BZOJ 2822 AHOI 2012 树屋阶梯 卡特兰数＋高精度

题目大意：高精度卡特兰数。

思路：上维基上看看，有一个模型和这个题一模一样，然后就剩下水水的高精度了。

（谁来教教我java...

CODE：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define BASE 10000#define MAX 100010using namespace std; struct BigInt{    int num[MAX],len;         BigInt(int _ = 0) {        num[len = 1] = _;    }    BigInt operator +(const BigInt &a)const {        BigInt re;        re.len = max(len,a.len);        int temp = 0;        for(int i = 1; i <= a.len; ++i) {            re.num[i] = temp + num[i] + a.num[i];            temp = re.num[i] / BASE;            re.num[i] %= BASE;        }        if(temp)    re.num[++re.len] = temp;        return re;    }    BigInt operator *(int a)const {        BigInt re;        re.len = len;        int temp = 0;        for(int i = 1; i <= len; ++i) {            re.num[i] = temp + num[i] * a;            temp = re.num[i] / BASE;            re.num[i] %= BASE;        }        while(temp) re.num[++re.len] = temp % BASE,temp /= BASE;        return re;    }    BigInt operator /(int a)const {        BigInt re;        re.len = len;        int temp = 0;        for(int i = len; i; --i) {            re.num[i] = (temp + num[i]) / a;            temp = (temp + num[i]) % a * BASE;        }        while(!re.num[re.len])  --re.len;        return re;    }}; int main(){    int x;    cin >> x;    BigInt ans(1);    for(int i = 2; i <= x; ++i)        ans = ans * (4 * i - 2) / (i + 1);    printf("%d",ans.num[ans.len]);    for(int i = ans.len - 1; i; --i)        printf("%04d",ans.num[i]);    return 0;}