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   "source": [
    "作业要求以学号命名，类似\"123456.ipynb\"，不需要加姓名。\n",
    "\n",
    "\n",
    "作业提交的方式见下面链接：\n",
    "https://blog.csdn.net/sheagu/article/details/122397816"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5d3dca5a",
   "metadata": {},
   "source": [
    "## 题目一\n",
    "\n",
    "有一个这样的DNA核酸序列\n",
    "\n",
    "“AGCTTTTCATTCTGACTGCAACGGGCAATATGTCTCTGTGTGGATTAAAAAAAGAGTGTCTGATAGCAGC”\n",
    "\n",
    "请把这个核酸序列存入一个list，并数一数A、G、C、T各有多少个。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "id": "e6521351",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "20 17 12 21\n"
     ]
    }
   ],
   "source": [
    "string = 'AGCTTTTCATTCTGACTGCAACGGGCAATATGTCTCTGTGTGGATTAAAAAAAGAGTGTCTGATAGCAGC'\n",
    "DNA = list(string)\n",
    "num_A,num_G,num_C,num_T = 0,0,0,0\n",
    "\n",
    "for dna in DNA:\n",
    "    if dna == 'A':\n",
    "        num_A += 1\n",
    "    elif dna == 'G':\n",
    "        num_G += 1\n",
    "    elif dna == 'C':\n",
    "        num_C += 1\n",
    "    else:\n",
    "        num_T += 1\n",
    "print(num_A,num_G,num_C,num_T)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "daf8500d",
   "metadata": {},
   "source": [
    "## 题目二\n",
    "\n",
    "一个花样滑冰运动员表演后，裁判给表演内容进行评分，分数从0.25分到10分，每次增加值为0.25分。\n",
    "\n",
    "试生成一个元组，把可能的得分存入元组，并遍历元组的每一项，打印“一个运动员可能得_____分”。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "90604ebb",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "一个运动员可能得0.25分。\n",
      "一个运动员可能得0.5分。\n",
      "一个运动员可能得0.75分。\n",
      "一个运动员可能得1.0分。\n",
      "一个运动员可能得1.25分。\n",
      "一个运动员可能得1.5分。\n",
      "一个运动员可能得1.75分。\n",
      "一个运动员可能得2.0分。\n",
      "一个运动员可能得2.25分。\n",
      "一个运动员可能得2.5分。\n",
      "一个运动员可能得2.75分。\n",
      "一个运动员可能得3.0分。\n",
      "一个运动员可能得3.25分。\n",
      "一个运动员可能得3.5分。\n",
      "一个运动员可能得3.75分。\n",
      "一个运动员可能得4.0分。\n",
      "一个运动员可能得4.25分。\n",
      "一个运动员可能得4.5分。\n",
      "一个运动员可能得4.75分。\n",
      "一个运动员可能得5.0分。\n",
      "一个运动员可能得5.25分。\n",
      "一个运动员可能得5.5分。\n",
      "一个运动员可能得5.75分。\n",
      "一个运动员可能得6.0分。\n",
      "一个运动员可能得6.25分。\n",
      "一个运动员可能得6.5分。\n",
      "一个运动员可能得6.75分。\n",
      "一个运动员可能得7.0分。\n",
      "一个运动员可能得7.25分。\n",
      "一个运动员可能得7.5分。\n",
      "一个运动员可能得7.75分。\n",
      "一个运动员可能得8.0分。\n",
      "一个运动员可能得8.25分。\n",
      "一个运动员可能得8.5分。\n",
      "一个运动员可能得8.75分。\n",
      "一个运动员可能得9.0分。\n",
      "一个运动员可能得9.25分。\n",
      "一个运动员可能得9.5分。\n",
      "一个运动员可能得9.75分。\n",
      "一个运动员可能得10.0分。\n"
     ]
    }
   ],
   "source": [
    "# 你的代码\n",
    "nums = []\n",
    "for i in range(1,41):\n",
    "    num = i/4\n",
    "    nums.append(num)\n",
    "nums = tuple(nums)\n",
    "for a in nums:    \n",
    "    print('一个运动员可能得'+str(a)+'分。')"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0c3a4e99",
   "metadata": {},
   "source": [
    "## 题目三\n",
    "\n",
    "创建一个字典，列出你所了解的地域美食，比如{'肠粉':{'城市':'广州','原料':'米'}}。当然，你可以做的更丰富一些。\n",
    "最后遍历你熟悉的美食，打印出，类似如下的句子：“肠粉是广州的一种美食，它的主要原料是米”。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "id": "b0026af6",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "肠粉是广州的一种美食，它的主要原料是米。\n",
      "热干面是武汉的一种美食，它的主要原料是面。\n",
      "鸭血粉丝汤是南京的一种美食，它的主要原料是肉。\n"
     ]
    }
   ],
   "source": [
    "# 你的代码\n",
    "delicacy = {\n",
    "    '肠粉':{'城市':'广州','原料':'米'},\n",
    "    '热干面':{'城市':'武汉','原料':'面'},\n",
    "    '鸭血粉丝汤':{'城市':'南京','原料':'肉'},\n",
    "}\n",
    "for name,information in delicacy.items():\n",
    "   print(f\"{name}是{information['城市']}的一种美食，它的主要原料是{information['原料']}。\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "4c3a6bc9",
   "metadata": {},
   "outputs": [],
   "source": []
  }
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