期末公共题.ipynb · 徐瑞博/邱英璞

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期末公共题.ipynb 11.72 KB

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<a id = Section1> </a>\n",
    "## 1.输入某年某月某日，判断这一天是这一年的第几天\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "2025年8月20日是这一年的第232天。\n"
     ]
    }
   ],
   "source": [
    "from datetime import datetime\n",
    "\n",
    "def day_of_year(year, month, day):\n",
    "    date = datetime(year, month, day)\n",
    "    day_num = date.timetuple().tm_yday\n",
    "    return day_num\n",
    "\n",
    "year = int(input(\"请输入年份: \"))\n",
    "month = int(input(\"请输入月份: \"))\n",
    "day = int(input(\"请输入日期: \"))\n",
    "\n",
    "print(f\"{year}年{month}月{day}日是这一年的第{day_of_year(year, month, day)}天。\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<a id = Section1> </a>\n",
    "## 3.判断1000中有多少个素数，并输出所有素数"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "在1到1000之间有168个素数:\n",
      "[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997]\n"
     ]
    }
   ],
   "source": [
    "def is_prime(n):\n",
    "    if n <= 1:\n",
    "        return False\n",
    "    for i in range(2, int(n**0.5) + 1):\n",
    "        if n % i == 0:\n",
    "            return False\n",
    "    return True\n",
    "\n",
    "def find_primes_in_range(limit):\n",
    "    primes = []\n",
    "    for num in range(2, limit + 1):\n",
    "        if is_prime(num):\n",
    "            primes.append(num)\n",
    "    return primes\n",
    "\n",
    "limit = 1000\n",
    "\n",
    "primes = find_primes_in_range(limit)\n",
    "print(f\"在1到{limit}之间有{len(primes)}个素数:\")\n",
    "print(primes)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<a id = Section1> </a>\n",
    "## 5.利用条件运算符的嵌套来完成此题：学习成绩>=90分的同学用A表示，60-89分之间的用B表示，60分以下的用C表示。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "分数: 50.0, 等级: C\n"
     ]
    }
   ],
   "source": [
    "def get_grade(score):\n",
    "    return 'A' if score >= 90 else ('B' if 60 <= score < 90 else 'C')\n",
    "\n",
    "try:\n",
    "    score = float(input(\"请输入学习成绩（0-100）：\"))\n",
    "    \n",
    "    if not (0 <= score <= 100):\n",
    "        print(\"分数必须在0到100之间。\")\n",
    "    else:\n",
    "        grade = get_grade(score)\n",
    "        print(f\"分数: {score}, 等级: {grade}\")\n",
    "except ValueError:\n",
    "    print(\"输入无效，请输入一个数字。\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<a id = Section1> </a>\n",
    "## 7.输入一行字符串，分别统计出其中英文字母、空格、数字和其它字符的个数"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "英文字母个数: 4\n",
      "空格个数: 0\n",
      "数字个数: 0\n",
      "其他字符个数: 0\n"
     ]
    }
   ],
   "source": [
    "def count_characters(input_string):\n",
    "    letters = 0\n",
    "    spaces = 0\n",
    "    digits = 0\n",
    "    others = 0\n",
    "\n",
    "    for char in input_string:\n",
    "        if char.isalpha():\n",
    "            letters += 1\n",
    "        elif char.isspace():\n",
    "            spaces += 1\n",
    "        elif char.isdigit():\n",
    "            digits += 1\n",
    "        else:\n",
    "            others += 1\n",
    "\n",
    "    return letters, spaces, digits, others\n",
    "\n",
    "input_string = input(\"请输入一行字符串：\")\n",
    "\n",
    "letters, spaces, digits, others = count_characters(input_string)\n",
    "\n",
    "print(f\"英文字母个数: {letters}\")\n",
    "print(f\"空格个数: {spaces}\")\n",
    "print(f\"数字个数: {digits}\")\n",
    "print(f\"其他字符个数: {others}\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<a id = Section1> </a>\n",
    "## 9.一个数如果恰好等于它的因子之和，这个数就称为“完数”。例如6=1＋2＋3.编程，找出1000以内的所有完数"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1到1000之间的完数有：[6, 28, 496]\n"
     ]
    }
   ],
   "source": [
    "def find_perfect_numbers(limit):\n",
    "    perfect_numbers = []\n",
    "    for num in range(1, limit + 1):\n",
    "        factors_sum = sum(i for i in range(1, num) if num % i == 0)\n",
    "        if factors_sum == num:\n",
    "            perfect_numbers.append(num)\n",
    "    return perfect_numbers\n",
    "\n",
    "limit = 1000\n",
    "\n",
    "perfect_numbers = find_perfect_numbers(limit)\n",
    "print(f\"1到{limit}之间的完数有：{perfect_numbers}\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<a id = Section1> </a>\n",
    "## 11.输入任意数N，求1+2!+3!+...+N!的值（注意实数范围）"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " 阶乘结果为153\n"
     ]
    }
   ],
   "source": [
    "def factorial(n):\n",
    "    if n == 0 or n == 1:\n",
    "        return 1\n",
    "    else:\n",
    "        return n * factorial(n - 1)\n",
    "\n",
    "def sum_of_factorials(N):\n",
    "    total = 0\n",
    "    for i in range(1, N + 1):\n",
    "        total += factorial(i)\n",
    "    return total\n",
    "\n",
    "try:\n",
    "    N = int(input(\"请输入一个正整数N：\"))\n",
    "    if N < 0:\n",
    "        raise ValueError(\"N必须是一个非负整数。\")\n",
    "    \n",
    "    result = sum_of_factorials(N)\n",
    "    print(f\" 阶乘结果为{result}\")\n",
    "except ValueError as e:\n",
    "    print(f\"输入无效：{e}\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<a id = Section1> </a>\n",
    "## 13.输入任意奇数N，按如下要求打印出图案（菱形）"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "  *\n",
      " ***\n",
      "*****\n",
      " ***\n",
      "  *\n"
     ]
    }
   ],
   "source": [
    "def print_diamond(N):\n",
    "    \"\"\"打印一个由奇数N控制的菱形图案\"\"\"\n",
    "    if N % 2 == 0:\n",
    "        print(\"N必须是一个奇数。\")\n",
    "        return\n",
    "    \n",
    "    \n",
    "    for i in range(1, N + 1):\n",
    "        if i <= (N + 1) // 2: \n",
    "            stars = 2 * i - 1\n",
    "        else:\n",
    "            stars = 2 * (N - i + 1) - 1 \n",
    "        \n",
    "        spaces = (N - stars) // 2\n",
    "        \n",
    "        print(' ' * spaces + '*' * stars)\n",
    "\n",
    "try:\n",
    "    N = int(input(\"请输入一个奇数N：\"))\n",
    "    if N <= 0 or N % 2 == 0:\n",
    "        raise ValueError(\"N必须是一个正奇数。\")\n",
    "    \n",
    "    print_diamond(N)\n",
    "except ValueError as e:\n",
    "    print(f\"输入无效：{e}\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<a id = Section1> </a>\n",
    "## 15.对10个数进行排序"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "原始数字列表: [16, 1, 84, 31, 5, 39, 85, 60, 19, 33]\n",
      "使用sorted()函数排序后的列表: [1, 5, 16, 19, 31, 33, 39, 60, 84, 85]\n"
     ]
    }
   ],
   "source": [
    "import random\n",
    "\n",
    "numbers = [random.randint(1, 100) for _ in range(10)]\n",
    "\n",
    "print(\"原始数字列表:\", numbers)\n",
    "\n",
    "sorted_numbers = sorted(numbers)\n",
    "print(\"使用sorted()函数排序后的列表:\", sorted_numbers)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<a id = Section1> </a>\n",
    "## 17.有一个已经排好序的数组（元素数量大于10）。从键盘任意输入一个数，要求按原来的规律将它插入数组中"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "插入后的新数组： [1, 3, 5, 7, 9, 10, 11, 13, 15, 17, 19, 21]\n"
     ]
    }
   ],
   "source": [
    "def insert_sorted_optimized(arr, num):\n",
    "    insert_point = next((i for i, x in enumerate(arr) if x > num), len(arr))\n",
    "    return arr[:insert_point] + [num] + arr[insert_point:]\n",
    "\n",
    "sorted_array = [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21]\n",
    "num_to_insert = 10\n",
    "new_sorted_array = insert_sorted_optimized(sorted_array, num_to_insert)\n",
    "print(\"插入后的新数组：\", new_sorted_array)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<a id = Section1> </a>\n",
    "## 19.有n个人围成一圈，顺序排号。从第一个人开始报数（从1到3报数），凡报到3的人退出圈子，问最后留下的是原来第几号的那位"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "最后留下的是原来第2号的那位。\n"
     ]
    }
   ],
   "source": [
    "def josephus_problem(n, k):\n",
    "\n",
    "    people = list(range(1, n + 1))\n",
    "    \n",
    "    step = k - 1\n",
    "    \n",
    "    while len(people) > 1:\n",
    "        index_to_remove = step % len(people)\n",
    "        \n",
    "        people.pop(index_to_remove)\n",
    "    \n",
    "    return people[0]\n",
    "\n",
    "n = 10  \n",
    "k = 3   \n",
    "result = josephus_problem(n, k)\n",
    "print(f\"最后留下的是原来第{result}号的那位。\")"
   ]
  }
 ],
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提交于 2024-12-01 15:58 . 期末

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