J-Easy Intergration

链接：https://ac.nowcoder.com/acm/contest/5666/J
来源：牛客网

题目描述:

Given n,find the value of
It can be proved that the value is a rational number .
Print the result as ( ) mod 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≤n≤1e6
The number of test cases does not exceed 1e5

输出描述：

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

solution:

先将函数变成x*(1-x)，然后将(1-x)的n次二项式展开，对展开后的函数进行积分，就可以将函数积分出来的公式写出来，然后进行找规律，找出的规律是在前一项上除以一个数，再乘以两个数（具体什么我已经忘了），找规律过程中不要化简，不然找不到规律

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
const int mod=998244353;
ll extend_gcd(ll a,ll b, ll&x,ll &y)
{
    if(a==0&&b==0)return -1;
    if(b==0){x=1;y=0;return a;}
    ll d=extend_gcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}
ll mod_reverse(ll a,ll n)
{
    ll x,y;
    ll d=extend_gcd(a,n,x,y);
    if(d==1)return (x%n+n)%n;
    else return -1;
}
ll a[2000005];
ll b[2000005];
ll c[2000005];
ll t=4;
int main()
{
    a[1]=1;
    b[1]=6;
    c[1]=1;
    for(int i=2;i<=1000000;i++)
    {
        a[i]=a[i-1]*i%mod;//cout<<a[i]<<endl;
        b[i]=b[i-1]*(t*(t+1)%mod)%mod;
        c[i]=c[i-1]*i%mod;
        t+=2;
    }
    while(~scanf("%d",&n))
    {
        ll s=mod_reverse(b[n],mod)*(a[n]*c[n]%mod)%mod;
        printf("%lld\n",s);
    }
}