Trees Made to Order ZOJ - 1062

题目链接:QAQ

题意：按题目意思构造顺序，输出第n个二叉树

思路：n个结点组成的二叉树个数满足卡特兰数，这样就可以找到第n个树含几个节点，然后递归构造即可

附上代码：

#include<cstdio>
long long num[25];
long long s[25];
void solve(int um, int x) {
	if (x == 1) {
		printf("X");
		return;
	}
	if (x == 0) {
		return;
	}
	int num1=um;
	int i, j;
	for (i = x-1, j = 0; i >= 0; i--, j++) {
		if (num1 > num[i] * num[j]) {
			num1 -= num[i] * num[j];
		}
		else
			break;
	}
	num1--;
	if (j != 0) {
		printf("(");
		solve(num1 / num[i]+1, j);//注意这里要加一
		printf(")");
	}
	printf("X");
	if (i != 0) {
		printf("(");
		solve(num1%num[i]+1, i);
		printf(")");
	}
	return;
}
int main(void) {
	long long tot;
	num[0] = 1;num[1] = 1;num[2] = 2;num[3] = 5;
	s[0] = 0; s[1] =  1; s[2] = 3; s[3] = 8;
	for (int z = 4; z <= 19; z++) {
		tot = 0;
		for (int i = z-1, j = 0; i >= 0; i--, j++) {
			tot += num[i] * num[j];
		}
		num[z] = tot;
		s[z] = s[z - 1] + tot;
	}
	int n;
	while (scanf("%d", &n) != EOF) {
		if (n == 0) break;
		int i;
		for (i = 1; i <= 19; i++) {
			if (s[i] >= n) break;
		}
		solve(n-s[i-1], i);
		printf("\n");
	}
	return 0;
}