bzoj1485 有趣的数列

传送门：https://www.lydsy.com/JudgeOnline/problem.php?id=1485

【题解】

Catalan数，注意不能直接用逆元，需要分解质因数。

# include <bits/stdc++.h>
using namespace std;

const int M = 5e5 + 10, N = 2e6 + 10;

int isp[N];
int p[M], pn;
int n, mod;
int times[M];

inline int pwr(int a, int b) {
    int ret = 1;
    while(b) {
        if(b&1) ret = 1ll * ret * a % mod;
        a = 1ll * a * a % mod;
        b >>= 1;
    }
    return ret;
}

int main() {
    cin >> n >> mod;
    for (int i=2; i<=n+n; ++i) isp[i] = 1;
    for (int i=2; i<=n+n; ++i) {
        if(isp[i]) {
            p[++pn] = i;
            isp[i] = pn;
            for (int j=i+i; j<=n+n; j+=i)
                isp[j] = 0;
        }
    }
    for (int i=n+1; i<=n+n; ++i) {
        int t = i;
        for (int j=1; j<=pn; ++j) {
            if(t % p[j] == 0) 
                while(t % p[j] == 0) ++times[j], t /= p[j];
            if(t == 1) break;
            if(isp[t]) {
                ++ times[isp[t]];
                break;
            }
        }
    }
//    for (int i=1; i<=5; ++i) 
//        cout << p[i] << ' ' << times[i] << endl;
    for (int i=2; i<=n+1; ++i) {
        int t = i;
        for (int j=1; j<=pn; ++j) {
            if(t % p[j] == 0) 
                while(t % p[j] == 0) --times[j], t /= p[j];
            if(t == 1) break;
            if(isp[t]) {
                -- times[isp[t]];
                break;
            }
        }
    }
    
    int ans = 1;
    for (int i=1; i<=pn; ++i) 
        if(times[i]) ans = 1ll * ans * pwr(p[i], times[i]) % mod;
    
    cout << ans;
    
    return 0;
}