uva401 - Palindromes结题报告

题目地址 ：
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=&problem=342&mosmsg=Submission+received+with+ID+13517947
好久没有刷题了 感觉题目比较简单 锻炼思维的严谨性 看来刷题还是很有必要的

贴上源码

#include<iostream>
#include<fstream>
#include<cstring>
#include<string>
using namespace std;
//Character	Reverse	Character	Reverse	Character	Reverse
//A	A	M	M	Y	Y
//B		N		Z	5
//C		O	O	1	1
//D		P		2	S
//E	3	Q		3	E
//F		R		4
//G		S	2	5	Z
//H	H	T	T	6
//I	I	U	U	7
//J	L	V	V	8	8
//K		W	W	9
//L	J	X	X
//判0与o的时候要注意哦;
char one[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789";
char two[] = "A   3  HIL JM O   2TUVWXY51SE Z  8 ";
char char_mirror[26]=
{
'A','*','*','*','3','*','*','H','I','L','*','J','M',
'*','O','*','*','R','2','T','U','V','W','X','Y','5'
};
char num_mirror[10]={'0','1','S','E','*','Z','*','*','8','*'};
int check_huiwen(char * str)
{
int len = strlen(str);
for (int i = 0; i < len/2; i++)
{
if (str[i] == '0' && str[len-i-1] == 'O' || str[i] == 'O' && str[len-i-1] == '0')
{
continue;
}
if (str[len-i-1] != str[i])
return -1;
}
return 0;
}

int check_mirror(char * str)
{
int len  = strlen(str);
for (int i = 0; i < len/2+1; i++)
{
if (isalpha(str[i]) && str[len-i-1] == char_mirror[str[i]-'A'] ||  isdigit(str[i])  && num_mirror[str[i]-'0'] == str[len-i-1])
{
continue;
}
else
{
if (str[i] == '0' &&  str[len - 1 - i] == 'O' || str[i] == 'O' && str[len - 1 - i] == '0')
continue;
return -1;
}

}
return 0;
}
int main()
{
#ifndef ONLINE_JUDGE
fstream cin("D://code//acm//txt//4.txt");
#endif
char str[25];
while (cin>>str)
{
int is_huiwen = 0;
int is_mirror = 0;
if ( check_huiwen(str) == 0 )
{
is_huiwen = 1;
}
if ( check_mirror(str) == 0 )
{
is_mirror = 1;
}
if (is_huiwen && is_mirror )
{
cout<<str<<" -- is a mirrored palindrome."<<endl<<endl;
}
else if (is_huiwen && ! is_mirror)
{
cout<<str<<" -- is a regular palindrome."<<endl<<endl;
}
else if (is_mirror && ! is_huiwen)
{
cout<<str<<" -- is a mirrored string."<<endl<<endl;
}
else if (!is_huiwen && ! is_mirror)
{
cout<<str<<" -- is not a palindrome."<<endl<<endl;
}

}
return 0;
}

这道题目不难却刷了很久

原因：

1.样例超过去之后 每一行都多复制了一些空格，以后注意

2.check_mirror函数for循环的终止条件i<len/2+1,中间点容易被忽视，这个也可以看作边界条件,以后注意

3.为什么会出现第二个问题呢 因为在写check_huiwen时 就写的 i<len/2 就顺手抄了过来 不多思考 两个问题还是不一样的 细节没把握好,以后参考以前的代码时一定注意边界条件

4.isdigit函数没有想到;

5.#ifndef ONLINE_JUDGE #endif

6.