UVA 10303 How Many Trees?

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题意：设f[i]表示一共有i个元素时二叉搜索树的个数，那么依次取1～n-1作为根节点，那么左右子树元素的个数就分别是(0，n-1),......,所以f[n] = f[0]*f[n-1]+f[1]*f[n-2]...+f[n-1]f[0],其实也就是Catalan数，高精度的计算，递推公式是f[n]=(4n-2)/(n+1)*f[n-1]

#include <iostream>#include <cstring>#include <cmath>#include <cstdio>#include <algorithm>using namespace std;const int MAXN = 4000;int f[MAXN][MAXN];int l[MAXN];void div(int x){    int *t = f[x],y=x+1,cur=0;    for (int i = l[x]-1; i >= 0; i--){        cur = cur*10+t[i];        t[i] = cur / y;        cur %= y;    }    int j;    for (j = l[x]-1; j > 0 && !t[j]; j--);    l[x] = j + 1; }void mul(int x){    int &i = l[x],c=0,y=4*x-2;    int *p=f[x-1],*t=f[x];    for (i = 0; i < l[x-1] || c; i++){        c += p[i] * y;        t[i] = c % 10;        c /= 10;    }}void print(int x){    int *p=f[x];    for (int i = l[x]-1; i >= 0; i--)        printf("%d",p[i]);    printf("\n");}int main(){    int n;    memset(f,0,sizeof(f));    f[1][0] = 1;    l[1] = 1;    for (int i = 2; i <= 1000; i++){        mul(i);        div(i);    }    while (scanf("%d",&n) != EOF){        print(n);    }    return 0;}