[2020牛客暑期多校训练营第一场] J题 Easy integration

题目链接：Easy Integration

题意：输入n，求解 $\int_0^1 {({x-x^2})^n }\,{\rm d}x的值，对998244353取模。$

题解：多次运用分部积分能找到规律，猜出答案，答案正确性可通过数学分析法求证。
$\int_0^1 {({x-x^2})^n }\,{\rm d}x$

⇒ $\int_0^1 {x^n*{(1-x)^n} }\,{\rm d}x$

= $\frac{x^{n+1}}{n+1}*(1-x)^n|^1_0-\int_0^1 (\frac{x^{n+1}}{n+1}*[(1-x)^n]') \,{\rm d}x$

= $\frac{n*\int_0^1[x^{n+1}*(1-x)^{n-1}]dx}{n+1}$
…

= $\frac{n*(n-1)*(n-2)*....*1*\int_0^1x^{2n}dx}{(n+1)*(n+2)*(n+3)*...*(2n)}$

= ${\frac{n!}{(n+1)*(n+2)*.....*(2n+1)}}$

= ${\frac{(n!)^2}{(2n+1)!}}$

对(2n+10)阶乘进行预处理，O(1)的时间复杂度可得结果，取模时用逆元即可。

#include <iostream>
#include <algorithm>
#include <sstream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
const long double PI = acos(-1);
const int maxn=1e5+5;
const int mod=998244353;
const int N = 2e6+10;
	
ll qpow(ll a,ll b)
{
	ll ans=1;
	while(b)
	{
		if(b&1) ans=ans*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ans%mod;
}

ll inv(ll a)
{ return (qpow(a, mod-2)+mod)%mod; }

ll f[N+10];
void init()
{
	f[0]=1;
	for(int i=1;i<N;i++)
		f[i]=i*f[i-1]%mod;
}

int main()
{
	int n;
	init();	
	while(scanf("%d",&n)!=EOF)
	{
		ll tmp=1;
		ll ans=(f[n]*f[n]%mod*inv(f[2*n+1])+mod)%mod;
		printf("%lld\n",ans);
	}
}