题面:
题意:
求给定的定积分。
题解,化成 ∫ xn (1-x)n dx 然后用分部积分法即可得。
分部积分法:∫ udv = uv - ∫ vdu
最终为 n!/((n+1)*(n+2) * … * (2n+1) )
代码:
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#include<bitset>
#include<map>
#include<unordered_map>
#include<set>
#define ui unsigned int
#define ll long long
#define llu unsigned ll
#define ld long double
#define pr make_pair
#define pb push_back
//#define lc (cnt<<1)
//#define rc (cnt<<1|1)
#define len(x) (t[(x)].r-t[(x)].l+1)
#define tmid ((l+r)>>1)
using namespace std;
const int inf=0x3f3f3f3f;
const ll lnf=0x3f3f3f3f3f3f3f3f;
const double dnf=1e18;
const int mod=998244353;
const double eps=1e-8;
const double pi=acos(-1.0);
const int hp=13331;
const int maxn=2000100;
const int maxm=100100;
const int up=1000;
ll fac[maxn];
ll mypow(ll a,ll b)
{
ll ans=1;
while(b)
{
if(b&1) ans=ans*a%mod;
a=a*a%mod;
b>>=1;
}
return ans;
}
int main(void)
{
fac[0]=1;
for(int i=1;i<maxn;i++)
fac[i]=fac[i-1]*i%mod;
int n;
while(scanf("%d",&n)!=EOF)
{
printf("%lld\n",fac[n]*mypow(fac[n*2+1]*mypow(fac[n],mod-2)%mod,mod-2)%mod);
}
return 0;
}