多项式重工业基地

2021-02-13 20:03:13 阅读：163 来源： 互联网

标签：int 多项式 fft lim modint inline mod 基地重工业

未完待续……（只是给自己存个板子）

快速傅里叶变换

#include <bits/stdc++.h>

using namespace std; const int maxn = 2e6 + 1e2;

struct Cp
{
	double x, y;

	inline Cp operator +(const Cp& b) const { return (Cp) { x + b.x, y + b.y }; }

	inline Cp operator -(const Cp& b) const { return (Cp) { x - b.x, y - b.y }; }

	inline Cp operator *(const Cp& b) const { return (Cp) { x * b.x - y * b.y, x * b.y + y * b.x }; }
};

namespace fft
{
	int l, r[maxn], lim = 1; long double Pi = acos(-1.0); inline void init(int n)
	{
		while (lim <= n) lim <<= 1, l++; for (int i = 0; i < lim; i++)
		{
			r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
		}
	}

	inline void fft(Cp A[], int v)
	{
		for (int i = 0; i < lim; i++) if (i < r[i]) swap(A[i], A[r[i]]);
		
		Cp Wn, w, x, y; for (int mid = 1; mid < lim; mid <<= 1)
		{
			Wn = (Cp) { cos(Pi / mid), v * sin(Pi / mid) }; for (int R = mid << 1, j = 0; j < lim; j += R)
			{
				w = (Cp) { 1, 0 }; for (int k = 0; k < mid; k++, w = w * Wn)
				{
					x = A[j + k], y = w * A[j + mid + k], A[j + k] = x + y, A[j + mid + k] = x - y;
				}
			}
		}
		for (int i = 0; i < lim; i++) (v > 0) ? (A[i].x /= lim, A[i].y /= lim) : (A[i].x *= lim, A[i].y /= lim);
	}

	inline void conv(Cp A[], Cp B[], Cp C[]) { for (int i = 0; i < lim; i++) C[i] = A[i] * B[i]; }
}

Cp A[maxn], B[maxn], C[maxn]; int n, m; int main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0), cin >> n >> m;
	for (int i = 0; i <= n; i++) cin >> A[i].x;
	for (int i = 0; i <= m; i++) cin >> B[i].x;
	fft::init(n + m), fft::fft(A, 1), fft::fft(B, 1), fft::conv(A, B, C), fft::fft(C, -1);
	for (int i = 0; i <= n + m; i++) cout << int(C[i].x + 0.5) << ' '; return 0;
	
}

快速数论变换：


#include <bits/stdc++.h>

using namespace std; const int maxn = 5e5 + 1e2, mod = 998244353, g = 3, h = 332748118;

namespace fft
{
	int l, r[maxn], lim = 1; inline void init(int n)
	{
		while (lim <= n) lim <<= 1, l++; for (int i = 0; i < lim; i++)
		{
			r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
		}
	}

	inline int fpow(int x, int k)
	{
		int res = 1; while (k) { if (k & 1) res = 1ll * res * x % mod; x = 1ll * x * x % mod; k >>= 1; } return res;
	}

	inline void fft(int A[], int v)
	{
		for (int i = 0; i < lim; i++) if (i < r[i]) swap(A[i], A[r[i]]);
		
		int Wn, w, x, y; for (int mid = 1; mid < lim; mid <<= 1)
		{
			Wn = fpow(v == 1 ? g : h, (mod - 1) / (mid << 1)); for (int R = mid << 1, j = 0; j < lim; j += R)
			{
				w = 1; for (int k = 0; k < mid; k++, w = 1ll * w * Wn % mod)
				{
					y = 1ll * w * A[j + mid + k] % mod, A[j + mid + k] = (A[j + k] - y) % mod, A[j + k] = (A[j + k] + y) % mod;
				}
			}
		}
		int inv = fpow(lim, mod - 2);
		for (int i = 0; i < lim; i++) (v > 0) ? (A[i] = A[i] * 1ll * inv % mod) : (A[i] = A[i] * 1ll * lim % mod);
	}

	inline void conv(int A[], int B[], int C[]) { for (int i = 0; i < lim; i++) C[i] = 1ll * A[i] * B[i] % mod; }
}

int A[maxn], B[maxn], C[maxn]; int n, m; int main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0), cin >> n >> m;
	for (int i = 0; i <= n; i++) cin >> A[i];
	for (int i = 0; i <= m; i++) cin >> B[i];
	fft::init(n + m), fft::fft(A, 1), fft::fft(B, 1), fft::conv(A, B, C), fft::fft(C, -1);
	for (int i = 0; i <= n + m; i++) cout << (C[i] + mod) % mod << ' '; return 0;
	
}

快速莫比乌斯/沃尔什变换：

namespace fwt

{
	int lim; inline void init(int n) { lim = 1; while (lim < n) lim <<= 1; }

	inline void OR(modint f[], modint t)
	{
		for (int mid = 1; mid < lim; mid <<= 1) for (int R = (mid << 1), j = 0; j < lim; j += R)
			for (int k = 0; k < mid; k++) f[j + k + mid] += f[j + k] * t;
	}

	inline void AND(modint f[], modint t)
	{
		for (int mid = 1; mid < lim; mid <<= 1) for (int R = (mid << 1), j = 0; j < lim; j += R)
			for (int k = 0; k < mid; k++) f[j + k] += f[j + k + mid] * t;
	}

	inline void XOR(modint f[], modint t)
	{
		for (int mid = 1; mid < lim; mid <<= 1) for (int R = (mid << 1), j = 0; j < lim; j += R)
			for (int k = 0; k < mid; k++)
			{
				f[j + k] += f[j + mid + k], f[j + mid + k] = f[j + k] - f[j + mid + k] - f[j + mid + k], f[j + k] *= t, f[j + mid + k] *= t;
			}
	}

	inline void CONV(modint A[], modint B[], modint C[]) { for (int i = 0; i < lim; i++) C[i] = A[i] * B[i]; }
}

modint 类（乱入）：

int ttmp; struct modint

{
	int x; modint(int o) { x = o; } modint() { x = 0; }

	inline modint &operator =(int o) { return x = o, *this; }

	inline modint &operator +=(modint b) { return x = (ttmp = x + b.x) >= mod ? ttmp - mod : ttmp, *this; }

	inline modint &operator -=(modint b) { return x = (ttmp = x - b.x) < 0 ? ttmp + mod : ttmp, *this; }

	inline modint &operator *=(modint b) { return x = 1ll * x * b.x % mod, *this; }

	inline modint operator ^(int k)
	{
		modint res = 1, xx = x; while (k) { if (k & 1) res *= xx; k >>= 1, xx = xx * xx; } return res;
	}

	inline modint operator /=(modint b) { return x = 1ll * x * ((b ^ (mod - 2)).x) % mod, *this; }

	inline modint operator /=(int b) { return x = 1ll * x * (modint(b) ^ (mod - 2)).x, *this; }

	inline modint operator +(modint b) const { return modint((ttmp = x + b.x) >= mod ? ttmp - mod : ttmp); }

	inline modint operator -(modint b) const { return modint((ttmp = x - b.x) < 0 ? ttmp + mod : ttmp); }

	inline modint operator *(modint b) const { return modint(1ll * x * b.x % mod); }

	inline modint operator /(modint b) const { return modint(1ll * x * ((b ^ (mod - 2)).x) % mod); }
};

未完待续……敬请期待（bushi

标签：int,多项式,fft,lim,modint,inline,mod,基地,重工业
来源： https://www.cnblogs.com/Linshey/p/14400905.html

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多项式重工业基地