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+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "DNA序列中A有20个\n",
+      "DNA序列中G有17个\n",
+      "DNA序列中C有12个\n",
+      "DNA序列中T有21个\n",
+      "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",
+      "肠粉是广州的一种美食，它的主要原料是米；\n",
+      "海蛎煎是泉州的一种美食，它的主要原料是海蛎；\n",
+      "糍粑是成都的一种美食，它的主要原料是米；\n"
+     ]
+    }
+   ],
+   "source": [
+    "\"\"\"\n",
+    "题目一\n",
+    "有一个这样的DNA核酸序列\n",
+    "“AGCTTTTCATTCTGACTGCAACGGGCAATATGTCTCTGTGTGGATTAAAAAAAGAGTGTCTGATAGCAGC”\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。\n",
+    "\"\"\"\n",
+    "# 初始化DNA序列与AGCT的个数\n",
+    "DNA_str = 'AGCTTTTCATTCTGACTGCAACGGGCAATATGTCTCTGTGTGGATTAAAAAAAGAGTGTCTGATAGCAGC'\n",
+    "DNA = []\n",
+    "DNA.extend(DNA_str)\n",
+    "    \n",
+    "elements = {'A': 0,\n",
+    "            'G': 0,\n",
+    "            'C': 0,\n",
+    "            'T': 0,\n",
+    "            }\n",
+    "\n",
+    "# 计算DNA序列中AGCT的个数             \n",
+    "for x in DNA:\n",
+    "    for key in elements:\n",
+    "        if x == key:\n",
+    "            elements[key] = elements[key]+1\n",
+    "            break\n",
+    "\n",
+    "# 输出DNA序列中AGCT的个数    \n",
+    "for key, number in elements.items():\n",
+    "    print(\"DNA序列中%s有%d个\" % (key,number))\n",
+    "    \n",
+    "    \n",
+    "\n",
+    "\"\"\"\n",
+    "题目二\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",
+    "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”。\n",
+    "\"\"\"\n",
+    "# 将可能的得分存入元组\n",
+    "score = 0.25\n",
+    "scores = []\n",
+    "while score <= 10:\n",
+    "    scores.append(score)\n",
+    "    score = score + 0.25\n",
+    "scores = tuple(scores)\n",
+    "\n",
+    "# 遍历并输出元组的每一项\n",
+    "for x in scores:\n",
+    "    print(\"一个运动员可能得\" + str(round(x,2)) + \"分\")\n",
+    "\n",
+    "\n",
+    "\n",
+    "\"\"\"\n",
+    "题目三\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'}}。\n",
+    "当然，你可以做的更丰富一些。 最后遍历你熟悉的美食，打印出，类似如下的句子：\n",
+    "“肠粉是广州的一种美食，它的主要原料是米”。\n",
+    "\"\"\"\n",
+    "# 列出地域美食\n",
+    "foods = {'肠粉':{'城市':'广州','原料':'米'},\n",
+    "         '海蛎煎':{'城市':'泉州','原料':'海蛎'},\n",
+    "         '糍粑':{'城市':'成都','原料':'米'},\n",
+    "        }\n",
+    "\n",
+    "# 遍历并打印\n",
+    "for name, information in foods.items():\n",
+    "    print(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,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
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+  "language_info": {
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+}