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254 lines
6.4 KiB
Python
Executable file
254 lines
6.4 KiB
Python
Executable file
"""#
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### 谜题描述
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You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
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An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
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You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
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Input
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First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
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Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
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Output
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Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
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Examples
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Input
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4
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0 1 3 2
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Output
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1 0
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Input
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9
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10 7 9 10 7 5 5 3 5
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Output
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4 14
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Input
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3
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8 10 3
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Output
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0 8
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Note
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In the first sample it is optimal to leave the array as it is by choosing x = 0.
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In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
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* i = 2, j = 3;
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* i = 2, j = 4;
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* i = 3, j = 4;
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* i = 8, j = 9.
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In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.
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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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#include <bits/stdc++.h>
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using namespace std;
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void solve() {
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int n;
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cin >> n;
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vector<int> a(n);
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for (int i = 0; i < n; i++) {
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cin >> a[i];
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}
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vector<string> data(n);
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for (int i = 0; i <= 29; i++) {
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long long tes = pow(2, i);
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for (int j = 0; j < n; j++) {
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if ((tes & a[j]) == tes) {
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data[j].push_back('1');
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} else {
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data[j].push_back('0');
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}
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}
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}
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vector<vector<int>> run(1);
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for (int i = 0; i < n; i++) {
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run[0].push_back(i);
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}
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long long b = 0;
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int ans = 0;
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vector<vector<int>> split;
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for (int i = 29; i >= 0; i--) {
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long long sum1 = 0, sum0 = 0;
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for (vector<int> j : run) {
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vector<int> zero;
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vector<int> one;
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int cnt0 = 0, cnt1 = 0;
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for (int k : j) {
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if (data[k][i] == '1') {
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sum0 += cnt0;
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cnt1++;
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one.push_back(k);
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} else {
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sum1 += cnt1;
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cnt0++;
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zero.push_back(k);
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}
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}
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if (one.size() != 0) split.push_back(one);
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if (zero.size() != 0) split.push_back(zero);
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}
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if (sum1 > sum0) {
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ans += pow(2, i);
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}
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b += min(sum1, sum0);
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swap(split, run);
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split.clear();
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}
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cout << b << \" \" << ans;
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}
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int main() {
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ios_base::sync_with_stdio(0);
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cin.tie(0);
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solve();
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return 0;
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}
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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 bootcamp import Basebootcamp
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import re
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import random
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def compute_min_inversion_xor(a):
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n = len(a)
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data = []
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for num in a:
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bits = []
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for i in range(30):
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if num & (1 << i):
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bits.append('1')
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else:
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bits.append('0')
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data.append(bits[::-1]) # Reverse to store MSB first
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run = [list(range(n))]
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ans = 0
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b_val = 0
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for i in range(30): # Now using correct bit order
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sum0 = 0
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sum1 = 0
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split = []
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for group in run:
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zero = []
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one = []
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cnt0 = 0
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cnt1 = 0
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for k in group:
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if data[k][i] == '1':
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sum0 += cnt0
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cnt1 += 1
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one.append(k)
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else:
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sum1 += cnt1
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cnt0 += 1
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zero.append(k)
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if zero:
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split.append(zero)
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if one:
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split.append(one)
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if sum1 > sum0:
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ans += (1 << (29 - i)) # Adjust for bit significance
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b_val += sum0
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else:
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b_val += sum1
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run = split
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return b_val, ans
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class Exorinversebootcamp(Basebootcamp):
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def __init__(self, n_min=1, n_max=10, max_value=10**9):
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self.n_min = n_min
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self.n_max = n_max
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self.max_value = max_value
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def case_generator(self):
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n = random.randint(self.n_min, self.n_max)
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a = [random.randint(0, self.max_value) for _ in range(n)]
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correct_inversions, correct_x = compute_min_inversion_xor(a)
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return {
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'n': n,
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'a': a,
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'correct_inversions': correct_inversions,
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'correct_x': correct_x
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}
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@staticmethod
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def prompt_func(question_case):
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a = question_case['a']
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n = question_case['n']
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prompt = f"""You are given an array of {n} non-negative integers. Your task is to choose a non-negative integer x such that the array b, formed by XORing each element of the array with x, has the minimum number of inversions. An inversion is a pair of indices i < j where b[i] > b[j]. If multiple x values yield the same minimal number of inversions, choose the smallest x.
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Input:
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The first line contains an integer n ({n} in this case).
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The second line contains {n} space-separated integers: {', '.join(map(str, a))}.
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Your goal is to determine the minimal number of inversions and the corresponding smallest x.
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Output two integers: the minimal number of inversions and the smallest x.
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Please provide your answer within [answer] and [/answer]. For example:
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[answer]42 7[/answer]
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Your answer should be two integers separated by a space, enclosed within the tags."""
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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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last_match = matches[-1].strip()
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parts = last_match.split()
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if len(parts) != 2:
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return None
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try:
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inv = int(parts[0])
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x = int(parts[1])
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return (inv, x)
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except ValueError:
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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correct_inv = identity['correct_inversions']
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correct_x = identity['correct_x']
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return solution == (correct_inv, correct_x)
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