CF1934B.Yet Another Coin Problem

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题目描述

You have 55 different types of coins, each with a value equal to one of the first 55 triangular numbers: 11 , 33 , 66 , 1010 , and 1515 . 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 nn .

We can show that the answer always exists.

输入格式

The first line contains one integer tt ( 1t1041 \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 nn ( 1n1091 \leq n \leq 10^9 ) — the target value.

输出格式

For each test case, output a single number — the minimum number of coins required.

输入输出样例

  • 输入#1

    14
    1
    2
    3
    5
    7
    11
    12
    14
    16
    17
    18
    20
    98
    402931328

    输出#1

    1
    2
    1
    3
    2
    2
    2
    3
    2
    3
    2
    2
    8
    26862090

说明/提示

In the first test case, for n=1n = 1 , the answer is 11 since only one 11 value coin is sufficient. 1=111 = 1 \cdot 1 .

In the fourth test case, for n=5n = 5 , the answer is 33 , which can be achieved using two 11 value coins and one 33 value coin. 5=21+135 = 2 \cdot 1 + 1 \cdot 3 .

In the seventh test case, for n=12n = 12 , the answer is 22 , which can be achieved using two 66 value coins.

In the ninth test case, for n=16n = 16 , the answer is 22 , which can be achieved using one 11 value coin and one 1515 value coin or using one 1010 value coin and one 66 value coin. 16=11+115=16+11016 = 1 \cdot 1 + 1 \cdot 15 = 1 \cdot 6 + 1 \cdot 10 .

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