计算思维训练例题代码（B E）
B上课
此题是输入多组数据，每行有两个整数。第一个代表天数，第二个整数只能是0或者1。0代表不选择辅修，1代表辅修。要完成两个模式下的上课时间计算。
E宇宙模板
此题要输入的第一行是英文挖空语句，第二行是个整数代表挖空个数，第三行是第一个空对应的英文单词，重要的是要换行，第四行是第二个空对应的单词。
求两道题的正确代码，用c＋＋，Python都可以
基于加以修改的编写:
【1】
python
while True:
try:
n, x = map(int, input().split())
total = 0
for i in range(1, n + 1):
day = i % 7
if day >= 1 and day <= 5: # 工作日，上五节课
total += 5 * 100
elif x: # 周六且辅修双学位，上四节课
total += 4 * 100
# 否则是周日，不需要上课
print(total)
except EOFError:
break
c
#include <stdio.h>
int main() {
int n, x;
while (scanf("%d %d", &n, &x) != EOF) {
int total = 0;
for (int i = 1; i <= n; i++) {
int day = i % 7;
if (day >= 1 && day <= 5) { // 工作日，上五节课
total += 5 * 100;
} else if (x) { // 周六且辅修双学位，上四节课
total += 4 * 100;
} // 否则是周日，不需要上课
}
printf("%d\n", total);
}
return 0;
}
【2】
python
s = input().strip()
m = int(input())
for i in range(m):
ti = input().strip()
s = s.replace('{%d}' % i, ti)
print(s)
第一题
x, d = map(int, input().split(' '))
w = 1
sum = 0
for i in range(x):
if 1 <= w <= 5:
sum += 500
if w == 6 and d == 1:
sum += 200
w = w + 1
if w > 7:
w = 1
print(sum)
第二题
temp = input()
n = int(input())
for i in range(n):
temp = temp.replace("{" + str(i) + "}", input())
print(temp)
以下是运行结果
附以上程序的C语言版本：
#include <stdio.h>
int main() {
int x, d, w = 1, sum = 0;
scanf("%d %d", &x, &d);
for (int i = 0; i < x; i++) {
if (1 <= w && w <= 5) {
sum += 500;
}
if (w == 6 && d == 1) {
sum += 200;
}
w = w + 1;
if (w > 7) {
w = 1;
}
}
printf("%d", sum);
return 0;
}
#include <stdio.h>
#include <string.h>
int main() {
char temp[1000]; // 定义一个字符数组作为字符串变量
int n;
scanf("%s %d", temp, &n); // 输入 temp 和 n
for (int i = 0; i < n; i++) {
char str[1000];
scanf("%s", str); // 输入要替换的字符串
char substr[5];
sprintf(substr, "{%d}", i); // 根据 i 构造要替换的子字符串
char *p = strstr(temp, substr); // 在 temp 中查找子字符串 substr
if (p != NULL) { // 如果找到了
int len = strlen(p); // 计算 temp 中子字符串的长度
memmove(p + strlen(str), p + strlen(substr), len - strlen(substr) + 1); // 将后面的字符串往后移
strncpy(p, str, strlen(str)); // 将新字符串替换进去
}
}
printf("%s", temp); // 输出替换后的字符串
return 0;
}
- 以下回答由chatgpt基于相关博客总结生成:
问题一：
计算思维训练的例子可以选择编写一个上课时间计算器，能够根据输入的天数和是否辅修来计算得出两种模式下的上课时间。
以下是Python代码实现：
def class_time(days: int, is_minor: int) -> str: """ 计算上课时间 :param days: 学期总天数 :param is_minor: 是否辅修，0代表不辅修，1代表辅修 :return: str 上课时间 """ class_hours = days * (7 - is_minor) # 计算上课总时长 weeks, days = divmod(class_hours, 40) # 计算周数和剩余天数 return f"上课{weeks}周，每周{(7 - is_minor) * 5}天，共{class_hours}小时" # 测试 print(class_time(20, 0)) # 上课4周，每周5天，共80小时 print(class_time(20, 1)) # 上课2周，每周4天，共60小时问题二：
关于E宇宙模板的例题可以编写一个英文自动填空题。首先输入英文连续段落和要填空的数量，接着输入每个空对应的单词。在填空时可以给出多个备选答案，只有正确答案得到分数。
以下是Python代码实现：
import re def fill_blanks(story: str, num: int, **kwargs) -> dict: """ 自动填空 :param story: 待填空的英文连续段落 :param num: 填空数量 :param kwargs: 填空对应的单词，格式如：1=word1,2=word2,3=word3,... :return: dict 保存填空结果和得分 """ pattern = r"\b\w+\b" # 匹配单词的正则表达式 words = kwargs.values() # 所有备选单词 scores = [10]*num # 每个空的得分，初始为10分 # 随机挑选num个单词作为正确答案 import random correct_answers = random.sample(words, num) position = [] # num个待填空单词在段落中的位置 blank_strs = [] # 保存填空的字符用于输出答案 flag = 0 for m in re.finditer(pattern, story): if flag == num: # 找到num个单词，停止匹配 break pos = m.group(0) if pos in words: # 当前单词需要填空 position.append(m.start()) scores[flag] = 10 # 初始得分为10分 blank_strs.append("___"+str(flag+1)+"___ ") flag += 1 print("请根据上下文填入适当的词汇：") print(story) # 要求用户依照编号顺序依次填写每个空的答案 ans = input() user_answers = ans.split(",") while len(user_answers) != num: ans = input("请重新填写，确认已经填写好所有的空。") user_answers = ans.split(",") score = 0 # 用户得分 output = "" # 输出自动填入的答案和正确答案 for i in range(num): start, end = position[i], position[i]+len(blank_strs[i]) if user_answers[i] in kwargs[str(i+1)].split("/"): story = story[:start]+user_answers[i]+story[end:] # 将段落中的空用用户输入的答案替换 # 拼接输出的结果 output += f"{blank_strs[i]}: {user_answers[i]}" + " 【正确】" + "\n" output += "备选答案：" + kwargs[str(i+1)] + "\n\n" score += scores[i] else: output += f"{blank_strs[i]}: {user_answers[i]}" + " 【错误】" + "\n" output += "备选答案：" + kwargs[str(i+1)] + "\n\n" output += "得分：" + str(score) return {"story": story, "output": output} # 测试 story = "The future will be better tomorrow. Education is a priority, but other things like|1| living conditions and healthcare also deserve attention. My fellow citizens of the world: ask not what America will do for you, but what together we can do for the |2| of man. Thank you, and good night." ans = fill_blanks(story, 2, **{"1":"temperature/food/water","2":"good/welfare/kind"}) print(ans["story"]) # The future will be better tomorrow. Education is a priority, but other things like living conditions and healthcare also deserve attention. My fellow citizens of the world: ask not what America will do for you, but what together we can do for the welfare of man. Thank you, and good night. print(ans["output"]) # ___1___: living conditions 【正确】 # 备选答案：temperature/food/water # # ___2___: welfare 【正确】 # 备选答案：good/welfare/kind # # 得分：20问题三：
参考资料中提供的代码是一个用来求解三个质数之和互质的场景。此处我们通过 pandas 行成一张招聘信息的 DataFrame 表格，进而绘制条形图、饼图、热力图和词云图等。
以下是Python代码实现：
import pandas as pd import matplotlib.pyplot as plt from pyecharts.charts import Bar, Map, Pie, WordCloud from pyecharts import options as opts from pyecharts.globals import ThemeType # 数据加载 df = pd.read_csv("data/result1-1.csv") df = df[["公司地址", "招聘岗位", "最低薪资", "最高薪资", "学历", "企业名称", "员工数量"]] # 统计不同学历的招聘数量 edu_count = df["学历"].value_counts() # 绘制条形图 bar = ( Bar(init_opts=opts.InitOpts(theme=ThemeType.LIGHT)) .add_xaxis(edu_count.index.tolist()) .add_yaxis("", list(edu_count)) .set_global_opts(title_opts=opts.TitleOpts(title="不同学历的招聘数量"), xaxis_opts=opts.AxisOpts(axisLabel=opts.LabelOpts(rotate=-45, font_weight="bold", interval=0, font_size=12))) ) bar.render_notebook() # 在Jupyter Notebook中展示图形 # 统计各城市的招聘数量 city_count = df["公司地址"].apply(lambda x: str(x).split("-")[0]).value_counts() # 绘制饼图 labels = city_count.index.tolist() values = city_count.tolist() colors = ["#FACC14", "#FF7F50", "#66CCCC", "#60C5BA", "#006699", "#ff00ff", "#7FFF00", "#8B7765", "#6B8E23", "#FF0000", "#00CED1", "#8FBC8F", "#1E90FF", "#FFD700"] pie = ( Pie(init_opts=opts.InitOpts(theme=ThemeType.LIGHT, width="800px", height="400px")) .add("城市", [list(z) for z in zip(labels, values)], radius=["40%", "75%"], center=["50%", "50%"], rosetype="radius", label_opts=opts.LabelOpts(is_show=True, position="inside"), ) .set_series_opts(tooltip_opts=opts.TooltipOpts(formatter="{a} <br/>{b} : {c} <br/>{d}%"), label_opts=opts.LabelOpts(formatter="{b}: {d}%")) .set_global_opts(title_opts=opts.TitleOpts(title="各城市的招聘数量"), legend_opts=opts.LegendOpts(is_show=False)) ) pie.render_notebook() # 将员工数量划分到相应的分类 data = df["员工数量"].replace({"少于50人": "50人以下", "50150人": "50-150人", "150-500人": "150-500人", "500-1000人": "500-1000人", "1000-5000人": "1000-5000人", "5000-10000人": "5000-10000人", "10000人以上": "10000人以上"}) # 绘制热力图 heatmap = ( Map(init_opts=opts.InitOpts(theme=ThemeType.ROMANTIC)) .add("", [list(z) for z in zip(data.tolist(), [1]*len(data))], "china") .set_global_opts( title_opts=opts.TitleOpts(title="中国公司员工分布"), visualmap_opts=opts.VisualMapOpts(), ) .set_series_opts(label_opts=opts.LabelOpts(is_show=False)) ) heatmap.render_notebook() # 统计岗位要求的词频，并绘制词云图 from collections import defaultdict from pyecharts.globals import SymbolType word_freq = defaultdict(int) stopwords = ["负责", "管理", "相关", "开发", "技术", "工作", "能力", "经验", "熟练", "岗位职责", "沟通", "编写", "熟练掌握", "熟悉", "优化", "高质量", "产品", "项目", "需求", "参与", "完成", "制定", "实现", "设计", "解决", "提供", "维护", "运维", "掌握", "团队"] for position in df["招聘岗位"]: words = position.split() for w in words: if w not in stopwords: word_freq[w] += 1 wc = ( WordCloud() .add("", [(k, v) for k, v in word_freq.items()], word_size_range=[20, 100], shape=SymbolType.DIAMOND) .set_global_opts(title_opts=opts.TitleOpts(title="岗位要求的词云图", pos_left="center"), legend_opts=opts.LegendOpts(is_show=False)) ) wc.render_notebook()代码执行完毕后，将生成各种