2020.3.1考试T1 多项式

标签：last   open   tmp   mat   else   lin   class   uid   pac   

出题人很凉心的把算法写成了题目名

首先我们可以发现每一维的贡献是独立的，这可以从 \(solve1\) 里看出来

然后我们可以考虑转化为 \(DP\) ，这可以从 \(solve2\) 里看出来

我们统计每一维能产生的贡献，就是 \(a\) 个 \(0\) 面， \(b\) 个 \(1\) 面， \(c\) 个 \(2\) 面这种形式，能写成一个多项式 \(ax^0+bx^1+cx^2\),而我们最终显然就是把所有的多项式都乘起来。

暴力一个一个乘就很 naive，分治 \(NTT\) 解决就好啦。

不透彻的话把每个 \(solve\) 都看一遍就好啦。

Warning：请不要学习本代码的分治 \(NTT\) 写法，考场上现想出来的，实现麻烦了不少，建议学一下其他大佬的写法。

#include<iostream>
#include<cstdio>
#define int long long
#define LL long long
using namespace std;
int n;
const int N = 400010, mod = 469762049, G = 3, Ginv = (mod + 1) / 3;
int a[N], b[N], c[N], ans[N];
inline int read()
{
	int res = 0; char ch = getchar(); bool XX = false;
	for (; !isdigit(ch); ch = getchar())(ch == ‘-‘) && (XX = true);
	for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
	return XX ? -res : res;
}
void solve1()
{
	int tmp;
	for (int i = 1; i <= a[1]; ++i)
	{
		tmp = 0;
		if (i == 1)++tmp; if (i == a[1])++tmp;
		++ans[tmp];
	}
	for (int i = 0; i <= 2 * n; ++i)printf("%lld\n", ans[i]);
}
void solve2()
{
	int tmp;
	for (int i = 1; i <= a[1]; ++i)
		for (int j = 1; j <= a[2]; ++j)
		{
			tmp = 0;
			if (i == 1)++tmp; if (i == a[1])++tmp;
			if (j == 1)++tmp; if (j == a[2])++tmp;
			++ans[tmp];
		}
	for (int i = 0; i <= 2 * n; ++i)printf("%lld\n", ans[i]);
}
void solve3()
{
	int tmp;
	for (int i = 1; i <= n; ++i)
	{
		tmp = a[i]; a[i] = b[i] = c[i] = 0;
		if (tmp == 1)a[i] = 1, c[i] = 0;
		else b[i] = 2, c[i] = tmp - b[i];
	}
	ans[0] = 1;
	for (int i = 1; i <= n; ++i)
	{
		for (int j = 2 * n; j >= 2; --j)
			ans[j] = ((LL)ans[j] * c[i] % mod + (LL)ans[j - 1] * b[i] % mod + (LL)ans[j - 2] * a[i] % mod) % mod;
		ans[1] = ((LL)ans[1] * c[i] % mod + (LL)ans[0] * b[i] % mod) % mod;
		ans[0] = (LL)ans[0] * c[i] % mod;
	}
	for (int i = 0; i <= 2 * n; ++i)printf("%lld\n", ans[i]);
}

/*下边 solve4*/

int last, top;
int r[N], zhan[30], tmp[500];
LL ksm(LL a, LL b, LL mod)
{
	LL res = 1;
	for (; b; b >>= 1, a = a * a % mod)
		if (b & 1)res = res * a % mod;
	return res;
}

void NTT(LL *A, int lim, int opt)
{
	if (last != lim)
	{
		last = lim;
		for (int i = 0; i < lim; ++i)
			r[i] = (r[i >> 1] >> 1) | (i & 1 ? (lim >> 1) : 0);
	}
	for (int i = 0; i < lim; ++i)
		if (i < r[i])swap(A[i], A[r[i]]);
	int len;
	LL wn, w, x, y;
	for (int mid = 1; mid < lim; mid <<= 1)
	{
		len = mid << 1;
		wn = ksm(opt == 1 ? G : Ginv, (mod - 1) / len, mod);
		for (int j = 0; j < lim; j += len)
		{
			w = 1;
			for (int k = j; k < j + mid; ++k, w = w * wn % mod)
			{
				x = A[k]; y = A[k + mid] * w % mod;
				A[k] = (x + y) % mod;
				A[k + mid] = (x - y + mod) % mod;
			}
		}
	}
	if (opt == 1)return;
	int ni = ksm(lim, mod - 2, mod);
	for (int i = 0; i < lim; ++i)A[i] = A[i] * ni % mod;
}
void MUL(LL *A, int n, LL *B, int m)
{
	if (n + m <= 115)
	{
		for (int i = 0, to = n + m; i <= to; ++i)tmp[i] = 0;
		for (int i = 0; i <= n; ++i)
			for (int j = 0; j <= m; ++j)
				(tmp[i + j] += A[i] * B[j] % mod) %= mod;
		for (int i = 0, to = n + m; i <= to; ++i)A[i] = tmp[i];
		for (int i = 0; i <= m; ++i)B[i] = 0;
	}
	else
	{
		int lim = 1;
		while (lim <= (n + m))lim <<= 1;
		NTT(A, lim, 1); NTT(B, lim, 1);
		for (int i = 0; i < lim; ++i)A[i] = A[i] * B[i] % mod, B[i] = 0;
		NTT(A, lim, -1);
	}
}
struct dxs
{
	int siz;
	LL v[N];
} A[30];
int newdxs()
{
	return zhan[top--];
}
void huidxs(int x)
{
	A[x].siz = 0;
	zhan[++top] = x;
}
int solve(int l, int r)
{
	if (l == r)
	{
		int k = newdxs();
		A[k].siz = 2;
		A[k].v[0] = c[l]; A[k].v[1] = b[l]; A[k].v[2] = a[l];
		return k;
	}
	int mid = (l + r) >> 1;
	int lson = solve(l, mid), rson = solve(mid + 1, r);
	MUL(A[lson].v, A[lson].siz, A[rson].v, A[rson].siz);
	A[lson].siz = A[lson].siz + A[rson].siz;
	huidxs(rson);
	return lson;
}
void solve4()
{
	int tmp;
	for (int i = 1; i <= n; ++i)
	{
		tmp = a[i]; a[i] = b[i] = c[i] = 0;
		if (tmp == 1)a[i] = 1, c[i] = 0;
		else b[i] = 2, c[i] = tmp - b[i];
	}
	for (int i = 1; i <= 25; ++i)zhan[++top] = i;
	int k = solve(1, n);
	for (int i = 0; i <= 2 * n; ++i)printf("%lld\n", A[k].v[i]);
}

/*上边 solve4*/

signed main()
{
	freopen("poly.in", "r", stdin);
	freopen("poly.out", "w", stdout);
	cin >> n;
	for (int i = 1; i <= n; ++i)
	{
		a[i] = read();
	}
	if (n == 1 && a[1] <= 1000)solve1();
	else if (n == 2 && a[1] <= 1000 && a[2] <= 1000)solve2();
	else if (n <= 5000)solve3();
	else solve4();
	fclose(stdin); fclose(stdout);
	return 0;
}

2020.3.1考试T1 多项式

标签：last   open   tmp   mat   else   lin   class   uid   pac   

原文地址：https://www.cnblogs.com/wljss/p/12628999.html