Easy Integration

文章目录

• 题目大意
• 输入描述:
• 输出描述:
• 输入样例
• 输出样例

Easy Integration

题目大意

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 ($p \cdot q^{-1}) \bmod$

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 $\leq n \leq 10^6$
• 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.

输入样例

1
2
3

输出样例

166374059
432572553
591816295

For n = 1, $\int_{0}^1 (x - x^2) \mathrm{d} x = \frac{x^2}{2} - \frac{x^3}{3} |_0^1$ = $\frac{1}{6}$
求一下逆元
X=166374059,a=1,b=6;
a/b mod 998244353==X
a mod 998244353 == b*X mod 998244353
oeis

#include <cstdio>
#include <algorithm>
#define rep(i,a,b) for(auto i=a;i<=b;++i)
using namespace std;
typedef long long ll;
const int mod=998244353,maxn=1e6+7;
ll a[maxn<<1],b[maxn<<1],inv[maxn<<1];
int main()
{
    int n;
    a[0]=b[0]=b[1]=inv[1]=inv[0]=1;
    rep(i,1,maxn<<1){
        a[i]=a[i-1]*i%mod;
        if(i>1) inv[i]=(mod-mod/i)*inv[mod%i]%mod,b[i]=b[i-1]*inv[i]%mod;
    }
    while(~scanf("%d",&n))
        printf("%lld\n",((a[n]%mod*a[n]%mod*b[(n<<1)+1]%mod)+mod)%mod);
    return 0;
}