2020牛客暑假多校第一场 【J题 Easy Integration】

题意：

思路：

开始是把式子用二项式定理展开，之后各项积分，复杂度为$O(n)$O(n).之后发现多组样例，时间复杂度就不可行了。之后就上OEIS了。。。
赛后参照网上网上的题解，需要用$n$n次分部积分。
分部积分公式：

$n$n次之后$(1 - x)$(1−x)的指数就变成了$0$0，那么积分就是$1$1了，然后只要考虑前面的系数就可以了。

/*
 * @file J.cpp
 * @path D:\code\ACM\牛客多校\第一场\J.cpp
 * @author Xiuchen
 * @date  2020-07-12 13:50:22
*/

#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<cmath>
#include<math.h>
#include<iostream>
#include<algorithm>
//#define DEBUG
#define dbg(x) cout << #x << " = "<< (x) << endl
#define dbg2(x1,x2) cout << #x1 << " = " << x1 << " " << #x2 << " = " << x2 << endl
#define dbg3(x1,x2,x3) cout<< #x1 << " = " << x1 << " " << #x2 << " = " << x2 << " " << #x3 << " = " << x3 <<endl
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fLL;
const ll mod = 998244353;
const int maxn = 2e6 + 100;
int gcd(int a, int b){
    return b ? gcd(b, a % b) : a;
}
ll qpow(ll a, ll b){
    ll ans = 1, base = a;
    while(b){
        if(b & 1) ans = ans * base % mod;
        base = base * base % mod;
        b >>= 1;
    }
    return ans;
}
int n;
ll jc[maxn], jc_inv[maxn]; 
ll inv[maxn], preans[maxn];
int main(){
#ifdef DEBUG
    freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
    jc[0] = 1, jc_inv[0] = 1;
    for(int i = 1; i <= 2000001; i++) inv[i] = qpow(i, mod - 2);
    for(int i = 1; i <= 2000001; i++) jc[i] = jc[i - 1] * i % mod;
    for(int i = 1; i <= 2000001; i++) jc_inv[i] = jc_inv[i - 1] * inv[i] % mod;
    // memset(preans, -1, sizeof(preans));

    while(~scanf("%d", &n)){
        ll ans = jc[n] * jc[n] % mod;
        ans = ans * jc_inv[2 * n + 1] % mod;
        printf("%lld\n", ans);
    }
    return 0;
}