Blog of RuSun

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

P6012 [P5087] 数学 加强版

P6012 [P5087] 数学 加强版

考虑构造生成函数,对于一个数 $k$ ,选择则答案乘 $k$ ,个数加一;不选择则不变:那么有 $f(x) = 1 + kx$ 。答案为 $[x ^ n]\prod f(x)$ 。

将所有函数乘起来,注意到每个多项式的次数很小,挨个乘代价很大。考虑分治,每次将左右一半合并。那么有 $T(n) = T(\frac n 2) + O(n \log n) = O(n \log ^ 2 n)$ 。

这里写了 3模数NTT ,需要卡一下。

查看代码
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#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
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(x % 10 + '0');
}
typedef __int128 L;
typedef long long LL;
const int N = 1e6 + 10, p1 = 998244353, p2 = 1004535809, p3 = 469762049, p = 1e9 + 7;
template <const int &mod>
struct NTT
{
int rev[N];
void adj (int &x) { x += x >> 31 & mod; }
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 = 0; 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;
adj((x[i | j] = p) += q - mod), adj((x[i | j | mid] = p) -= q);
}
}
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, vector <int> &f, int m, vector <int> &g, int nm, vector <int> &w)
{
static int A[N], B[N], C[N];
for (int i = 0; i <= n; ++i)
adj(A[i] = f[i] - mod);
for (int i = 0; i <= m; ++i)
adj(B[i] = g[i] - mod);
int bit = 1;
while (n + m + 1 > 1 << bit) ++bit;
int tot = 1 << bit;
for (int i = n + 1; i < tot; ++i) A[i] = 0;
for (int i = m + 1; i < tot; ++i) B[i] = 0;
ntt(A, bit, 1), ntt(B, bit, 1);
for (int i = 0; i < tot; ++i)
C[i] = (LL)A[i] * B[i] % mod;
ntt(C, bit, -1);
for (int i = 0; i <= nm; ++i) w[i] = C[i];
}
};
NTT <p1> q1; NTT <p2> q2; NTT <p3> q3;
int CRT (int x1, int x2, int x3)
{
L res = 0, mul = (L)p1 * p2 * p3;
res += (L)x1 * p2 * p3 * q1.binpow((LL)p2 * p3 % p1);
res += (L)x2 * p1 * p3 * q2.binpow((LL)p1 * p3 % p2);
res += (L)x3 * p1 * p2 * q3.binpow((LL)p1 * p2 % p3);
return res % mul % p;
}
void solve (int l, int r, int &k, vector <int> &g)
{
if (l == r) return g[0] = 1, read(g[1]);
int mid = l + r >> 1;
int nm = min(k, r - l + 1), n = min(k, mid - l + 1), m = min(k, r - mid);
vector <int> A, B, C, D, E;
A.resize(n + 1); B.resize(m + 1);
C.resize(nm + 1), D.resize(nm + 1), E.resize(nm + 1);
solve(l, mid, k, A), solve(mid + 1, r, k, B);
q1.PolyMul(n, A, m, B, nm, C);
q2.PolyMul(n, A, m, B, nm, D);
q3.PolyMul(n, A, m, B, nm, E);
for (int i = 0; i <= nm; ++i)
g[i] = CRT(C[i], D[i], E[i]);
}
int main ()
{
int n, k;
read(n, k);
vector <int> A; A.resize(min(k, n) + 1);
solve(1, n, k, A);
write(A[k]);
return 0;
}