题干：
Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev decided to build a house of cards. For that they've already found a hefty deck of n playing cards. Let's describe the house they want to make:
- The house consists of some non-zero number of floors.
- Each floor consists of a non-zero number of rooms and the ceiling. A room is two cards that are leaned towards each other. The rooms are made in a row, each two adjoining rooms share a ceiling made by another card.
- Each floor besides for the lowest one should contain less rooms than the floor below.
Please note that the house may end by the floor with more than one room, and in this case they also must be covered by the ceiling. Also, the number of rooms on the adjoining floors doesn't have to differ by one, the difference may be more.
While bears are practicing to put cards, Horace tries to figure out how many floors their house should consist of. The height of the house is the number of floors in it. It is possible that you can make a lot of different houses of different heights out of n cards. It seems that the elephant cannot solve this problem and he asks you to count the number of the distinct heights of the houses that they can make using exactly n cards.
Input
The single line contains integer n (1 ≤ n ≤ 1012) — the number of cards.
Output
Print the number of distinct heights that the houses made of exactly n cards can have.
Examples
Input
13
Output
1
Input
6
Output
0
Note
In the first sample you can build only these two houses (remember, you must use all the cards):
Thus, 13 cards are enough only for two floor houses, so the answer is 1.
The six cards in the second sample are not enough to build any house.
题目大意：
现在有n个卡片，问能叠成多少种高度的房子。（叠法按照题意和图片就看出来了）
解题报告：
直接枚举，因为可以发现推出来的公式中是平方级别的，所以复杂度就是1e6左右，可以直接暴力。推公式是这么推的：先按照最少的方式看看可不可以叠出当前枚举的层数，然后再看剩下的卡片数是否是3的倍数，如果是3的倍数那就直接都加在最下面一层就ok了，否则就是这一层做不到。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
ll n;
int main()
{cin >> n;ll ans = 0;for (ll i = 1;; i++) {ll tmp = (i+1)*i/2 * 2;//房子 tmp += (i-1)*i/2;//天花板 if (tmp>n) break;if ((n-tmp)%3 == 0) ans++;}printf("%d\n",ans);return 0;
}
)
