模板 - 数学 - 快速傅里叶变换/快速数论变换（FFT/NTT）

先看看。

通常模数常见的有998244353,1004535809,469762049，这几个的原根都是3。
所求的项数还不能超过2的23次方（因为998244353的分解）。

感觉没啥用。

#include <cstdio>
#include <cstring>
 
template <class T>
inline void swap(T &a, T &b)
{
    T c;
    c = a;
    a = b;
    b = c;
}
 
const int siz = ;
 
const int P = , G = ;
 
inline int pow(int a, int b)
{
    int r = ;
 
    while (b)
    {
        if (b & )
            r = 1LL * r * a % P;
 
        b >>= , a = 1LL * a * a % P;
    }
 
    return r;
}
 
inline void calculateNTT(int *s, int n, int f)
{
    {
        int cnt = ;
 
        static int rev[siz];
 
        while (n >> cnt)++cnt; --cnt;
 
        memset(rev, , sizeof rev);
 
        for (int i = ; i < n; ++i)
        {
            rev[i] |= rev[i >> ] >> ;
            rev[i] |= (i & ) << (cnt - );
        }
 
        for (int i = ; i < n; ++i)if (i < rev[i])swap(s[i], s[rev[i]]);
    }
 
    {
        for (int i = ; i < n; i <<= )
        {
            int wn = pow(G, (P - ) / (i * ));
 
            if (f == -)wn = pow(wn, P - );
 
            for (int j = ; j < n; j += (i << ))
            {
                int wk = ;
 
                for (int k = ; k < i; ++k, wk = 1LL * wk * wn % P)
                {
                    int x = s[j + k];
                    int y = 1LL * s[i + j + k] * wk % P;
 
                    s[j + k] = x + y;
                    s[i + j + k] = x - y;
 
                    s[j + k] = (s[j + k] % P + P) % P;
                    s[i + j + k] = (s[i + j + k] % P + P) % P;
                }
            }
        }
    }
 
    {
        if (f == -)
        {
            int inv = pow(n, P - );
 
            for (int i = ; i < n; ++i)
                s[i] = 1LL * s[i] * inv % P;
        }
    }
}
 
signed main(void)
{
    static char sa[siz];
    static char sb[siz];
 
    scanf("%s", sa);
    scanf("%s", sb);
 
    static int la, a[siz];
    static int lb, b[siz];
 
    la = strlen(sa);
    lb = strlen(sb);
 
    for (int i = ; i < la; ++i)a[i] = sa[la - i - ] - '';
    for (int i = ; i < lb; ++i)b[i] = sb[lb - i - ] - '';
 
    int len; for (len = ; len < la || len < lb; len <<= );
 
    calculateNTT(a, len << , +);
    calculateNTT(b, len << , +);
 
    for (int i = ; i < len << ; ++i)a[i] = 1LL * a[i] * b[i] % P;
 
    calculateNTT(a, len << , -); 
 
    for (int i = ; i < len << ; ++i)a[i + ] += a[i] / , a[i] = a[i] % ;
 
    len <<= ; while (!a[len])--len;
 
    for (int i = len; ~i; --i)printf("%d", a[i]); puts("");
}

快速傅里叶变换FFT