The media have and will continue to play a significant role in the development and application of many scientific disciplines, including astronomy. With the increasing global awareness about climate change and environment protection, astronomy, as a key scientific discipline, now faces the challenge of changing societal views on the importance and urgency of environmental issues.

First, it is important to understand the role of astronomy in our lives. Understanding the universe around us, particularly our place within it, is fundamental to our existence as conscious beings. Astronomy helps us answer fundamental questions about the origins of the universe, the nature of matter and energy, and the possibility of extraterrestrial life. It also has enormous practical applications, including navigation and communication technologies, remote sensing, and space weather prediction.

However, the success of astronomy depends on the preservation of dark skies. City lights and air pollution wash out the stars from the night sky, which can be a problem for both professional and amateur astronomers. The presence of bright lights can interfere with astronomical observations and limit our ability to study the early universe, black holes, and other phenomena.

As such, it is important to protect areas of the country from the effects of light pollution. Light pollution can be mitigated by methods such as directing light downwards or shielding it, replacing traditional streetlights with LED alternatives, and setting up observatories in remote areas. This would allow astronomers to make observations of the objects they study without the complications of light pollution.

In addition to the benefits for astronomy, reducing light pollution has advantages for human health and animal life. Exposure to artificial light at night can disrupt the natural sleep cycle of animals and humans, leading to negative impacts on health as well as feeding and breeding patterns. Light pollution can also disrupt ecosystems by altering the behavior of nocturnal animals and insects.

To address light pollution and its environmental impacts, communities can implement night sky protection measures, such as requiring light fixtures to have less output or using shielded lights to direct light downwards. People can also take personal responsibility for reducing their own light pollution by switching to wildlife-friendly outdoor lighting, closing curtains at night, and using blackout shades to prevent streetlights from shining into their windows.

Overall, we must all take steps to reduce light pollution and preserve the beauty of the night sky for future generations to enjoy. Only by working together can we protect the world around us from the harmful impacts of artificial light.

5,5,5,5,5,5,5,5,5,5,5,5])-30): print("Test 6 passe") else: print("Test 6 failed")

14001\11001$ 은 둘째줄시 1가 나옵니다.

이번 문제는 위와 비슷한 문제입니다.

병사들은 능력치를 각각 조금씩만 올리면 되므로 이를 더해가면서 처리합니다.

class Solution: def findPairs(self, nums: List[int], k: int) -> int: nums=set(nums) nums = list(nums) # 중복제거 check = set([]) cnt = 0 for i in nums: # enumerate의 경우 if i+k in nums: cnt+=1 #cnt에다가 더해줌 if i+k in check : cnt-=1 else : check.add(i)

    return cnt

Runtime: 72 ms, faster than 92.82% of Python3 online submissions for K-diff Pairs in an Array.

Memory Usage: 15.9 MB, less than 23.08% of Python3 online submissions for K-diff Pairs in an Array.

Solution 2. 배열에서 순서대로 탐색하여 같은 수가 나오면 통과.

그리고 바로 직전에 큰수와 비교하여 k보다 크면 이후는 절대 성립될 수 없다.

def findPairs(self, nums: List[int], k: int) -> int: nums = sorted(nums) ans = 0 lastnumi = None for i in range(len(nums)): if i and nums[i] == nums[i-1]: continue if i < len(nums) - 1 and nums[i]+k in nums[i+1:]: ans += 1 while i < len(nums) - 1 and nums[i] == nums[i+1]: i += 1 lastnumi = i return ans

Runtime: 56 ms, faster than 31.1 % of Python3 online submissions for K-diff Pairs in an Array.

Memory Usage: 15.4 MB, less than 69.23% of Python3 online submissions for K-diff Pairs in an Array.

Solution with dictionary

class Solution: def findPairs(self, nums: List[int], k: int) -> int: count = collections.Counter(nums) ans = 0 if k > 0: ans = sum(1 for x in count if x + k in count) elif k < 0: ans = sum(1 for x in count if x-k in count and x-k!=x) else: ans = sum(1 for x in count if count[x]>1) return ans

""" Test case """ s = Solution() print(s.findPairs([3, 1, 4, 1, 5], 2)) # 2 print(s.findPairs([1, 2, 3, 4, 5], 1)) # 4 print(s.findPairs([1, 3, 1, 5, 4], 0)) # 1 print(s.findPairs([1,2,4,4,3,3,0,9,2,3], 3)) # 2