一、题目链接
二、题目大意
在一个二维平面内，有 n n n 个起点 ( 0 , a i ) (0, a_i) (0,ai) 要走到对应的终点 ( i , 0 ) (i, 0) (i,0)，每次可以向下走或向左走，问不相交路径组的方案数.
1 ≤ n ≤ 5 × 1 0 5 , 0 ≤ a i ≤ 1 0 6 , a i < a i + 1 1 \leq n \leq 5 \times 10^5, 0 \leq a_i \leq 10^6, a_i < a_{i+1} 1≤n≤5×105,0≤ai≤106,ai<ai+1.
三、分析
易知从
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\binom{a_i + j}{j}
(jai+j).
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M = \left [ \begin{matrix} \binom{a_1 + 1}{1} & \binom{a_1 + 2}{2} & \cdots & \binom{a_1 + n}{n} \\ \binom{a_2 + 1}{1} & \binom{a_2 + 2}{2} & \cdots & \binom{a_2 + n}{n} \\ \vdots & \vdots & \cdots & \vdots \\ \binom{a_n + 1}{1} & \binom{a_n + 2}{2} & \cdots & \binom{a_n + n}{n} \end{matrix} \right]
M=⎣⎢⎢⎢⎡(1a1+1)(1a2+1)⋮(1an+1)(2a1+2)(2a2+2)⋮(2an+2)⋯⋯⋯⋯(na1+n)(na2+n)⋮(nan+n)⎦⎥⎥⎥⎤
则由 LGV 引理可知
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∣M∣ 即为答案.
由于
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max(n) = 5 \times 10^5
max(n)=5×105，直接高斯消元是无法承受的，考虑简便计算
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∣M∣ 的方法.
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\begin{aligned} |M| &= \left | \begin{matrix} \binom{a_1 + 1}{1} & \binom{a_1 + 2}{2} & \cdots & \binom{a_1 + n}{n} \\ \binom{a_2 + 1}{1} & \binom{a_2 + 2}{2} & \cdots & \binom{a_2 + n}{n} \\ \vdots & \vdots & \cdots & \vdots \\ \binom{a_n + 1}{1} & \binom{a_n + 2}{2} & \cdots & \binom{a_n + n}{n} \end{matrix} \right| \\ &= \left | \begin{matrix} \frac{(a_1 + 1)!}{1!a_1!} & \frac{(a_1 + 2)!}{2!a_1!} & \cdots & \frac{(a_1 + n)!}{n!a_1!} \\ \frac{(a_2 + 1)!}{1!a_2!} & \frac{(a_2 + 2)!}{2!a_2!} & \cdots & \frac{(a_2 + n)!}{n!a_2!} \\ \vdots & \vdots & \cdots & \vdots \\ \frac{(a_n + 1)!}{1!a_n!} & \frac{(a_n + 2)!}{2!a_n!} & \cdots & \frac{(a_n + n)!}{n!a_n!} \\ \end{matrix} \right| \\ &= \prod_{i=1}^n{\frac{1}{i!}}\left | \begin{matrix} \frac{(a_1 + 1)!}{a_1!} & \frac{(a_1 + 2)!}{a_1!} & \cdots & \frac{(a_1 + n)!}{a_1!} \\ \frac{(a_2 + 1)!}{a_2!} & \frac{(a_2 + 2)!}{a_2!} & \cdots & \frac{(a_2 + n)!}{a_2!} \\ \vdots & \vdots & \cdots & \vdots \\ \frac{(a_n + 1)!}{a_n!} & \frac{(a_n + 2)!}{a_n!} & \cdots & \frac{(a_n + n)!}{a_n!} \\ \end{matrix} \right| \\ &= \prod_{i=1}^n{\frac{1}{i!}}\left | \begin{matrix} (a_1 + 1) & (a_1 + 1)（a_1 + 2) & \cdots & \prod_{i=1}^n{(a_1 + i)} \\ (a_2 + 1) & (a_2 + 1)（a_2 + 2) & \cdots & \prod_{i=1}^n{(a_2 + i)} \\ \vdots & \vdots & \cdots & \vdots \\ (a_n + 1) & (a_n + 1)（a_n + 2) & \cdots & \prod_{i=1}^n{(a_n + i)} \\ \end{matrix} \right| \\ &= \prod_{i=1}^n{\frac{1}{i!}}\left | \begin{matrix} (a_1 + 1) & (a_2 + 1) & \cdots & (a_n + 1) \\ (a_1 + 1)（a_1 + 2) & (a_2 + 1)（a_2 + 2) & \cdots & (a_n + 1)（a_n + 2) \\ \vdots & \vdots & \cdots & \vdots \\ \prod_{i=1}^n{(a_1 + i)} & \prod_{i=1}^n{(a_2 + i)} & \cdots & \prod_{i=1}^n{(a_n + i)} \\ \end{matrix} \right| \\ &= \prod_{i=1}^n{\frac{1}{i!}}\left | \begin{matrix} (a_1 + 1) & (a_2 + 1) & \cdots & (a_n + 1) \\ (a_1 + 1)^2 & (a_2 + 1)^2 & \cdots & (a_n + 1)^2 \\ \vdots & \vdots & \cdots & \vdots \\ (a_1 + 1)^n & (a_2 + 1)^n & \cdots & (a_n + 1)^n \\ \end{matrix} \right| \\ &= \prod_{i=1}^n{\frac{a_i + 1}{i!}}\left | \begin{matrix} 1 & 1 & \cdots & 1 \\ a_1 + 1 & a_2 + 1 & \cdots & a_n + 1 \\ \vdots & \vdots & \cdots & \vdots \\ (a_1 + 1)^{n-1} & (a_2 + 1)^{n-1} & \cdots & (a_n + 1)^{n-1} \\ \end{matrix} \right| \\ &= \prod_{i=1}^n{\frac{a_i + 1}{i!} \prod_{j=1}^{i-1}{(a_i - a_j)}} \end{aligned}
∣M∣=∣∣∣∣∣∣∣∣∣(1a1+1)(1a2+1)⋮(1an+1)(2a1+2)(2a2+2)⋮(2an+2)⋯⋯⋯⋯(na1+n)(na2+n)⋮(nan+n)∣∣∣∣∣∣∣∣∣=∣∣∣∣∣∣∣∣∣∣1!a1!(a1+1)!1!a2!(a2+1)!⋮1!an!(an+1)!2!a1!(a1+2)!2!a2!(a2+2)!⋮2!an!(an+2)!⋯⋯⋯⋯n!a1!(a1+n)!n!a2!(a2+n)!⋮n!an!(an+n)!∣∣∣∣∣∣∣∣∣∣=i=1∏ni!1∣∣∣∣∣∣∣∣∣∣a1!(a1+1)!a2!(a2+1)!⋮an!(an+1)!a1!(a1+2)!a2!(a2+2)!⋮an!(an+2)!⋯⋯⋯⋯a1!(a1+n)!a2!(a2+n)!⋮an!(an+n)!∣∣∣∣∣∣∣∣∣∣=i=1∏ni!1∣∣∣∣∣∣∣∣∣(a1+1)(a2+1)⋮(an+1)(a1+1)（a1+2)(a2+1)（a2+2)⋮(an+1)（an+2)⋯⋯⋯⋯∏i=1n(a1+i)∏i=1n(a2+i)⋮∏i=1n(an+i)∣∣∣∣∣∣∣∣∣=i=1∏ni!1∣∣∣∣∣∣∣∣∣(a1+1)(a1+1)（a1+2)⋮∏i=1n(a1+i)(a2+1)(a2+1)（a2+2)⋮∏i=1n(a2+i)⋯⋯⋯⋯(an+1)(an+1)（an+2)⋮∏i=1n(an+i)∣∣∣∣∣∣∣∣∣=i=1∏ni!1∣∣∣∣∣∣∣∣∣(a1+1)(a1+1)2⋮(a1+1)n(a2+1)(a2+1)2⋮(a2+1)n⋯⋯⋯⋯(an+1)(an+1)2⋮(an+1)n∣∣∣∣∣∣∣∣∣=i=1∏ni!ai+1∣∣∣∣∣∣∣∣∣1a1+1⋮(a1+1)n−11a2+1⋮(a2+1)n−1⋯⋯⋯⋯1an+1⋮(an+1)n−1∣∣∣∣∣∣∣∣∣=i=1∏ni!ai+1j=1∏i−1(ai−aj)
至此计算
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\begin{aligned} \prod_{j=1}^{i-1}{a_i - a_j} \end{aligned}
j=1∏i−1ai−aj.
记 f ( x ) = ∑ i = 1 n x a i \begin{aligned} f(x) = \sum_{i=1}^{n}{x^{a_i}}\end{aligned} f(x)=i=1∑nxai， g ( x ) = ∑ i = 1 n x − a i \begin{aligned}g(x) = \sum_{i=1}^{n}{x^{-a_i}} \end{aligned} g(x)=i=1∑nx−ai.
用 NTT 处理出 h = f ∗ g = ∑ c i x i \begin{aligned} h = f*g = \sum{c_ix^i} \end{aligned} h=f∗g=∑cixi.
按 ( a i − a j ) (a_i - a_j) (ai−aj) 计算贡献，则 ∏ i = 1 n ∏ j = 1 i − 1 ( a i − a j ) = ∏ i > 0 i c i \begin{aligned} \prod_{i=1}^n{\prod_{j=1}^{i-1}(a_i - a_j)} = \prod_{i > 0}{i^{c_i}} \end{aligned} i=1∏nj=1∏i−1(ai−aj)=i>0∏ici.
因此答案为 ∏ i = 1 n a i + 1 i ! × ∏ i > 0 i c i \begin{aligned} \prod_{i=1}^{n}{\frac{a_i + 1}{i!}} \times \prod_{i>0}{i^{c_i}} \end{aligned} i=1∏ni!ai+1×i>0∏ici.
四、代码实现
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T& x)
{
x = 0; int f = 1; char ch = getchar();
while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
while(isdigit(ch)) {x = x * 10 + ch - '0', ch = getchar();}
x *= f;
}
template <typename T>
void print(T x)
{
if(x < 0) putchar('-'), x = -x;
if(x > 9) print(x / 10);
putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char ch)
{
print(x), putchar(ch);
}
typedef double db;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
template <ui M_> struct ModInt {
static constexpr ui M = M_;
ui x;
constexpr ModInt(): x(0U){}
constexpr ModInt(ui x_): x(x_ % M){}
constexpr ModInt(ull x_): x(x_ % M){}
constexpr ModInt(int x_): x(((x_ %= static_cast<int>(M)) < 0) ? (x_ + static_cast<int>(M)) : x_){}
constexpr ModInt(ll x_): x(((x_ %= static_cast<ll>(M)) < 0) ? (x_ + static_cast<ll>(M)) : x_){}
ModInt &operator+=(const ModInt &a) {x = ((x += a.x) >= M) ? (x - M) : x; return *this;}
ModInt &operator-=(const ModInt &a) {x = ((x -= a.x) >= M) ? (x + M) : x; return *this;}
ModInt &operator*=(const ModInt &a) {x = (static_cast<ull>(x) * a.x) % M; return *this;}
ModInt &operator/=(const ModInt &a) {return (*this *= a.inv());}
ModInt pow(ll e) const {
if(e < 0) return inv().pow(-e);
ModInt a = *this, b = 1U; for(; e; e >>= 1) {if(e & 1) b *= a; a *= a;} return b;
}
ModInt inv() const {
ui a = M, b = x; int y = 0, z = 1;
while(b) {const ui q = a / b; const ui c = a - q * b; a = b; b = c; const int w = y - static_cast<int>(q) * z; y = z; z = w;}
return ModInt(y);
}
ModInt operator+() const {return *this;}
ModInt operator-() const {ModInt a; a.x = x ? (M - x) : 0U; return a;}
ModInt operator+(const ModInt &a) const {return (ModInt(*this) += a);}
ModInt operator-(const ModInt &a) const {return (ModInt(*this) -= a);}
ModInt operator*(const ModInt &a) const {return (ModInt(*this) *= a);}
ModInt operator/(const ModInt &a) const {return (ModInt(*this) /= a);}
template <class T> friend ModInt operator+(T a, const ModInt &b) {return (ModInt(a) += b);}
template <class T> friend ModInt operator-(T a, const ModInt &b) {return (ModInt(a) -= b);}
template <class T> friend ModInt operator*(T a, const ModInt &b) {return (ModInt(a) *= b);}
template <class T> friend ModInt operator/(T a, const ModInt &b) {return (ModInt(a) /= b);}
explicit operator bool() const {return x;}
bool operator==(const ModInt &a) const {return (x == a.x);}
bool operator!=(const ModInt &a) const {return (x != a.x);}
friend std::ostream &operator<<(std::ostream &os, const ModInt &a) {return os << a.x;}
};
constexpr unsigned MO = 998244353U;
constexpr unsigned MO2 = 2U * MO;
constexpr int FFT_MAX = 23;
using Mint = ModInt<MO>;
constexpr Mint FFT_ROOTS[FFT_MAX + 1] = {1U, 998244352U, 911660635U, 372528824U, 929031873U, 452798380U, 922799308U, 781712469U, 476477967U, 166035806U, 258648936U, 584193783U, 63912897U, 350007156U, 666702199U, 968855178U, 629671588U, 24514907U, 996173970U, 363395222U, 565042129U, 733596141U, 267099868U, 15311432U};
constexpr Mint INV_FFT_ROOTS[FFT_MAX + 1] = {1U, 998244352U, 86583718U, 509520358U, 337190230U, 87557064U, 609441965U, 135236158U, 304459705U, 685443576U, 381598368U, 335559352U, 129292727U, 358024708U, 814576206U, 708402881U, 283043518U, 3707709U, 121392023U, 704923114U, 950391366U, 428961804U, 382752275U, 469870224U};
constexpr Mint FFT_RATIOS[FFT_MAX] = {911660635U, 509520358U, 369330050U, 332049552U, 983190778U, 123842337U, 238493703U, 975955924U, 603855026U, 856644456U, 131300601U, 842657263U, 730768835U, 942482514U, 806263778U, 151565301U, 510815449U, 503497456U, 743006876U, 741047443U, 56250497U, 867605899U};
constexpr Mint INV_FFT_RATIOS[FFT_MAX] = {86583718U, 372528824U, 373294451U, 645684063U, 112220581U, 692852209U, 155456985U, 797128860U, 90816748U, 860285882U, 927414960U, 354738543U, 109331171U, 293255632U, 535113200U, 308540755U, 121186627U, 608385704U, 438932459U, 359477183U, 824071951U, 103369235U};
void fft(Mint *as, int n) {
int m = n;
if(m >>= 1) {
for(int i = 0; i < m; ++i) {
const ui x = as[i + m].x;
as[i + m].x = as[i].x + MO - x;
as[i].x += x;
}
}
if(m >>= 1) {
Mint prod = 1U;
for(int h = 0, i0 = 0; i0 < n; i0 += (m<<1)) {
for(int i = i0; i < i0 + m; ++i) {
const ui x = (prod * as[i + m]).x;
as[i + m].x = as[i].x + MO - x;
as[i].x += x;
}
prod *= FFT_RATIOS[__builtin_ctz(++h)];
}
}
while(m) {
if(m >>= 1) {
Mint prod = 1U;
for(int h = 0, i0 = 0; i0 < n; i0 += (m<<1)) {
for(int i = i0; i < i0 + m; ++i) {
const ui x = (prod * as[i + m]).x;
as[i + m].x = as[i].x + MO - x;
as[i].x += x;
}
prod *= FFT_RATIOS[__builtin_ctz(++h)];
}
}
if(m >>= 1) {
Mint prod = 1U;
for(int h = 0, i0 = 0; i0 < n; i0 += (m<<1)) {
for(int i = i0; i < i0 + m; ++i) {
const ui x = (prod * as[i + m]).x;
as[i].x = (as[i].x >= MO2) ? (as[i].x - MO2) : as[i].x;
as[i + m].x = as[i].x + MO - x;
as[i].x += x;
}
prod *= FFT_RATIOS[__builtin_ctz(++h)];
}
}
}
for(int i = 0; i < n; ++i) {
as[i].x = (as[i].x >= MO2) ? (as[i].x - MO2) : as[i].x;
as[i].x = (as[i].x >= MO) ? (as[i].x - MO) : as[i].x;
}
}
void invFft(Mint *as, int n) {
int m = 1;
if(m < n>>1) {
Mint prod = 1U;
for(int h = 0, i0 = 0; i0 < n; i0 += (m<<1)) {
for(int i = i0; i < i0 + m; ++i) {
const ull y = as[i].x + MO - as[i + m].x;
as[i].x += as[i + m].x;
as[i + m].x = (prod.x * y) % MO;
}
prod *= INV_FFT_RATIOS[__builtin_ctz(++h)];
}
m <<= 1;
}
for(; m < n>>1; m <<= 1) {
Mint prod = 1U;
for(int h = 0, i0 = 0; i0 < n; i0 += (m<<1)) {
for(int i = i0; i < i0 + (m>>1); ++i) {
const ull y = as[i].x + MO2 - as[i + m].x;
as[i].x += as[i + m].x;
as[i].x = (as[i].x >= MO2) ? (as[i].x - MO2) : as[i].x;
as[i + m].x = (prod.x * y) % MO;
}
for(int i = i0 + (m>>1); i < i0 + m; ++i) {
const ull y = as[i].x + MO - as[i + m].x;
as[i].x += as[i + m].x;
as[i + m].x = (prod.x * y) % MO;
}
prod *= INV_FFT_RATIOS[__builtin_ctz(++h)];
}
}
if(m < n) {
for(int i = 0; i < m; ++i) {
const ui y = as[i].x + MO2 - as[i + m].x;
as[i].x += as[i + m].x;
as[i + m].x = y;
}
}
const Mint invN = Mint(n).inv();
for(int i = 0; i < n; ++i) {
as[i] *= invN;
}
}
void fft(vector<Mint> &as) {
fft(as.data(), as.size());
}
void invFft(vector<Mint> &as) {
invFft(as.data(), as.size());
}
vector<Mint> convolve(vector<Mint>& as, vector<Mint>& bs) {
if(as.empty() || bs.empty()) return {};
const int len = as.size() + bs.size() - 1;
int n = 1;
for(; n < len; n <<= 1) {}
as.resize(n); fft(as);
bs.resize(n); fft(bs);
for(int i = 0; i < n; ++i) as[i] *= bs[i];
invFft(as);
as.resize(len);
return as;
}
const int M = (int)5e5;
const int N = (int)1e5;
const int S = (int)1e6;
const int inf = 0x3f3f3f3f;
const ll mod = (ll)998244353;
int fac[M + 5], inv[M + 5], invfac[M + 5];
void init()
{
fac[0] = fac[1] = inv[1] = invfac[0] = invfac[1] = 1;
for(int i = 2; i <= M; ++i)
{
fac[i] = 1ll * fac[i - 1] * i % mod;
inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
invfac[i] = 1ll * invfac[i - 1] * inv[i] % mod;
}
}
vector<Mint> as, bs;
ll quick(ll a, ll b, ll p = mod)
{
ll s = 1;
while(b)
{
if(b & 1) s = s * a % p;
a = a * a % p;
b >>= 1;
}
return s % p;
}
void work()
{
int n; read(n);
int ans = 1;
as.resize(S + 1), bs.resize(S + 1);
for(int i = 1, a; i <= n; ++i)
{
read(a);
as[a] = 1, bs[S - a] = 1;
ans = 1ll * ans * (a + 1) % mod * invfac[i] % mod;
}
convolve(as, bs);
int k = as.size();
for(int i = S + 1; i < k; ++i) if(as[i].x) ans = 1ll * ans * quick(i - S, as[i].x) % mod;
print(ans, '\n');
}
int main()
{
// ios::sync_with_stdio(0);
// cin.tie(0); cout.tie(0);
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
// int T; read(T);
// for(int ca = 1; ca <= T; ++ca)
// {
// work();
// }
init();
work();
// cerr << 1.0 * clock() / CLOCKS_PER_SEC << "\n";
return 0;
}
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