{
  "question_id": 7455446,
  "question_title": "优雅的字符串",
  "question_content": "<p>对于一个字符串&#xff0c;如果这个字符串在ASCII码的意义上是有序的&#xff0c;即升序或降序&#xff0c;则称该字符串为“优雅的字符串”。一个长为n的字符串s&#xff0c;对于1&lt;&#61; i &lt;&#61; n - 1&#xff0c;如果总有s[i] &lt;&#61; s[i&#43;1]&#xff0c;则字符串为升序&#xff1b;如果总有s[i] &gt;&#61; s[i&#43;1]&#xff0c;则字符串降序。现给定一个字符串&#xff0c;请你判断该字符串是否“优雅”&#xff0c;如果该字符串是“优雅”的&#xff0c;则判断是“正优雅”&#xff08;升序&#xff09;还是“负优雅”&#xff08;降序&#xff09;。\n输入描述\n一个非空&#xff0c;不包含空格的字符串s&#xff0c;保证字符串中的字符都是ASCII标准中的字符。数据保证不会出现所有字符相同的字符串。&#xff08;2&lt;&#61; length(s) &lt;&#61;1e5&#xff09;\n输出描述\n如果字符串“正优雅”&#xff0c;则输出“Positive elegance”&#xff0c;如果“负优雅”&#xff0c;则输出“Negative elegance”&#xff0c;否则输出“Non elegance”&#xff08;均不含引号&#xff09;。\n样例输入 (*&#43;12356ASdfz\n样例输出\nPositive elegance</p>",
  "difficulty": "困难",
  "answer_id": 53438399,
  "answer_content": "<p>代码如下&#xff0c;如有帮助&#xff0c;请采纳一下&#xff0c;谢谢。</p>\n\n<pre>\n<code>#include &lt;stdio.h&gt;\n#include &lt;string.h&gt;\n\nint main()\n{\n\tchar a[100] &#61; {0};\n\tint i;\n\tint zyy &#61; 1;\n\tint fyy &#61; 1;\n\tprintf(&#34;请输入字符串:&#34;);\n\tgets(a);\n\t//判断是否正优雅\n\tfor (i&#61;1;i&lt;strlen(a);i&#43;&#43;)\n\t{\n\t\tif(a[i] &lt; a[i-1])\n\t\t{\n\t\t\tzyy &#61; 0;\n\t\t\tbreak;\n\t\t}\n\t}\n\t//判断是否负优雅\n\tfor (i&#61;1;i&lt;strlen(a);i&#43;&#43;)\n\t{\n\t\tif(a[i] &gt; a[i-1])\n\t\t{\n\t\t\tfyy &#61; 0;\n\t\t\tbreak;\n\t\t}\n\t}\n\n\tif (zyy &amp;&amp; !fyy)\n\t{\n\t\tprintf(&#34;Positive elegance\\n&#34;);\n\t}else if (!zyy &amp;&amp; fyy)\n\t{\n\t\tprintf(&#34;Negative elegance\\n&#34;);\n\t}else\n\t\tprintf(&#34;Non elegance\\n&#34;);\n\treturn 0;\n\n}</code></pre>\n\n<p> </p>\n",
  "tag_name": "c语言",
  "cpp": "#include <stdio.h>\n#include <string.h>\nint main()\n{\n\tchar a[100] = {0};\n\tint i;\n\tint zyy = 1;\n\tint fyy = 1;\n\tprintf(\"请输入字符串:\");\n\tgets(a);\n\tfor (i=1;i<strlen(a);i++)\n\t{\n\t\tif(a[i] < a[i-1])\n\t\t{\n\t\t\tzyy = 0;\n\t\t\tbreak;\n\t\t}\n\t}\n\tfor (i=1;i<strlen(a);i++)\n\t{\n\t\tif(a[i] > a[i-1])\n\t\t{\n\t\t\tfyy = 0;\n\t\t\tbreak;\n\t\t}\n\t}\n\tif (zyy && !fyy)\n\t{\n\t\tprintf(\"Positive elegance\\n\");\n\t}else if (!zyy && fyy)\n\t{\n\t\tprintf(\"Negative elegance\\n\");\n\t}else\n\t\tprintf(\"Non elegance\\n\");\n\treturn 0;\n}",
  "topic_link": "https://bbs.csdn.net/topics/600470165",
  "status": 1,
  "keywords": "算法高阶,字符串匹配,算法问题选编,利用有限自动机进行字符串匹配",
  "license": "csdn.net",
  "notebook": {
    "cpp": "https://codechina.csdn.net/csdn/csdn-daily-code/-/jupyter/master/data/notebook/answer/ipynb/cpp/73.ipynb?type=file"
  },
  "notebook_enable": 1,
  "author": "qq_45807915"
}