【2020年牛客暑假第一场】J题 Easy integration

题目链接：https://ac.nowcoder.com/acm/contest/5666/J

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

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.

*$1\le n\le 10^6$.
*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.

输入
$1$
$2$
$3$

输出
$166374059$
$432572553$
$591816295$

思路

根据${\int }_0^1(x-x^2{)}^{n}dx$可得${\int }_0^1{x}^n(1-x{)}^{n}dx$。
利用n次分部积分法可得计算式$$\frac{1\ast 2\ast ...\ast (n-1)\ast n}{(n+1)\ast (n+2)\ast ....\ast (2n)\ast (2n+1)}$$
上下同乘$n!$得
$$\frac{(n!{)}^2}{(2n+1)!}$$

先对阶乘做一个预处理，然后就成分子除以分母，可以用逆元（费马小定理）求。

Code

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;

#define INF 0x7f7f7f
#define mem(a,b) memset(a , b , sizeof(a))
#define FOR(i, x, n) for(int i = x;i <= n; i++)
//const ll mod = 1e9 + 7;
// const int maxn = 1e5 + 10;

const ll mod = 998244353;
int n;

ll f[2000005];

ll FPM(ll x,ll power)
{
    
    
    ll ans = 1;
    while(power)
    {
    
    
        if(power & 1)
        {
    
    
            ans = (ans * x) % mod;
        }
        x = (x * x) % mod;
        power >>= 1;
    }
    return ans % mod;
}

void F()
{
    
    
    f[0] = 1;
    for(int i = 1;i <= 2000001; i++)
    {
    
    
        f[i] = f[i - 1] * i % mod;
    }
}

void solve()
{
    
    
    F();
    while(scanf("%d",&n)!=EOF)
    {
    
    
        ll ans = f[n] * f[n] % mod;
        ans = (ans * FPM(f[2 * n + 1], mod - 2)) % mod;
        printf("%lld\n",ans % mod);
    }
}

signed main()
{
    
    
    //ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
#ifdef FZT_ACM_LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#else
    //ios::sync_with_stdio(false);
    int T = 1;
    //cin >> T;
    while(T--)
        solve();
#endif
    return 0;
}