卡特兰数取模

普通递推的方法不能用来取模，因为(4*n-2)/(n+1)可能是小数。

可以利用另外一个公式：(2*n)!/n!*(n+1)!

可以利用素数的幂来求阶乘，然后就可以在计算的过程中取模。
http://suo.im/cj3mc

#include 
#include 
#include 
#include 

using namespace std;

typedef long long ll;

int dp[1000005];
bool isprime[1000005];
ll prime[200005], top;

void Prime(int n)
{
    top = 0;
    memset(isprime, true, sizeof(isprime));
    for(int i = 2; i < n; i++){
        if(isprime[i])
            prime[top++] = i;
        for(int k = 0; k < top && i * prime[k] < n; k++){
            isprime[i*prime[k]] = false;
            if(i % prime[k] == 0)
                break;
        }
    }
}

ll quickpow(ll m, ll n, ll mod)
{
    ll b = 1;
    while(n){
          if (n & 1)
             b = (b * m) % mod;
          n = n >> 1 ;
          m = (m * m) % mod;
    }
    return b;
}

void fac_1(ll n)
{
    ll len = -1;
    while(prime[++len] <= n){
        ll sum = 0;
        dp[prime[len]] = 0;
        ll num = prime[len];
        while(num <= n){
            sum += n / num;
            num = num * prime[len];
        }
        dp[prime[len]] += sum;
    }
}

void fac_2(ll n)
{
    ll len = -1;
    while(prime[++len] <= n){
        ll sum = 0;
        ll num = prime[len];
        while(num <= n){
            sum += n / num;
            num = num * prime[len];
        }
        dp[prime[len]] -= sum;

    }
}

int main()
{
    ll n, m;
    Prime(1000000);
    while(cin>>n>>m){
        n = n - 2;
        if(m == 1){
            cout<<"0"<continue;
        }
        if(n == 1){
            cout<<"1"<continue;
        }
        fac_1(2 * n);
        fac_2(n);
        fac_2(n+1);
        int resl = 1;
        for(int i = 0; prime[i] <= 2*n; i++){
            if(dp[prime[i]])
                resl = (resl * quickpow(prime[i], dp[prime[i]], m)) % m;
        }
        cout<return 0;
}