tyvj/joyoi 1374 火车进出栈问题(水水版)

我受不了了。

Catalan数第100项，30000项，50000项，cnm

这tm哪里是在考数学，分明是在考高精度，FFT......

有剧毒！

我只得写高精度，只能过100的那个题，两个进化版超时......

 #include <cstdio>
 #include <string>
 using namespace std;
 const int N = ;

 int p[N], tp, sum[N];
 bool vis[N];

 struct LL {
     string n;
     LL operator * (const long long &x) const {
         int a[N];
         int t = n.size();
         for(int i = ; i < t; i++) {
             a[t - i - ] = ((int)(n[i] - '')) * x;
         }
         for(int i = ; i < t; i++) {
             if(a[i] > ) {
                 a[i + ] += a[i] / ;
                 a[i] %= ;
                 if(i +  == t) t++;
             }
         }
         string f = "";
         for(int i = ; i < t; i++) {
             f = (char)(a[i] + '') + f;
         }
         LL ans;
         ans.n = f;
         return ans;
     }
     void out() {
         for(int i = ; i < n.size(); i++) {
             putchar(n[i]);
         }
         return;
     }
 };

 void getprime(int b) {
     for(int i = ; i <= b; i++) {
         if(!vis[i]) {
             p[++tp] = i;
         }
         for(int j = ; j <= tp && p[j] * i <= b; j++) {
             vis[p[j] * i] = ;
             if(i % p[j] == ) {
                 break;
             }
         }
     }
     return;
 }

 inline long long pow(int a, int b) {
     if(b < ) {
         printf("ERROR!!!\n");
         return ;
     }
     long long ans = ;
     while(b) {
         if(b & ) {
             ans = ans * a;
         }
         a = a * a;
         b = b >> ;
     }
     return ans;
 }

 int main() {
     int n;
     scanf("%d", &n);
     getprime(n << );
     int n2 = n << ;

     for(int i = ; i <= tp; i++) {
         int _i = p[i];
         while(_i <= n2) {
             sum[i] += n2 / _i;
             _i *= p[i];
         }
     }
     for(int i = ; i <= tp && p[i] <= n; i++) {
         int _i = p[i];
         while(_i <= n) {
             sum[i] -= n / _i;
             _i *= p[i];
         }
     }
     n++;
     for(int i = ; i <= tp && p[i] <= n; i++) {
         int _i = p[i];
         while(_i <= n) {
             sum[i] -= n / _i;
             _i *= p[i];
         }
     }

     LL ans;
     ans.n = '';
     for(int i = ; i <= tp; i++) {
         ans = ans * pow(p[i], sum[i]);
     }
     ans.out();
     return ;
 }

AC代码