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 + , N = 2e6 + ;

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

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

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

     int ans = ;
     for (int i=; i<=pn; ++i)
         if(times[i]) ans = 1ll * ans * pwr(p[i], times[i]) % mod;

     cout << ans;

     return ;
 }