FZU 1064 教授的测试

递归构造答案。

根据当前整颗树的编号，可以计算左右子树有几个节点以及编号。因此，不断dfs下去就可以了。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

long long c[] = { , , , , , , , , , , , ,
, , , , ,  };
long long n;

void dfs(long long num, long long p)
{
    long long left=, right=, sum = ,left_p, right_p;
    int level = (int)num;
    for (int i = ; i <= level-; i++)
    {
        sum = sum + c[i] * c[level -  - i];
        if (sum >= p)
        {
            sum = sum - c[i] * c[level -  - i];
            p = p - sum;
            left = (long long)i;
            right = (long long)(level -  - i);
            if (c[level -  - i] == ) left_p = p / c[level -  - i];
            else left_p = p / c[level -  - i] + ;
            right_p = p% c[level -  - i];
            if (right_p == ) right_p = c[level -  - i];
            break;
        }
    }

    if (left != ){printf("(");dfs(left, left_p);printf(")");}
    printf("X");
    if (right != ){printf("(");dfs(right, right_p);printf(")");}
}

void work()
{
    long long p, left, right, sum = ,left_p=, right_p=;
    int level;
    for (int i = ; i <= ; i++)
    {
        sum = sum + c[i];
        if (sum >= n){ sum = sum - c[i]; level = i; p = n - sum; break; }
    }
    dfs((long long)level, p);
    printf("\n");
}

int main()
{
    while (~scanf("%lld", &n))
    {
        if (!n) break;
        work();
    }
    return ;
}