Yet Another Coin Problem
You have $$$5$$$ different types of coins, each with a value equal to one of the first $$$5$$$ triangular numbers: $$$1$$$, $$$3$$$, $$$6$$$, $$$10$$$, and $$$15$$$. These coin types are available in abundance. Your goal is to find the minimum number of these coins required such that their total value sums up to exactly $$$n$$$.We can show that the answer always exists.
In the first test case, for $$$n = 1$$$, the answer is $$$1$$$ since only one $$$1$$$ value coin is sufficient. $$$1 = 1 \cdot 1$$$.In the fourth test case, for $$$n = 5$$$, the answer is $$$3$$$, which can be achieved using two $$$1$$$ value coins and one $$$3$$$ value coin. $$$5 = 2 \cdot 1 + 1 \cdot 3$$$.In the seventh test case, for $$$n = 12$$$, the answer is $$$2$$$, which can be achieved using two $$$6$$$ value coins.In the ninth test case, for $$$n = 16$$$, the answer is $$$2$$$, which can be achieved using one $$$1$$$ value coin and one $$$15$$$ value coin or using one $$$10$$$ value coin and one $$$6$$$ value coin. $$$16 = 1 \cdot 1 + 1 \cdot 15 = 1 \cdot 6 + 1 \cdot 10$$$.
The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The description of the test cases follows.The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^9$$$) — the target value.
For each test case, output a single number — the minimum number of coins required.
for n in [*map(int, open(0))][1:]:
    print(n // 15 + int('301212112221222'[(n + 1) % 15 * (5 != n != 8)]))
b = (1, 3, 6, 10, 15)
a = [9] * 14 + [0]
for _ in [0] * 10:
    a += (min((a[-x] for x in b)) + 1,)
for n in [*map(int, open(0))][1:]:
    print(n < 10 and a[14 + n] or (n % 15 == 14) + 1 - (n % 15 in (*b, 0, 5)) - -n // 15)
for n in [*map(int, open(0))][1:]:
    print(n // 15 + int('301212112221222'[(n + 1) % 15 * (5 != n != 8)]))