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60 lines
981 B
Markdown
60 lines
981 B
Markdown
# 6-6 判断回文字符串
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本题要求编写函数,判断给定的一串字符是否为“回文”。所谓“回文”是指顺读和倒读都一样的字符串。如“XYZYX”和“xyzzyx”都是回文。
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### 函数接口定义:
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```c++
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bool palindrome( char *s );
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```
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函数`palindrome`判断输入字符串`char *s`是否为回文。若是则返回`true`,否则返回`false`。
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### 裁判测试程序样例:
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```c++
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#include <stdio.h>
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#include <string.h>
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#define MAXN 20
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typedef enum {false, true} bool;
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bool palindrome( char *s );
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int main()
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{
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char s[MAXN];
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scanf("%s", s);
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if ( palindrome(s)==true )
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printf("Yes\n");
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else
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printf("No\n");
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printf("%s\n", s);
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return 0;
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}
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/* 你的代码将被嵌在这里 */
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```
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### 输入样例1:
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```in
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thisistrueurtsisiht
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```
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### 输出样例1:
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```out
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Yes
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thisistrueurtsisiht
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```
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### 输入样例2:
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```
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thisisnottrue
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```
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### 输出样例2:
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```
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No
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thisisnottrue
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```
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