Codeforces Round 931 (Div. 2) |
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Finished |
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.
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.
14123571112141617182098402931328
1 2 1 3 2 2 2 3 2 3 2 2 8 26862090
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. $$$16 = 1 \cdot 1 + 1 \cdot 15 = 1 \cdot 6 + 1 \cdot 10$$$.
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