BZOJ 2287. [HZOI 2015]疯狂的机器人 [FFT 组合计数]

2287. [HZOI 2015]疯狂的机器人

题意：从原点出发，走n次，每次上下左右不动，只能在第一象限，最后回到原点方案数

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这不煞笔提，组合数写出来发现卷积NTT，然后没考虑第一象限gg




其实就是卡特兰数

只不过这里\(C(i)\)是第\(\frac{i}{2}\)项，奇数为0

令\(f[n]\)为走n次回到原点方案数，$$

f[n]=\sum_{i=0}^{{n}C(i)C(n-i)\binom{n}{i}=n!\sum_{i=0}}{n}C(i)\frac{1}{i!}C(n-i)\frac{1}{(n-i)!}

\[
注意阶乘和阶乘逆元别乘错了，别丢东西！！！

```cpp
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N=(1<<18)+5, INF=1e9;
const ll P=998244353;
inline int read(){
char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}

ll Pow(ll a, ll b, ll P) {
ll ans=1;
for(; b; b>>=1, a=a*a%P)
if(b&1) ans=ans*a%P;
return ans;
}
namespace NTT{
int n, rev[N], g;
void ini(int lim) {
g=3;
n=1; int k=0;
while(n<lim) n<<=1, k++;
for(int i=0; i<n; i++) rev[i] = (rev[i>>1]>>1) | ((i&1)<<(k-1));
}
void dft(ll *a, int flag) {
for(int i=0; i<n; i++) if(i<rev[i]) swap(a[i], a[rev[i]]);
for(int l=2; l<=n; l<<=1) {
int m=l>>1;
ll wn=Pow(g, flag==1 ? (P-1)/l : P-1-(P-1)/l, P);
for(ll *p=a; p!=a+n; p+=l) {
ll w=1;
for(int k=0; k<m; k++) {
ll t = w*p[k+m]%P;
p[k+m] = (p[k] - t + P)%P;
p[k] = (p[k] + t)%P;
w = w*wn%P;
}
}
}
if(flag==-1) {
ll inv=Pow(n, P-2, P);
for(int i=0; i<n; i++) a[i] = a[i]*inv%P;
}
}
void mul(ll *a, ll *b) {
dft(a, 1);
for(int i=0; i<n; i++) a[i]=a[i]*a[i]%P;
dft(a, -1);
}
}using NTT::dft; using NTT::ini; using NTT::mul;

int n;
ll inv[N], fac[N], facInv[N];
ll a[N], b[N];
ll C(int n, int m) {return fac[n]*facInv[m]%P*facInv[n-m]%P;}
int main() {
//freopen("in","r",stdin);
freopen("crazy_robot.in","r",stdin);
freopen("crazy_robot.out","w",stdout);
n=read(); ini(n+n+1);
inv[1]=1; fac[0]=facInv[0]=1;
for(int i=1; i<=n; i++) {
if(i!=1) inv[i] = (P-P/i)*inv[P%i]%P;
fac[i] = fac[i-1]*i%P;
facInv[i] = facInv[i-1]*inv[i]%P;
}
a[0]=b[0]=1;
for(int i=2; i<=n; i+=2) a[i]=b[i]= C(i, i>>1) * inv[(i>>1)+1] %P * facInv[i] %P;

mul(a, b);
for(int i=0; i<=n; i++) a[i]=a[i]*fac[i]%P;
ll ans=0;
for(int m=0; m<=n; m+=2) (ans += C(n, m) * a[m]%P) %=P;
printf("%lld\n", ans);
}
```\]