使用数论进行快速傅立叶变换（快速数论变换）（附UOJ - 34 多项式乘法代码）

Fast-Number-Theoretic-Transform.cpp

@@ -0,0 +1,112 @@
+#include <cstdio>
+#include <cstring>
+
+#define MOD 998244353
+#define ROOT 3
+
+typedef long long ll;
+
+using namespace std;
+
+int n, n0, n1, k;
+ll a0[500000], a1[500000];
+
+inline void swap(ll &x, ll &y)
+{
+	ll temp = x;
+	x = y;
+	y = temp;
+}
+
+ll qpow(ll x, int y)
+{
+	if (y == 1) return x;
+	ll t = qpow(x, y >> 1);
+	t = t * t % MOD;
+	if ((y & 1) == 0) return t;
+	else return t * x % MOD;
+}
+
+void pre(ll *a)
+{
+	static ll temp[500000];
+	memcpy(temp, a, sizeof(temp));
+	for (int i = 0; i < n; i++)
+	{
+		int x = 0;
+		for (int p = 0; p < k; p++)
+		{
+			if ((i & (1 << p)) != 0)
+			{
+				x += (1 << (k - p - 1));
+			}
+		}
+		a[i] = temp[x];
+	}
+}
+
+void dft(ll *a, int rev)
+{
+	ll _g = rev == 1 ? ROOT : qpow(ROOT, MOD - 2);
+	for (int i = 2; i <= n; i <<= 1)
+	{
+		ll gn = qpow(_g, (MOD - 1) / i);
+		for (int p = 0; p < n; p += i)
+		{
+			ll g = 1LL;
+			for (int q = p; q < p + (i >> 1); q++)
+			{
+				ll temp = a[q];
+				a[q] = (a[q] + g * a[q + (i >> 1)]) % MOD;
+				a[q + (i >> 1)] = (temp - g * a[q + (i >> 1)]) % MOD;
+				g = g * gn % MOD;
+			}
+		}
+	}
+}
+
+inline ll read()
+{
+	char c;
+	do
+	{
+		c = getchar();
+	} while (c < '0' || c > '9');
+	return (ll)(c - '0');
+}
+
+int main()
+{
+	scanf("%d%d", &n0, &n1);
+	for (int i = 0; i <= n0; i++)
+	{
+		a0[i] = read();
+	}
+	for (int i = 0; i <= n1; i++)
+	{
+		a1[i] = read();
+	}
+	n = n0 + n1;
+	while (n > 0)
+	{
+		k++;
+		n >>= 1;
+	}
+	n = 1 << k;
+	pre(a0);
+	pre(a1);
+	dft(a0, 1);
+	dft(a1, 1);
+	for (int i = 0; i < n; i++)
+	{
+		a0[i] = a0[i] * a1[i] % MOD;
+	}
+	pre(a0);
+	dft(a0, -1);
+	ll inv = qpow(n, MOD - 2);
+	for (int i = 0; i <= n0 + n1; i++)
+	{
+		printf("%lld ", (a0[i] * inv % MOD + MOD) % MOD);
+	}
+	return 0;
+}