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185 lines
5.2 KiB
Python
Executable file
185 lines
5.2 KiB
Python
Executable file
"""#
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### 谜题描述
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The weight of a sequence is defined as the number of unordered pairs of indexes (i,j) (here i < j) with same value (a_{i} = a_{j}). For example, the weight of sequence a = [1, 1, 2, 2, 1] is 4. The set of unordered pairs of indexes with same value are (1, 2), (1, 5), (2, 5), and (3, 4).
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You are given a sequence a of n integers. Print the sum of the weight of all subsegments of a.
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A sequence b is a subsegment of a sequence a if b can be obtained from a by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
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Input
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Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^5). Description of the test cases follows.
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The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
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The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
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It is guaranteed that the sum of n over all test cases does not exceed 10^5.
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Output
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For each test case, print a single integer — the sum of the weight of all subsegments of a.
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Example
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Input
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2
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4
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1 2 1 1
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4
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1 2 3 4
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Output
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6
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0
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Note
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* In test case 1, all possible subsegments of sequence [1, 2, 1, 1] having size more than 1 are:
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1. [1, 2] having 0 valid unordered pairs;
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2. [2, 1] having 0 valid unordered pairs;
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3. [1, 1] having 1 valid unordered pair;
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4. [1, 2, 1] having 1 valid unordered pairs;
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5. [2, 1, 1] having 1 valid unordered pair;
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6. [1, 2, 1, 1] having 3 valid unordered pairs.
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Answer is 6.
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* In test case 2, all elements of the sequence are distinct. So, there is no valid unordered pair with the same value for any subarray. Answer is 0.
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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from sys import stdin
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from itertools import repeat
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from collections import defaultdict
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def solve():
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n = int(stdin.readline())
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a = map(int, stdin.readline().split(), repeat(10, n))
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s = defaultdict(int)
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ans = 0
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for i, x in enumerate(a):
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ans += s[x] * (n - i)
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s[x] += i + 1
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print ans
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T = int(stdin.readline())
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for t in xrange(T):
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solve()
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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from collections import defaultdict
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from random import randint
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import re
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from bootcamp import Basebootcamp
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class Csequencepairweightbootcamp(Basebootcamp):
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def __init__(self, max_t=2, max_n=5, a_max=5):
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self.max_t = max_t
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self.max_n = max_n
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self.a_max = a_max # 控制元素的范围
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def case_generator(self):
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t = randint(1, self.max_t)
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cases = []
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for _ in range(t):
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n = randint(1, self.max_n)
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a = [randint(1, self.a_max) for _ in range(n)]
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output = self._calculate_single_case(n, a)
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cases.append({
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'n': n,
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'a': a,
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'output': output
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})
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identity = {
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't': t,
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'cases': cases
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}
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return identity
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@staticmethod
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def _calculate_single_case(n, a):
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s = defaultdict(int)
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ans = 0
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for i, x in enumerate(a):
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ans += s[x] * (n - i)
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s[x] += (i + 1)
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return ans
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@staticmethod
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def prompt_func(question_case):
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input_lines = [str(question_case['t'])]
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for case in question_case['cases']:
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input_lines.append(str(case['n']))
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input_lines.append(' '.join(map(str, case['a'])))
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input_str = '\n'.join(input_lines)
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prompt = f"""你是编程竞赛的参赛者，请解决以下问题：
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问题描述：
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给定多个测试用例，每个测试用例要求计算数组所有子段的权重总和。权重定义为子段中相同值的无序对（i, j）的数量（i < j 并且a_i等于a_j）。子段是原数组的连续子序列。
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输入格式：
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输入的第一行是测试用例数目t。每个测试用例包含两行：第一行是整数n（数组长度），第二行是n个整数a_1到a_n。
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输出格式：
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对每个测试用例，输出一个整数，表示所有子段的权重总和。
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示例：
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输入：
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2
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4
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1 2 1 1
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4
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1 2 3 4
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输出：
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6
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0
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现在，请解决以下输入中的测试用例：
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输入：
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{input_str}
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请将答案放入[answer]标签内，每个测试用例的结果各占一行。例如：
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[answer]
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答案1
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答案2
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[/answer]
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请确保您的答案正确且格式正确。"""
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return prompt
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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if not matches:
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return None
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content = matches[-1].strip()
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lines = [line.strip() for line in content.split('\n')]
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lines = [line for line in lines if line]
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try:
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solution = list(map(int, lines))
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except:
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return None
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return solution
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@classmethod
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def _verify_correction(cls, solution, identity):
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if not isinstance(solution, list):
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return False
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expected = [case['output'] for case in identity['cases']]
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return solution == expected
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