2020牛客暑期多校训练营（第一场）J.Easy Integration

2020牛客暑期多校训练营（第一场）J.Easy Integration

题目链接

题目描述

Given n, find the value of ${\int }_0^1(x-x^2{)}^n\mathrm{d}x$
It can be proved that the value is a rational number $\frac{p}{q}$.
Print the result as $(p\cdot q^{-1})\mathrm{m}\mathrm{o}\mathrm{d}998244353$.

输入描述:

The input consists of several test cases and is terminated by end-of-file.
Each test case contains an integer n.

$1\le n\le 10^6$

The number of test cases does not exceed $10^5$

输出描述:

For each test case, print an integer which denotes the result.

示例1

输入

1
2
3

输出

166374059
432572553
591816295

这道题比较考验思维，不是说你求积的思维而是你找答案的思维，坦白说，这个定积分我一个高数上下满绩，数竞二等奖的人都不会，你们就千万别傻傻地死磕积分妄图求原函数了(数学大佬当我没说)，赶紧，赶紧，赶紧找规律！！！
把 $n=1,2,3,4$ 时的定积分一求，分子为 $1$，分母为 $6,30,140,630$，然后赶紧上 $OEIS$ 搜一下公式，发现就是 $\frac{(2n+1)!}{(n!{)}^2}$，所以说网络赛脑子要灵光一些，学会利用网络的优势，输出就是求个逆元，阶乘预处理一下即可，AC代码如下：

#include
using namespace std;
typedef long long  ll;
const ll N=2e6+5;
const ll mod=998244353;
ll power(ll a,ll b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}
ll f[N],n;
void pre(){
    f[0]=1;
    for(ll i=1;i<=N;i++){
        f[i]=i*f[i-1]%mod;
    }
}

int main(){
    pre();
    while(~scanf("%lld",&n)){
        printf("%lld\n",power(f[2*n+1]*power(power(f[n],2),mod-2)%mod,mod-2));
    }
	return 0;
}