UVa 401 - Palindromes -AC

来源：互联网发布：七天网络查分怎样查分编辑：程序博客网时间：2024/05/29 16:10

这道题看完题思路大致清晰，总体分开两个函数写，一个判断回文串，一个判断镜像串。

回文串判断好写，循环判断前半截与后半截是否一致即可。

关键在判断镜像串，一开始就也没多想，简单的判断呗~虽然知道要写好多，但也懒得细想。先判断字符串中有没有不是镜像字符的，有就直接返回0，没有再依次判断对称位置上的镜像字符是否对应，有不对应的就直接返回0，全都对应最后返回1.

虽然直接AC了，但这样写完了190行，突然感觉我好傻~
先贴出来吧~

#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAX_SIZE 10005int pal(char str[MAX_SIZE])//palindrome{    int i,j,slen;    slen=strlen(str);    for (i=0,j=slen-1;i<slen/2;i++,j--)        if (str[i]!=str[j])            break;    if (i==slen/2)        return 1;    return 0;}int mir (char str[MAX_SIZE])//mirrored{    int i,j,slen;    slen=strlen(str);    for (i=0;i<slen;i++)    {        switch (str[i])        {            case 'B':                return 0;            case 'C':                return 0;            case 'D':                return 0;            case 'F':                return 0;            case 'G':                return 0;            case 'K':                return 0;            case 'N':                return 0;            case 'P':                return 0;            case 'Q':                return 0;            case 'R':                return 0;            case '4':                return 0;            case '6':                return 0;            case '7':                return 0;            case '9':                return 0;        }    }    for (i=0;i<slen/2;i++)    {        switch(str[i])        {            case 'A':                if (str[slen-i-1]=='A')                    continue;                else                    return 0;            case 'E':                if (str[slen-i-1]=='3')                    continue;                else                    return 0;            case 'H':                if (str[slen-i-1]=='H')                    continue;                else                    return 0;            case 'I':                if (str[slen-i-1]=='I')                    continue;                else                    return 0;            case 'J':                if (str[slen-i-1]=='L')                    continue;                else                    return 0;            case 'L':                if (str[slen-i-1]=='J')                    continue;                else                    return 0;            case 'M':                if (str[slen-i-1]=='M')                    continue;                else                    return 0;            case 'O':                if (str[slen-i-1]=='O')                    continue;                else                    return 0;            case 'S':                if (str[slen-i-1]=='2')                    continue;                else                    return 0;            case 'T':                if (str[slen-i-1]=='T')                    continue;                else                    return 0;            case 'U':                if (str[slen-i-1]=='U')                    continue;                else                    return 0;            case 'V':                if (str[slen-i-1]=='V')                    continue;                else                    return 0;            case 'W':                if (str[slen-i-1]=='W')                    continue;                else                    return 0;            case 'X':                if (str[slen-i-1]=='X')                    continue;                else                    return 0;            case 'Y':                if (str[slen-i-1]=='Y')                    continue;                else                    return 0;            case 'Z':                if (str[slen-i-1]=='5')                    continue;                else                    return 0;            case '1':                if (str[slen-i-1]=='1')                    continue;                else                    return 0;            case '2':                if (str[slen-i-1]=='S')                    continue;                else                    return 0;            case '3':                if (str[slen-i-1]=='E')                    continue;                else                    return 0;            case '5':                if (str[slen-i-1]=='Z')                    continue;                else                    return 0;            case '8':                if (str[slen-i-1]=='8')                    continue;                else                    return 0;        }    }    return 1;}int main(){    char str[MAX_SIZE];    while (gets(str)!=NULL)    {        int pa,mi;        pa=pal(str);        mi=mir(str);        if (pa==0&&mi==0)            printf("%s -- is not a palindrome.\n",str);        else if (pa==1&&mi==0)            printf("%s -- is a regular palindrome.\n",str);        else if (pa==0&&mi==1)            printf("%s -- is a mirrored string.\n",str);        else if (pa&&mi)            printf("%s -- is a mirrored palindrome.\n",str);        printf("\n");    }    return 0;}

AC之后感觉这题这么写实在太狠了，然后又看了一下网上代码，看到

首先可以用两个数组把题目中的表格存起来

char *ch = "AEHIJLMOSTUVWXYZ12358";
char *re = "A3HILJMO2TUVWXY51SEZ8";

这才受到启发，把原来代码进行了改进，但这样要注意判断镜像串的时候要讨论字符串长度是否为1，否则会WA

修改后AC的代码：

#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAX_SIZE 10005int pal(char str[MAX_SIZE]){    int i,j,slen;    slen=strlen(str);    for (i=0,j=slen-1;i<slen/2;i++,j--)        if (str[i]!=str[j])            break;    if (i==slen/2)        return 1;    return 0;}int mir (char str[MAX_SIZE]){    char *ch="AEHIJLMOSTUVWXYZ12358";    char *re="A3HILJMO2TUVWXY51SEZ8";    int i,j,slen;    slen=strlen(str);    if (slen==1)    {        for (j=0;j<strlen(ch);j++)            if (str[0]==ch[j]&&str[0]==re[j])                return 1;        return 0;    }    else    {        for (i=0;i<slen/2;i++)        {            for (j=0;j<strlen(ch);j++)                if (str[i]==ch[j]&&str[slen-i-1]==re[j])                {                    j=-1;                    break;                }            if (j!=-1)                return 0;        }        return 1;    }}int main(){    char str[MAX_SIZE];    while (gets(str)!=NULL)    {        int pa,mi;        pa=pal(str);        mi=mir(str);        if (pa==0&&mi==0)            printf("%s -- is not a palindrome.\n",str);        else if (pa==1&&mi==0)            printf("%s -- is a regular palindrome.\n",str);        else if (pa==0&&mi==1)            printf("%s -- is a mirrored string.\n",str);        else if (pa&&mi)            printf("%s -- is a mirrored palindrome.\n",str);        printf("\n");    }    return 0;}

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