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FZU 1064 教授的测试

2016-03-17 21:24 337 查看
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[20] = { 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786,
208012, 742900, 2674440, 9694845, 35357670, 129644790 };
long long n;

void dfs(long long num, long long p)
{
long long left=0, right=0, sum = 0,left_p, right_p;
int level = (int)num;
for (int i = 0; i <= level-1; i++)
{
sum = sum + c[i] * c[level - 1 - i];
if (sum >= p)
{
sum = sum - c[i] * c[level - 1 - i];
p = p - sum;
left = (long long)i;
right = (long long)(level - 1 - i);
if (c[level - 1 - i] == 1) left_p = p / c[level - 1 - i];
else left_p = p / c[level - 1 - i] + 1;
right_p = p% c[level - 1 - i];
if (right_p == 0) right_p = c[level - 1 - i];
break;
}
}

if (left != 0){printf("(");dfs(left, left_p);printf(")");}
printf("X");
if (right != 0){printf("(");dfs(right, right_p);printf(")");}
}

void work()
{
long long p, left, right, sum = 0,left_p=0, right_p=0;
int level;
for (int i = 1; i <= 17; 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 0;
}
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