华为机试题：字符转换Univesity

迷南。 2022-08-20 07:29 126阅读 0赞

<table style="margin:0px; padding:0px; width:970px; border:0px; border-collapse:collapse; border-spacing:0px; empty-cells:show; color:rgb(0,0,0); font-family:'Times New Roman',Arial,微软雅黑,宋体,Tahoma,Helvetica,SimSun,sans-serif,宋体"> 
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 <td style="margin:0px; padding:6px 3px 3px; word-break:break-all; word-wrap:break-word; border-right-style:solid; border-right-width:1px; border-right-color:rgb(238,238,238); font-size:14px; border-bottom-style:solid; border-bottom-width:1px; border-bottom-color:rgb(238,238,238)"> 描述:&nbsp;</td> 
 <td style="margin:0px; padding:6px 3px 3px; word-break:break-all; word-wrap:break-word; font-size:14px; border-bottom-style:solid; border-bottom-width:1px; border-bottom-color:rgb(238,238,238)"> 简要描述 字符替换，按照替换规则，将输入的字符串做转换;其它字符不作处理。 &nbsp; 详细描述： 题目背景 字母的替换规则，如下所示，A被E替换，B被C替换，依次类推。&nbsp; A&nbsp;B&nbsp;C&nbsp;D&nbsp;E&nbsp;F&nbsp;G&nbsp;H&nbsp;I&nbsp;J&nbsp;K&nbsp;L&nbsp;M&nbsp;N&nbsp;O&nbsp;P&nbsp;Q&nbsp;R&nbsp;S&nbsp;T&nbsp;U&nbsp;V&nbsp;W&nbsp;X&nbsp;Y&nbsp;Z E&nbsp;C&nbsp;F&nbsp;A&nbsp;J&nbsp;K&nbsp;L&nbsp;B&nbsp;D&nbsp;G&nbsp;H&nbsp;I&nbsp;V&nbsp;W&nbsp;Z&nbsp;Y&nbsp;M&nbsp;N&nbsp;O&nbsp;P&nbsp;Q&nbsp;R&nbsp;S&nbsp;T&nbsp;U&nbsp;X&nbsp; a&nbsp;b&nbsp;c&nbsp;d&nbsp;e&nbsp;f&nbsp;g&nbsp;h&nbsp;i&nbsp;j&nbsp;k&nbsp;l&nbsp;m&nbsp;n&nbsp;o&nbsp;p&nbsp;q&nbsp;r&nbsp;s&nbsp;t&nbsp;u&nbsp;v&nbsp;w&nbsp;x&nbsp;y&nbsp;z e&nbsp;r&nbsp;w&nbsp;q&nbsp;t&nbsp;y&nbsp;g&nbsp;h&nbsp;b&nbsp;n&nbsp;u&nbsp;i&nbsp;o&nbsp;p&nbsp;s&nbsp;j&nbsp;k&nbsp;d&nbsp;l&nbsp;f&nbsp;a&nbsp;z&nbsp;x&nbsp;c&nbsp;v&nbsp;m&nbsp; &nbsp; &nbsp; </td> 
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#include <stdlib.h>
 #include "oj.h"
 
 
 
 char* CharReplace(char* pSourceChar)
 {
 	if (pSourceChar == NULL)
 	{
 		return 0;
 	}
 
 	int temp = 0;
 
 	char bigchar[] = { "ECFAJKLBDGHIVWZYMNOPQRSTUX" };
 	char smallchar[] = { "erwqtyghbnuiopsjkdlfazxcvm" };
 	int i = 0;
 
 	for (i = 0; pSourceChar[i] != '\0'; i++)
 	{
 		if (pSourceChar[i] >= 'A'&&pSourceChar[i] <= 'Z')
 		{
 			temp = (pSourceChar[i] - 'A');
 			pSourceChar[i] = bigchar[temp];
 		}
 		else if (pSourceChar[i] >= 'a'&&pSourceChar[i] <= 'z')
 		{
 			temp = (pSourceChar[i] - 'a');
 			pSourceChar[i] = smallchar[temp];
 		}
 
 	}
 	pSourceChar[i] = '\0';
 	return pSourceChar;
 }