评价一下蜘蛛侠平行宇宙插曲scared of the dark

蜘蛛侠平行宇宙插曲"Scared of the Dark"是一首非常出色的歌曲。它充满了情感和能量，旋律简单易记，歌词也很有意义。这首歌曲描绘了主角蜘蛛侠在面对恐惧和不确定性时的内心挣扎和成长。歌曲中的音乐、歌词和情感都很好地配合了电影的主题和情节，使得观众更加深入地理解了蜘蛛侠的内心世界。

总之，"Scared of the Dark"是一首非常出色的电影插曲，它为电影增添了很多情感和能量，也是一首非常值得听的歌曲。

相关问题

爬取豆瓣top250的电影信息以及几个评价，然后用jieba库分词看一下情感极性，把top250的电影分词几个情感类，分别写入Angry、Disgusted、Happy、Neutral、Sad、Scared、Surpriser的xlsx文件中

好的，这是一个比较复杂的任务，需要用到爬虫、文本处理、情感分析等技术。我将分步骤地介绍如何完成这个任务。

1. 爬取豆瓣top250的电影信息

首先，我们需要爬取豆瓣top250的电影信息。可以使用Python的requests和BeautifulSoup库来实现。

import requests
from bs4 import BeautifulSoup

url = 'https://movie.douban.com/top250'
headers = {
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.3'}
movies = []

for i in range(10):
    params = {'start': i * 25, 'filter': ''}
    response = requests.get(url, headers=headers, params=params)
    soup = BeautifulSoup(response.text, 'html.parser')
    items = soup.find_all('div', class_='item')
    for item in items:
        movie = {}
        movie['rank'] = item.find('em').text
        movie['title'] = item.find('span', class_='title').text
        movie['score'] = item.find('span', class_='rating_num').text
        movie['quote'] = item.find('span', class_='inq').text if item.find('span', class_='inq') else ''
        movies.append(movie)

这段代码会爬取豆瓣top250的前10页电影信息，每页有25个电影。爬取到的电影信息会以字典的形式保存在一个列表中。

2. 爬取电影的评价

接下来，我们需要爬取每个电影的评价。可以使用Python的requests和正则表达式来实现。

import re

for movie in movies:
    movie_url = 'https://movie.douban.com/subject/' + movie['rank']
    response = requests.get(movie_url, headers=headers)
    pattern = re.compile('<span class="short">(.*?)</span>', re.S)
    comments = re.findall(pattern, response.text)
    movie['comments'] = comments[:3] if comments else []

这段代码会遍历每个电影，爬取该电影的评价。由于豆瓣的评价是异步加载的，所以不能直接使用BeautifulSoup来解析。我们可以使用正则表达式来匹配评价的内容。爬取到的评价会以列表的形式保存在电影的字典中。

3. 使用jieba库分词

接下来，我们需要使用jieba库分词。jieba库是Python中常用的中文分词库，可以将一段中文文本分割成单词的形式。

import jieba

for movie in movies:
    comments = '\n'.join(movie['comments'])
    words = jieba.cut(comments)
    movie['words'] = list(words)

这段代码会遍历每个电影的评价，使用jieba库将评价分词，并将分词结果以列表的形式保存在电影的字典中。

4. 情感分析

最后，我们需要对每个电影的评价进行情感分析，并将结果保存到xlsx文件中。情感分析可以使用Python的TextBlob库来实现。

from textblob import TextBlob
import openpyxl

emotions = ['Angry', 'Disgusted', 'Happy', 'Neutral', 'Sad', 'Scared', 'Surprised']
workbook = openpyxl.Workbook()
worksheet = workbook.active
worksheet.append(['Rank', 'Title'] + emotions)

for movie in movies:
    row = [movie['rank'], movie['title']]
    for emotion in emotions:
        count = 0
        for word in movie['words']:
            if TextBlob(word).sentiment.polarity > 0 and emotion == 'Happy':
                count += 1
            elif TextBlob(word).sentiment.polarity < 0 and emotion == 'Angry':
                count += 1
            elif TextBlob(word).sentiment.polarity < -0.5 and emotion == 'Disgusted':
                count += 1
            elif TextBlob(word).sentiment.polarity == 0 and emotion == 'Neutral':
                count += 1
            elif TextBlob(word).sentiment.polarity < 0 and emotion == 'Sad':
                count += 1
            elif TextBlob(word).sentiment.polarity < 0 and emotion == 'Scared':
                count += 1
            elif TextBlob(word).sentiment.polarity > 0.5 and emotion == 'Surprised':
                count += 1
        row.append(count)
    worksheet.append(row)

workbook.save('top250.xlsx')

这段代码会遍历每个电影的分词结果，使用TextBlob库对每个单词进行情感分析，并根据情感极性将单词归类到不同的情感类别中。然后，我们将每个电影的情感类别数量保存到xlsx文件中。

注意，由于对情感极性的判断是基于单个单词的，所以这种情感分析方法的准确性可能不够高。如果需要更准确的情感分析结果，可以考虑使用基于机器学习的情感分析方法。

Kars is tired and resentful of the narrow mindset of his village since they are content with staying where they are and are not trying to become the perfect life form. Being a top-notch inventor, Kars wishes to enhance his body and become the perfect life form. Unfortunately, n of the villagers have become suspicious of his ideas. The i -th villager has a suspicion of ai on him. Individually each villager is scared of Kars, so they form into groups to be more powerful. The power of the group of villagers from l to r be defined as f(l,r) where f(l,r)=|al−al+1|+|al+1−al+2|+…+|ar−1−ar|. Here |x−y| is the absolute value of x−y . A group with only one villager has a power of 0 . Kars wants to break the villagers into exactly k contiguous subgroups so that the sum of their power is minimized. Formally, he must find k−1 positive integers 1≤r1<r2<…<rk−1<n such that f(1,r1)+f(r1+1,r2)+…+f(rk−1+1,n) is minimised. Help Kars in finding the minimum value of f(1,r1)+f(r1+1,r2)+…+f(rk−1+1,n) . Input The first line contains a single integer t (1≤t≤100) — the number of test cases. The description of test cases follows. The first line of each test case contains two integers n,k (1≤k≤n≤100) — the number of villagers and the number of groups they must be split into. The second line of each test case contains n integers a1,a2,…,an (1≤ai≤500) — the suspicion of each of the villagers. Output For each test case, output a single integer — the minimum possible value of sum of power of all the groups i. e. the minimum possible value of f(1,r1)+f(r1+1,r2)+…+f(rk−1+1,n) . Example inputCopy 3 4 2 1 3 5 2 6 3 1 9 12 4 7 2 12 8 1 9 8 2 3 3 1 8 7 7 9 2 outputCopy 4 11 2 Note In the first test case, we will group the villagers with suspicion (1,3,5,2) into (1,3,5) and (2) . So, f(1,3)+f(4,4)=(|1−3|+|3−5|)+0=4+0=4 . In the second test case, we will group the villagers with suspicion (1,9,12,4,7,2) into (1),(9,12),(4,7,2) . So, f(1,1)+f(2,3)+f(4,6)=0+3+8=11 .

这是一个关于分组问题的题目，题目给出了每个村民的怀疑程度，要求将村民分成 k 个连续的子组，使得他们的总怀疑程度最小。我们可以使用动态规划来解决这个问题。

首先，我们定义一个二维数组 dp，其中 dp[i][j] 表示将前 i 个村民分成 j 个连续子组时的最小怀疑程度。

接下来，我们可以使用状态转移方程来计算 dp[i][j]。假设当前村民为第 i 个，我们可以考虑将其与前面的村民组成一个子组，或者单独作为一个子组。如果将其与前面的村民组成一个子组，则 dp[i][j] = dp[k][j-1] + f(k+1, i)，其中 k < i。如果单独作为一个子组，则 dp[i][j] = dp[i-1][j-1] + f(i, i)，即当前村民的怀疑程度。我们取这两种情况中的最小值作为 dp[i][j] 的值。

最后，答案就是 dp[n][k]。

下面是一个具体的实现示例：

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int solve(vector<int>& suspicions, int n, int k) {
    vector<vector<int>> dp(n + 1, vector<int>(k + 1, 0));
    vector<vector<int>> prefixSum(n + 1, vector<int>(n + 1, 0));

    for (int i = 1; i <= n; i++) {
        prefixSum[i][i] = suspicions[i - 1];
        for (int j = i + 1; j <= n; j++) {
            prefixSum[i][j] = prefixSum[i][j - 1] + suspicions[j - 1];
        }
    }

    for (int i = 1; i <= n; i++) {
        dp[i][1] = prefixSum[1][i];
    }

    for (int j = 2; j <= k; j++) {
        for (int i = j; i <= n; i++) {
            dp[i][j] = INT_MAX;
            for (int l = j - 1; l < i; l++) {
                dp[i][j] = min(dp[i][j], dp[l][j - 1] + prefixSum[l + 1][i]);
            }
        }
    }

    return dp[n][k];
}

int main() {
    int t;
    cin >> t;

    while (t--) {
        int n, k;
        cin >> n >> k;

        vector<int> suspicions(n);
        for (int i = 0; i < n; i++) {
            cin >> suspicions[i];
        }

        int result = solve(suspicions, n, k);
        cout << result << endl;
    }

    return 0;
}

希望这个解答对你有帮助！如果还有其他问题，请随时问我。