Description
现在在二维平面内原点上有一只机器人,他每次可以选择向右走,向左走,向下走,向上走和不走(每次如果走只能走一格)。机器人不能走到横坐标是负数或者纵坐标是负数的点上。
给定操作次数 (n) ,求有多少种不同的操作序列使得机器人在操作后会回到原点,输出答案模 (998244353) 后的结果。
(1leq nleq 100000)
Solution
应该不难想吧...
显然我们先考虑前四种走法...不走的情况可以组合数求出来。
对于一类操作(向上向下或向左向右),显然是成组出现的。更具体地,这就是 (Catalan) 数。
记 (Catalan) 数的第 (i) 项为 (C_i) 。
记多项式
[A(x)=sum_{i=0}^infty frac{[2mid i]cdot C_{leftlfloorfrac{i}{2} ight floor}}{i!}x^i]
那么答案就是 (sum_{i=0}^n [2mid i]cdot i!cdot A^2(i)cdot {nchoose i}) 。 ( ext{NTT}) 优化即可 。
Code
#include <bits/stdc++.h>
using namespace std;
const int N = 4*100000, yzh = 998244353;
int n, inv[N+5], fac[N+5], ifac[N+5], a[N+5], len, R[N+5], L;
int quick_pow(int a, int b) {
int ans = 1;
while (b) {
if (b&1) ans = 1ll*ans*a%yzh;
b >>= 1, a = 1ll*a*a%yzh;
}
return ans;
}
void NTT(int *A, int o) {
for (int i = 0; i < len; i++) if (i < R[i]) swap(A[i], A[R[i]]);
for (int i = 1; i < len; i <<= 1) {
int gn = quick_pow(3, (yzh-1)/(i<<1)), x, y;
if (o == -1) gn = quick_pow(gn, yzh-2);
for (int j = 0; j < len; j += (i<<1)) {
int g = 1;
for (int k = 0; k < i; k++, g = 1ll*g*gn%yzh) {
x = A[j+k], y = 1ll*g*A[j+k+i]%yzh;
A[j+k] = (x+y)%yzh, A[j+k+i] = (x-y)%yzh;
}
}
}
}
int C(int n, int m) {return 1ll*fac[n]*ifac[m]%yzh*ifac[n-m]%yzh; }
void work() {
scanf("%d", &n); inv[0] = inv[1] = fac[0] = ifac[0] = 1;
for (int i = 1; i <= n; i++) fac[i] = 1ll*i*fac[i-1]%yzh;
for (int i = 2; i <= n+1; i++) inv[i] = -1ll*yzh/i*inv[yzh%i]%yzh;
for (int i = 1; i <= n; i++) ifac[i] = 1ll*inv[i]*ifac[i-1]%yzh;
for (int i = 0; i <= n; i += 2) a[i] = 1ll*C(i, i/2)*inv[i/2+1]%yzh*ifac[i]%yzh;
for (len = 1; len <= (n<<1); len <<= 1) ++L;
for (int i = 0; i < len; i++) R[i] = (R[i>>1]>>1)|((i&1)<<(L-1));
NTT(a, 1);
for (int i = 0; i < len; i++) a[i] = 1ll*a[i]*a[i]%yzh;
NTT(a, -1);
for (int i = 0, inv = quick_pow(len, yzh-2); i < len; i++)
a[i] = 1ll*a[i]*inv%yzh*fac[i]%yzh;
int ans = 0;
for (int i = 0; i <= n; i += 2)
(ans += 1ll*a[i]*C(n, i)%yzh) %= yzh;
printf("%d
", (ans+yzh)%yzh);
}
int main() {work(); return 0; }