Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

CF1251F Red-White Fence

LuoGu: CF1251F Red-White Fence

CF: F. Red-White Fence

注意到红色的数量很少,考虑对于每一个红色统计答案。如果所有 $k$ 个白色长度互不相同,那么选择 $i$ 个白色答案为 $\binom k i 2 ^ i$ 。如果有相同的,大于等于 $2$ 个实际上是等价于 $2$ 个的,对于一组 $2$ 个,左侧可以选择是否选择,右侧可以选择是否选择,如果有 $k$ 组,即 $\binom {2k} i$ 。两种情况背包问题,卷积即可。

查看代码
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#include <cstdio>
#include <vector>
#include <algorithm>
#define pb push_back
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
flag |= c == '-';
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x); read(y...); }
template <class Type>
void write (Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar('0' + x % 10);
}
typedef long long LL;
const int N = 5e6 + 10, mod = 998244353, inv2 = mod + 1 >> 1;
int rev[N];
int binpow (int b, int k = mod - 2)
{
int res = 1;
for (; k; k >>= 1, b = (LL)b * b % mod)
if (k & 1) res = (LL)res * b % mod;
return res;
}
void ntt (int *x, int bit, int op)
{
int tot = 1 << bit;
for (int i = 1; i < tot; ++i)
if ((rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1) > i)
swap(x[rev[i]], x[i]);
for (int mid = 1; mid < tot; mid <<= 1)
{
int w1 = binpow(3, (mod - 1) / (mid << 1));
if (!~op) w1 = binpow(w1);
for (int i = 0; i < tot; i += mid << 1)
for (int j = 0, k = 1; j < mid; ++j, k = (LL)k * w1 % mod)
{
int p = x[i | j], q = (LL)k * x[i | j | mid] % mod;
x[i | j] = (p + q) % mod, x[i | j | mid] = (p - q) % mod;
}
}
if (~op) return;
int itot = binpow(tot);
for (int i = 0; i < tot; ++i)
x[i] = (LL)x[i] * itot % mod;
}
void PolyMul (int n, int *f, int m, int *g, int nm, int *res)
{
int bit = 0;
while (1 << bit < n + m - 1) ++bit;
int tot = 1 << bit;
for (int i = n; i < tot; ++i) f[i] = 0;
for (int i = m; i < tot; ++i) g[i] = 0;
ntt(f, bit, 1), ntt(g, bit, 1);
for (int i = 0; i < tot; ++i)
res[i] = (LL)f[i] * g[i] % mod;
ntt(res, bit, -1);
for (int i = nm; i < tot; ++i) res[i] = 0;
}
vector <int> h1, h2;
int fact[N], ifact[N], p2[N], h[N], ans[N];
void init ()
{
fact[0] = 1;
for (int i = 1; i < N; ++i)
fact[i] = (LL)fact[i - 1] * i % mod;
ifact[N - 1] = binpow(fact[N - 1]);
for (int i = N - 1; i; --i)
ifact[i - 1] = (LL)ifact[i] * i % mod;
p2[0] = 1;
for (int i = 1; i < N; ++i)
p2[i] = p2[i - 1] * 2 % mod;
}
int C (int a, int b) { return a < b ? 0 : (LL)fact[a] * ifact[b] % mod * ifact[a - b] % mod; }
int main ()
{
init();
int n, m, q;
read(n, m);
for (int i = 1; i <= n; ++i) read(h[i]);
sort(h + 1, h + n + 1);
for (int i = 1; i <= n; ++i)
if (i == n || h[i] ^ h[i + 1])
h1.pb(h[i]);
else
{
h2.pb(h[i]);
while (i < n && h[i] == h[i + 1]) ++i;
}
for (int k; m; --m)
{
static int A[N], B[N];
read(k);
int t = lower_bound(h1.begin(), h1.end(), k) - h1.begin();
for (int i = 0; i <= t; ++i)
A[i] = (LL)C(t, i) * p2[i] % mod;
int s = lower_bound(h2.begin(), h2.end(), k) - h2.begin() << 1;
for (int i = 0; i <= s; ++i)
B[i] = C(s, i);
PolyMul(t + 1, A, s + 1, B, s + t + 1, A);
for (int i = 0; i <= s + t; ++i)
(ans[i + 1 + k] += A[i]) %= mod;
}
read(q);
for (int a; q; --q, puts(""))
read(a), write((ans[a >> 1] + mod) % mod);
return 0;
}