{
   "question_id": 1090014,
   "question_title": "有序表的折半查找",
   "question_content": "问题描述：\n用有序表表示静态查找表时，通常检索函数可以用折半查找来实现。\n折半查找的查找过程是：首先确定待查记录所在的范围，然后逐步缩小范围直到找到或者确定找不到相应的记录为止。而每次需要缩小的范围均为上一次的一半，这样的查找过程可以被称为折半查找。  \n第二行包含n个用空格隔开的正整数，表示n个有序的整数。输入保证这n个整数是从小到大递增的。\n第三行包含k个用空格隔开的正整数，表示k次查询的目标。\n输出：\n只有1行，包含k个整数，分别表示每一次的查询结果。如果在查询中找到了对应的整数，则输出其相应的位置，否则输出-1。\n请在每个整数后输出一个空格，并请注意行尾输出换行。",
   "difficulty": "困难",
   "answer_id": 1296683,
   "answer_content": "\n```\n#include <stdio.h>\n\nint binary( int *a, int key, int n )\n{\n    int left = 0, right = n - 1, mid = 0;\n    mid = ( left + right ) / 2;\n    while( left < right && a[mid] != key )\n    {\n        if( a[mid] < key )\n        left = mid + 1;\n        else if( a[mid] > key )\n        right = mid - 1;\n        mid = ( left + right ) / 2;\n    }\n    if( a[mid] == key )   return mid;\n    return -1;\n}\n\nint main (void)\n{\n    int Base_a[20] = {1,3,5,8,9,40,120,123,125,150,199,200,1250,1255,1900,2000,2001,3000,3950,5000};\n\tint Search_a[5] = {12,199,9,2001,3500};\n\t\n\tint result = 0x00;\n\n\tfor(int i = 0;i < sizeof(Search_a)/sizeof(Search_a[0]);i++)\n\t{\n\t\tresult = binary(Base_a,Search_a[i],sizeof(Base_a)/sizeof(Base_a[0]));\n\t\tprintf(\"[%d %d] \",Search_a[i],result);\n\t}\n\tprintf(\"\\n\");\n\t\n\treturn 0;\n}\n```",
   "tag_name": "c语言",
   "cpp": "#include <stdio.h>\nint binary( int *a, int key, int n )\n{\n\tint left = 0, right = n - 1, mid = 0;\n\tmid = ( left + right ) / 2;\n\twhile( left < right && a[mid] != key )\n\t{\n\t\tif( a[mid] < key )\n\t\tleft = mid + 1;\n\t\telse if( a[mid] > key )\n\t\tright = mid - 1;\n\t\tmid = ( left + right ) / 2;\n\t}\n\tif( a[mid] == key )   return mid;\n\treturn -1;\n}\nint main (void)\n{\n\tint Base_a[20] = {1,3,5,8,9,40,120,123,125,150,199,200,1250,1255,1900,2000,2001,3000,3950,5000};\n\tint Search_a[5] = {12,199,9,2001,3500};\n\tint result = 0x00;\n\tfor(int i = 0;i < sizeof(Search_a)/sizeof(Search_a[0]);i++)\n\t{\n\t\tresult = binary(Base_a,Search_a[i],sizeof(Base_a)/sizeof(Base_a[0]));\n\t\tprintf(\"[%d %d] \",Search_a[i],result);\n\t}\n\tprintf(\"\\n\");\n\treturn 0;\n}",
   "topic_link": "https://bbs.csdn.net/topics/600470249",
   "status": 1,
   "keywords": "动态表,表扩张,算法中阶,摊还分析,高级设计和分析技术",
   "license": "csdn.net",
   "notebook": {
      "cpp": "https://codechina.csdn.net/csdn/csdn-daily-code/-/jupyter/master/data/notebook/answer/ipynb/cpp/13.ipynb?type=file"
   },
   "notebook_enable": 1
}