火车进出栈问题

Posted 2021-03-05 hhyx

# 题意

1~n 分别进栈，出栈序列可能有多少种

# 题解

进出栈序列即catalan数

C(n,2n)/n+1

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <vector>
 5 using namespace std;
 6 typedef long long LL;
 7 const int N=120010;
 8 LL primes[N],cnt,sum[N];
 9 LL ans[N],tt;
10 bool st[N];
11 LL n;
12 void get_primes(LL n){
13     for (int i = 2; i <=n ; ++i) {
14         if(!st[i]) primes[cnt++]=i;
15         for (int j = 0; primes[j]<=n/i ; ++j) {
16             st[primes[j]*i]=true;
17             if(i%primes[j]==0) break;
18         }
19     }
20 }
21 int get(LL n,LL p){
22     int res=0;
23     while(n){
24         res+=n/p;
25         n/=p;
26     }
27     return res;
28 }
29 void multi(int b)
30 {
31     LL t = 0;
32     for (int i = 0; i <= tt; i ++ )
33     {
34         ans[i] = ans[i] * b + t;
35         t = ans[i] / 1000000000;
36         ans[i] %= 1000000000;
37     }
38     while (t)
39     {
40         ans[++tt] = t % 1000000000;
41         t /= 1000000000;
42     }
43 }
44 void out()
45 {
46     printf("%lld", ans[tt]);
47     for (int i = tt - 1; i >= 0; i -- ) printf("%09lld", ans[i]);
48     cout << endl;
49 }
50 int main(){
51     scanf("%lld",&n);
52     get_primes(2*n);
53 
54     for (int i = 0; i <cnt ; ++i) {
55         LL p=primes[i];
56         sum[i]=get(2*n,p)-get(n,p)-get(n+1,p);
57     }
58     ans[0]=1;
59     for (int i = 0; i <cnt ; ++i)
60         for (int j = 1; j <=sum[i] ; ++j)
61             multi(primes[i]);
62 
63     out();
64 }