返回信息流大一新生菜鸟。写一道UVaOJ的题目。不知道为什么是wrong answer?求大神指点。
原题如下:
题号是401
A regular palindrome is a string of numbers or letters that is the same forward as backward. For example, the string "ABCDEDCBA" is a palindrome because it is the same when the string is read from left to right as when the string is read from right to left.
A mirrored string is a string for which when each of the elements of the string is changed to its reverse (if it has a reverse) and the string is read backwards the result is the same as the original string. For example, the string "3AIAE" is a mirrored string because "A" and "I" are their own reverses, and "3" and "E" are each others' reverses.
A mirrored palindrome is a string that meets the criteria of a regular palindrome and the criteria of a mirrored string. The string "ATOYOTA" is a mirrored palindrome because if the string is read backwards, the string is the same as the original and because if each of the characters is replaced by its reverse and the result is read backwards, the result is the same as the original string. Of course, "A", "T", "O", and "Y" are all their own reverses.
A list of all valid characters and their reverses is as follows.
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
Note that O (zero) and 0 (the letter) are considered the same character and therefore ONLY the letter "0" is a valid character.
Input
Input consists of strings (one per line) each of which will consist of one to twenty valid characters. There will be no invalid characters in any of the strings. Your program should read to the end of file.
Output
For each input string, you should print the string starting in column 1 immediately followed by exactly one of the following strings.
STRING CRITERIA
" -- is not a palindrome." if the string is not a palindrome and is not a mirrored string
" -- is a regular palindrome." if the string is a palindrome and is not a mirrored string
" -- is a mirrored string." if the string is not a palindrome and is a mirrored string
" -- is a mirrored palindrome." if the string is a palindrome and is a mirrored string
Note that the output line is to include the -'s and spacing exactly as shown in the table above and demonstrated in the Sample Output below.
In addition, after each output line, you must print an empty line.
Sample Input
NOTAPALINDROME
ISAPALINILAPASI
2A3MEAS
ATOYOTA
Sample Output
NOTAPALINDROME -- is not a palindrome.
ISAPALINILAPASI -- is a regular palindrome.
2A3MEAS -- is a mirrored string.
ATOYOTA -- is a mirrored palindrome.
我写的代码如下:
#include <stdio.h>
#include <string.h>
int main()
{
char a[100];
int i,j,m,n,t;
char s;
int c;
while(scanf("%s",a)&&((c=getchar())!=EOF))
{
int flag1=1,flag2=1;
n=strlen(a);
m=n-1;
t=m;
for(i=0;i<=t;i++,t--)
{
if(a[i]!=a[t])
{
flag1=0;
}
}
for(j=0;j<=m&&flag2;j++,m--)
{
switch(a[j])
{
case 'A':
if(a[m]=='A')
flag2=1;
break;
case 'M':
if(a[m]=='M')
flag2=1;
break;
case 'O':
if(a[m]=='O')
flag2=1;
break;
case 'W':
if(a[m]=='W')
flag2=1;
break;
case 'U':
if(a[m]=='U')
flag2=1;
break;
case 'H':
if(a[m]=='H')
flag2=1;
break;
case 'Y':
if(a[m]=='Y')
flag2=1;
break;
case 'X':
if(a[m]=='X')
flag2=1;
break;
case 'T':
if(a[m]=='T')
flag2=1;
break;
case '8':
if(a[m]=='8')
flag2=1;
break;
case 'V':
if(a[m]=='V')
flag2=1;
break;
case '2':
if(a[m]=='S')
flag2=1;
break;
case 'S':
if(a[m]=='2')
flag2=1;
break;
case '1':
if(a[m]=='1')
flag2=1;
break;
case 'Z':
if(a[m]=='5')
flag2=1;
break;
case '5':
if(a[m]=='Z')
flag2=1;
break;
case '3':
if(a[m]=='E')
flag2=1;
break;
case 'E':
if(a[m]=='3')
flag2=1;
break;
case 'L':
if(a[m]=='J')
flag2=1;
break;
case 'J':
if(a[m]=='L')
flag2=1;
break;
default:
flag2=0;
break;
}
}
if(flag1&&flag2)
printf("%s -- is a mirrored palindrome.\n\n",a);
if(flag1&&!flag2)
printf("%s -- is a regular palindrome.\n\n",a);
if(!flag1&&flag2)
printf("%s -- is a mirrored string.\n\n",a);
if(!flag2&&!flag1)
printf("%s -- is not a palindrome.\n\n",a);
}
return 0;
}
试了几个输入感觉都没问题,然后溢出的情况也没有的。不知道是哪里出了问题?
这是一条镜像帖。来源:北邮人论坛 / cpp / #89284同步于 2015/10/29
该镜像源已超过 30 天没有更新,可能在源站已被删除。
CPP机器人发帖
[问题]【急需请教】求大神指教
CcZhome
2015/10/29镜像同步4 回复
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4 条回复
已经解决
#include <stdio.h>
#include <string.h>
int main()
{
char a[100];
int i,j,m,n,t;
char s;
int c;
while(scanf("%s",a)&&((c=getchar())!=EOF))
{
int flag1=1,flag2=1;
n=strlen(a);
m=n-1;
t=m;
for(i=0;i<=t;i++,t--)
{
if(a[i]!=a[t])
{
flag1=0;
}
}
for(j=0;j<=m&&flag2;j++,m--)
{
switch(a[j])
{
case 'A':
if(a[m]=='A')
flag2=1;
else flag2=0;
break;
case 'M':
if(a[m]=='M')
flag2=1;
else flag2=0;
break;
case 'O':
if(a[m]=='O')
flag2=1;
else flag2=0;
break;
case 'W':
if(a[m]=='W')
flag2=1;
else flag2=0;
break;
case 'U':
if(a[m]=='U')
flag2=1;
else flag2=0;
break;
case 'H':
if(a[m]=='H')
flag2=1;
else flag2=0;
break;
case 'Y':
if(a[m]=='Y')
flag2=1;
else flag2=0;
break;
case 'X':
if(a[m]=='X')
flag2=1;
else flag2=0;
break;
case 'T':
if(a[m]=='T')
flag2=1;
else flag2=0;
break;
case '8':
if(a[m]=='8')
flag2=1;
else flag2=0;
break;
case 'V':
if(a[m]=='V')
flag2=1;
else flag2=0;
break;
case '2':
if(a[m]=='S')
flag2=1;
else flag2=0;
break;
case 'S':
if(a[m]=='2')
flag2=1;
else flag2=0;
break;
case '1':
if(a[m]=='1')
flag2=1;
else flag2=0;
break;
case 'Z':
if(a[m]=='5')
flag2=1;
else flag2=0;
break;
case '5':
if(a[m]=='Z')
flag2=1;
else flag2=0;
break;
case '3':
if(a[m]=='E')
flag2=1;
else flag2=0;
break;
case 'E':
if(a[m]=='3')
flag2=1;
else flag2=0;
break;
case 'L':
if(a[m]=='J')
flag2=1;
else flag2=0;
break;
case 'J':
if(a[m]=='L')
flag2=1;
else flag2=0;
break;
default:
flag2=0;
break;
}
}
if(flag1&&flag2)
printf("%s -- is a mirrored palindrome.\n\n",a);
if(flag1&&!flag2)
printf("%s -- is a regular palindrome.\n\n",a);
if(!flag1&&flag2)
printf("%s -- is a mirrored string.\n\n",a);
if(!flag2&&!flag1)
printf("%s -- is not a palindrome.\n\n",a);
}
return 0;
}
char table[256];
table['A'] = 'A';
table['E'] = '3';
table['J'] = 'L';
table['L'] = 'J';
...
// look up
char in = 'A';
char out = table[in];