AcWing 889. 满足条件的01序列 卡特兰数

Posted 2021-06-02 幽殇默

https://www.acwing.com/problem/content/891/
$$卡特兰数的公式：C_{2n}^n-C_{2n}^{n-1}=\frac{C_{2n}^n}{n+1}$$

上图摘自：https://www.acwing.com/solution/content/8907/

#include<cstdio>
#include<iostream>
using namespace std;
const int mod=1e9+7; 
typedef  long long int LL;
LL qmi(LL a,LL b, LL p)
{
	LL res=1;
	while(b)
	{
		if(b&1) res=res*a%mod;
		a=a*a%mod;
		b=b>>1;
	}
	return res;
}
int main(void)
{
	LL n; cin>>n;
	LL sum=1;
	for(int i=1;i<=n;i++)// n! 
	{
		sum=sum*i%mod;
	}
	LL temp=sum*sum%mod;
	for(int i=n+1;i<=2*n;i++)// (2n)!
	{
		sum=sum*i%mod;
	}
	cout<<sum*qmi(temp,mod-2,mod)%mod*qmi(n+1,mod-2,mod)%mod<<endl;
	return 0;
} 

y总代码：

#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 100010, mod = 1e9 + 7;


int qmi(int a, int k, int p)
{
    int res = 1;
    while (k)
    {
        if (k & 1) res = (LL)res * a % p;
        a = (LL)a * a % p;
        k >>= 1;
    }
    return res;
}


int main()
{
    int n;
    cin >> n;

    int a = n * 2, b = n;
    int res = 1;
    for (int i = a; i > a - b; i -- ) res = (LL)res * i % mod;

    for (int i = 1; i <= b; i ++ ) res = (LL)res * qmi(i, mod - 2, mod) % mod;

    res = (LL)res * qmi(n + 1, mod - 2, mod) % mod;

    cout << res << endl;

    return 0;
}