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 = 100010;

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 = 0; i < t; i++) {
            a[t - i - 1] = ((int)(n[i] - '0')) * x;
        }
        for(int i = 0; i < t; i++) {
            if(a[i] > 9) {
                a[i + 1] += a[i] / 10;
                a[i] %= 10;
                if(i + 1 == t) t++;
            }
        }
        string f = "";
        for(int i = 0; i < t; i++) {
            f = (char)(a[i] + '0') + f;
        }
        LL ans;
        ans.n = f;
        return ans;
    }
    void out() {
        for(int i = 0; i < n.size(); i++) {
            putchar(n[i]);
        }
        return;
    }
};

void getprime(int b) {
    for(int i = 2; i <= b; i++) {
        if(!vis[i]) {
            p[++tp] = i;
        }
        for(int j = 1; j <= tp && p[j] * i <= b; j++) {
            vis[p[j] * i] = 1;
            if(i % p[j] == 0) {
                break;
            }
        }
    }
    return;
}

inline long long pow(int a, int b) {
    if(b < 0) {
        printf("ERROR!!!\n");
        return 1;
    }
    long long ans = 1;
    while(b) {
        if(b & 1) {
            ans = ans * a;
        }
        a = a * a;
        b = b >> 1;
    }
    return ans;
}

int main() {
    int n;
    scanf("%d", &n);
    getprime(n << 1);
    int n2 = n << 1;

    for(int i = 1; i <= tp; i++) {
        int _i = p[i];
        while(_i <= n2) {
            sum[i] += n2 / _i;
            _i *= p[i];
        }
    }
    for(int i = 1; i <= tp && p[i] <= n; i++) {
        int _i = p[i];
        while(_i <= n) {
            sum[i] -= n / _i;
            _i *= p[i];
        }
    }
    n++;
    for(int i = 1; i <= tp && p[i] <= n; i++) {
        int _i = p[i];
        while(_i <= n) {
            sum[i] -= n / _i;
            _i *= p[i];
        }
    }

    LL ans;
    ans.n = '1';
    for(int i = 1; i <= tp; i++) {
        ans = ans * pow(p[i], sum[i]);
    }
    ans.out();
    return 0;
}