mirror of
https://github.com/InternLM/InternBootcamp.git
synced 2026-04-26 17:13:14 +00:00
254 lines
6.4 KiB
Python
Executable file
254 lines
6.4 KiB
Python
Executable file
"""#
|
|
|
|
### 谜题描述
|
|
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)).
|
|
|
|
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.
|
|
|
|
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.
|
|
|
|
Input
|
|
|
|
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
|
|
|
|
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.
|
|
|
|
Output
|
|
|
|
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
|
|
|
|
Examples
|
|
|
|
Input
|
|
|
|
|
|
4
|
|
0 1 3 2
|
|
|
|
|
|
Output
|
|
|
|
|
|
1 0
|
|
|
|
|
|
Input
|
|
|
|
|
|
9
|
|
10 7 9 10 7 5 5 3 5
|
|
|
|
|
|
Output
|
|
|
|
|
|
4 14
|
|
|
|
|
|
Input
|
|
|
|
|
|
3
|
|
8 10 3
|
|
|
|
|
|
Output
|
|
|
|
|
|
0 8
|
|
|
|
Note
|
|
|
|
In the first sample it is optimal to leave the array as it is by choosing x = 0.
|
|
|
|
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:
|
|
|
|
* i = 2, j = 3;
|
|
* i = 2, j = 4;
|
|
* i = 3, j = 4;
|
|
* i = 8, j = 9.
|
|
|
|
|
|
|
|
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.
|
|
|
|
Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
|
|
```python
|
|
#include <bits/stdc++.h>
|
|
using namespace std;
|
|
void solve() {
|
|
int n;
|
|
cin >> n;
|
|
vector<int> a(n);
|
|
for (int i = 0; i < n; i++) {
|
|
cin >> a[i];
|
|
}
|
|
vector<string> data(n);
|
|
for (int i = 0; i <= 29; i++) {
|
|
long long tes = pow(2, i);
|
|
for (int j = 0; j < n; j++) {
|
|
if ((tes & a[j]) == tes) {
|
|
data[j].push_back('1');
|
|
} else {
|
|
data[j].push_back('0');
|
|
}
|
|
}
|
|
}
|
|
vector<vector<int>> run(1);
|
|
for (int i = 0; i < n; i++) {
|
|
run[0].push_back(i);
|
|
}
|
|
long long b = 0;
|
|
int ans = 0;
|
|
vector<vector<int>> split;
|
|
for (int i = 29; i >= 0; i--) {
|
|
long long sum1 = 0, sum0 = 0;
|
|
for (vector<int> j : run) {
|
|
vector<int> zero;
|
|
vector<int> one;
|
|
int cnt0 = 0, cnt1 = 0;
|
|
for (int k : j) {
|
|
if (data[k][i] == '1') {
|
|
sum0 += cnt0;
|
|
cnt1++;
|
|
one.push_back(k);
|
|
} else {
|
|
sum1 += cnt1;
|
|
cnt0++;
|
|
zero.push_back(k);
|
|
}
|
|
}
|
|
if (one.size() != 0) split.push_back(one);
|
|
if (zero.size() != 0) split.push_back(zero);
|
|
}
|
|
if (sum1 > sum0) {
|
|
ans += pow(2, i);
|
|
}
|
|
b += min(sum1, sum0);
|
|
swap(split, run);
|
|
split.clear();
|
|
}
|
|
cout << b << \" \" << ans;
|
|
}
|
|
int main() {
|
|
ios_base::sync_with_stdio(0);
|
|
cin.tie(0);
|
|
solve();
|
|
return 0;
|
|
}
|
|
```
|
|
|
|
|
|
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。
|
|
"""
|
|
|
|
from bootcamp import Basebootcamp
|
|
from bootcamp import Basebootcamp
|
|
import re
|
|
import random
|
|
|
|
def compute_min_inversion_xor(a):
|
|
n = len(a)
|
|
data = []
|
|
for num in a:
|
|
bits = []
|
|
for i in range(30):
|
|
if num & (1 << i):
|
|
bits.append('1')
|
|
else:
|
|
bits.append('0')
|
|
data.append(bits[::-1]) # Reverse to store MSB first
|
|
|
|
run = [list(range(n))]
|
|
ans = 0
|
|
b_val = 0
|
|
|
|
for i in range(30): # Now using correct bit order
|
|
sum0 = 0
|
|
sum1 = 0
|
|
split = []
|
|
for group in run:
|
|
zero = []
|
|
one = []
|
|
cnt0 = 0
|
|
cnt1 = 0
|
|
for k in group:
|
|
if data[k][i] == '1':
|
|
sum0 += cnt0
|
|
cnt1 += 1
|
|
one.append(k)
|
|
else:
|
|
sum1 += cnt1
|
|
cnt0 += 1
|
|
zero.append(k)
|
|
if zero:
|
|
split.append(zero)
|
|
if one:
|
|
split.append(one)
|
|
if sum1 > sum0:
|
|
ans += (1 << (29 - i)) # Adjust for bit significance
|
|
b_val += sum0
|
|
else:
|
|
b_val += sum1
|
|
run = split
|
|
|
|
return b_val, ans
|
|
|
|
class Exorinversebootcamp(Basebootcamp):
|
|
def __init__(self, n_min=1, n_max=10, max_value=10**9):
|
|
self.n_min = n_min
|
|
self.n_max = n_max
|
|
self.max_value = max_value
|
|
|
|
def case_generator(self):
|
|
n = random.randint(self.n_min, self.n_max)
|
|
a = [random.randint(0, self.max_value) for _ in range(n)]
|
|
correct_inversions, correct_x = compute_min_inversion_xor(a)
|
|
return {
|
|
'n': n,
|
|
'a': a,
|
|
'correct_inversions': correct_inversions,
|
|
'correct_x': correct_x
|
|
}
|
|
|
|
@staticmethod
|
|
def prompt_func(question_case):
|
|
a = question_case['a']
|
|
n = question_case['n']
|
|
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.
|
|
|
|
Input:
|
|
The first line contains an integer n ({n} in this case).
|
|
The second line contains {n} space-separated integers: {', '.join(map(str, a))}.
|
|
|
|
Your goal is to determine the minimal number of inversions and the corresponding smallest x.
|
|
|
|
Output two integers: the minimal number of inversions and the smallest x.
|
|
|
|
Please provide your answer within [answer] and [/answer]. For example:
|
|
[answer]42 7[/answer]
|
|
|
|
Your answer should be two integers separated by a space, enclosed within the tags."""
|
|
return prompt
|
|
|
|
@staticmethod
|
|
def extract_output(output):
|
|
matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
|
|
if not matches:
|
|
return None
|
|
last_match = matches[-1].strip()
|
|
parts = last_match.split()
|
|
if len(parts) != 2:
|
|
return None
|
|
try:
|
|
inv = int(parts[0])
|
|
x = int(parts[1])
|
|
return (inv, x)
|
|
except ValueError:
|
|
return None
|
|
|
|
@classmethod
|
|
def _verify_correction(cls, solution, identity):
|
|
correct_inv = identity['correct_inversions']
|
|
correct_x = identity['correct_x']
|
|
return solution == (correct_inv, correct_x)
|