POJ 1095 Trees Made to Order（计数问题）

Trees Made to Order

Description

We can number binary trees using the following scheme:

The empty tree is numbered 0.

The single-node tree is numbered 1.

All binary trees having m nodes have numbers less than all those having m+1 nodes.

Any binary tree having m nodes with left and right subtrees L and R is numbered n such that all trees having m nodes numbered > n have either Left subtrees numbered higher than L, or A left subtree = L and a right subtree numbered higher than R.

The first 10 binary trees and tree number 20 in this sequence are shown below:

Your job for this problem is to output a binary tree when given its order number.

Input

Input consists of multiple problem instances. Each instance consists of a single integer n, where 1 <= n <= 500,000,000. A value of n = 0 terminates input. (Note that this means you will never have to output the empty tree.)

Output

For each problem instance, you should output one line containing the tree corresponding to the order number for that instance. To print out the tree, use the following scheme:

A tree with no children should be output as X.

A tree with left and right subtrees L and R should be output as (L‘)X(R‘), where L‘ and R‘ are the representations of L and R.

If L is empty, just output X(R‘).

If R is empty, just output (L‘)X.

Sample Input

1
20
31117532
0

Sample Output

X
((X)X(X))X
(X(X(((X(X))X(X))X(X))))X(((X((X)X((X)X)))X)X)

Source

East Central North America 2001

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i < b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 20

int dp[N];

void inint()
{
	int i,j;
	dp[0]=dp[1]=1;
	for(i=2;i<N;i++)
	{
		fre(j,0,i)
		  dp[i]+=dp[j]*dp[i-1-j];  //i个节点的情况有列举左边放j，因为根节点需要一个，那么右边i-1-j
	}

}

void dfs(int n,int k)    //n个节点,第k大
{
	if(n==1)    //开始是k==1,一直错
	{
		pf("X");
		return ;
	}

	int i=0;
    for(i=0;;i++)
		if(dp[i]*dp[n-i-1]>=k) break;  //左边有dp[i]种情况,右边有dp[n-i-1],因为有一个点当根节点
	 else  k-=dp[i]*dp[n-i-1];

    if(i)       //左边有节点
	{
		pf("(");
		dfs(i,(k-1)/dp[n-i-1]+1);
		pf(")");
	}
	pf("X");
	if(n-i-1)    //右边有节点
	{
		pf("(");
		dfs(n-1-i,(k-1)%dp[n-1-i]+1);
		pf(")");
	}
}
int main()
{
	int i,j,n;
	inint();
	while(sf(n),n)
	{
		for(i=1;;i++)
		  if(dp[i]>=n) break; //已经锁定i个节点是满足条件的
		 else n-=dp[i];
		dfs(i,n);
      pf("\n");
	}
    return 0;
}