CF1096G.Lucky Tickets[DP+卷积+多项式求幂] Educational Codeforces Round 57

给定一个偶数长度n和字符集(0..9中的一些数字) 问有多少个串的前 \(\frac{n}{2}\) 位的位数和跟后 \(\frac{n}{2}\) 位相等

\[f\left( i,j \right) \text{表示}i\text{个数的和是}j\text{的方案数} \\ \text{答案是}\sum_y{f\left( \frac{n}{2},y \right)}^2\text{（前半部分和后半部分都是}\sum_y{f\left( \frac{n}{2},y \right)}\text{）} \\ \text{所以只需要考虑怎么求}f \\ f\left( x+\text{1,}y \right) =\sum_{i=0}^9{f\left( x,y-i \right)} \\ \text{可以把上面的柿子补成卷积形式} \\ \text{令}H=f\left( x+1 \right) \text{，}F=f\left( x \right) \text{，}H\left( y \right) \text{表示的就是}f\left( x+\text{1,}y \right) \text{，}F\text{同理} \\ H\left( k \right) =\sum_{i=0}^{\min \left( k,9 \right)}{F\left( i \right)}\ast G\left( k-i \right) \\ \text{即}f\left( x+1 \right) \left( k \right) =\sum_{i=0}^{\min \left( k,9 \right)}{f\left( x \right) \left( i \right)}\ast G\left( k-i \right) \\ \text{比较容易发现没有字符集限制的情况下}G=1 \\ \text{有限制的时候}G=\sum_{i=0}^9{a_ix^i}\text{，}a_i\text{表示}i\text{是否在允许的字符集中，}x\text{没有实际意义} \\ \text{所以要做的就是求出}G\left( x \right) ^{\frac{n}{2}}\text{然后统计每一项系数的平方和} \]

贴个Tutorial里的代码

#include<bits/stdc++.h>

using namespace std;

const int LOGN = 21;
const int N = (1 << LOGN);
const int MOD = 998244353;
const int g = 3;

#define forn(i, n) for(int i = 0; i < int(n); i++)

inline int mul(int a, int b)
{
	return (a * 1ll * b) % MOD;
}

inline int norm(int a) 
{
	while(a >= MOD)
		a -= MOD;
	while(a < 0)
		a += MOD;
	return a;
}

inline int binPow(int a, int k) 
{
	int ans = 1;
	while(k > 0) 
	{
		if(k & 1)
			ans = mul(ans, a);
		a = mul(a, a);
		k >>= 1;
	}
	return ans;
}

inline int inv(int a) 
{
	return binPow(a, MOD - 2);
}

vector<int> w[LOGN];
vector<int> iw[LOGN];
vector<int> rv[LOGN];

void precalc() 
{
	int wb = binPow(g, (MOD - 1) / (1 << LOGN));
	
	for(int st = 0; st < LOGN; st++) 
	{
		w[st].assign(1 << st, 1);
		iw[st].assign(1 << st, 1);
		
		int bw = binPow(wb, 1 << (LOGN - st - 1));
		int ibw = inv(bw);
		
		int cw = 1;
		int icw = 1;
		
		for(int k = 0; k < (1 << st); k++) 
		{
			w[st][k] = cw;
			iw[st][k] = icw;
			
			cw = mul(cw, bw);
			icw = mul(icw, ibw);
		}
		
		rv[st].assign(1 << st, 0);
		
		if(st == 0) 
		{
			rv[st][0] = 0;
			continue;
		}
		int h = (1 << (st - 1));
		for(int k = 0; k < (1 << st); k++)
			rv[st][k] = (rv[st - 1][k & (h - 1)] << 1) | (k >= h);
	}
}

inline void fft(int a[N], int n, int ln, bool inverse) 
{	
	for(int i = 0; i < n; i++) 
	{
		int ni = rv[ln][i];
		if(i < ni)
			swap(a[i], a[ni]);
	}
	
	for(int st = 0; (1 << st) < n; st++) 
	{
		int len = (1 << st);
		for(int k = 0; k < n; k += (len << 1)) 
		{
			for(int pos = k; pos < k + len; pos++) 
			{
				int l = a[pos];
				int r = mul(a[pos + len], (inverse ? iw[st][pos - k] : w[st][pos - k]));
				
				a[pos] = norm(l + r);
				a[pos + len] = norm(l - r);
			}
		}
	}
	
	if(inverse) 
	{
		int in = inv(n);
		for(int i = 0; i < n; i++)
			a[i] = mul(a[i], in);
	}
}

int aa[N], bb[N], cc[N];

inline void multiply(int a[N], int sza, int b[N], int szb, int c[N], int &szc) 
{
	int n = 1, ln = 0;
	while(n < (sza + szb))
		n <<= 1, ln++;
	for(int i = 0; i < n; i++)
		aa[i] = (i < sza ? a[i] : 0);
	for(int i = 0; i < n; i++)
		bb[i] = (i < szb ? b[i] : 0);
		
	fft(aa, n, ln, false);
	fft(bb, n, ln, false);
	
	for(int i = 0; i < n; i++)
		cc[i] = mul(aa[i], bb[i]);
		
	fft(cc, n, ln, true);
	
	szc = n;
	for(int i = 0; i < n; i++)
		c[i] = cc[i];
}

vector<int> T[N];

int a[N];
int b[N];
int c[N];

#define sz(a) (int(a.size()))

int main()
{
	precalc();
	int n, k;
	scanf("%d %d", &n, &k);
	for(int i = 0; i < k; i++)
	{
	    int x;
	    scanf("%d", &x);
	    a[x] = 1;
	}
	int nn = 1, ln = 0;
	int nw = (n * 5) + 1;
	while(nn < nw)
	{
		nn *= 2;
		ln++;
	}
	fft(a, nn, ln, false);
	forn(i, nn)
		a[i] = binPow(a[i], n / 2);
	fft(a, nn, ln, true);
	int ans = 0;
	forn(i, nn)
		ans = norm(ans + binPow(a[i], 2));
	printf("%d\n", ans);
	return 0;	
}