1, pos = 0; pos < len; pos += j) {
            LL w = 1;
            for (int k = 0; k < mid; ++k, w = (w * wn) % p) {
                int x = A[pos + k], y = w * A[pos + mid + k] % p;
                A[pos + k] = (x + y) % p;
                A[pos + k + mid] = (x - y + p) % p;

            }
        }
    }
    if (type == -1) {
        LL limit_inv = inv(len);
        for (int i = 0; i < len; ++i)
            A[i] = (A[i] * limit_inv) % p;
    }
}
void poly_mul(LL* a, LL* b, int deg)//多项式乘法
{
    for (len = 1, L = 0; len <= deg; len <<= 1) L++;
    for (int i = 0; i < len; ++i) {
        RR[i] = (RR[i >> 1] >> 1) | ((i & 1) << (L - 1));
    }
    NTT(a, 1);
    NTT(b, 1);
    for (int i = 0; i < len; ++i) a[i] = a[i] * b[i] % p;
    NTT(a, -1);
}
int main()
{
    n = qread(), m = qread();
    for (int i = 0; i <= n; ++i) a[i] = (qread() + p) % p;//取模
    for (int i = 0; i <= m; ++i) b[i] = (qread() + p) % p;
    poly_mul(a, b, n + m);
    for (int i = 0; i <= n + m; ++i)
        printf("%d ", a[i]);
    return 0;
}

附带原根表

$p=r \cdot 2^k + 1$	$r$	$k$	$g$
3	1	1	2
5	1	2	2
17	1	4	3
97	3	5	5
193	3	6	5
257	1	8	3
7681	15	9	17
12289	3	12	11
40961	5	13	3
65537	1	16	3
786433	3	18	10
5767169	11	19	3
7340033	7	20	3
23068673	11	21	3
104857601	25	22	3
167772161	5	25	3
469762049	7	26	3
998244353	119	23	3
1004535809	479	21	3
2013265921	15	27	31
2281701377	17	27	3
3221225473	3	30	5
75161927681	35	31	3
77309411329	9	33	7
206158430209	3	36	22
2061584302081	15	37	7
2748779069441	5	39	3
6597069766657	3	41	5
39582418599937	9	42	5
79164837199873	9	43	5
263882790666241	15	44	7
1231453023109121	35	45	3
1337006139375617	19	46	3
3799912185593857	27	47	5
4222124650659841	15	48	19
7881299347898369	7	50	6
31525197391593473	7	52	3
180143985094819841	5	55	6
1945555039024054273	27	56	5
4179340454199820289	29	57	3