解题思路:先用预处理处理保存下镜像的数据,再细心模拟,at the last but not least,In addition, after each output line(一个), you must print an empty line(又一个),即是两个换行。
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=105;
char a[37];
char r[37];
void pre(){
memset(a,0,sizeof(a));
memset(r,0,sizeof(r));
for(int i=0;i<26;i++){
a[i]='A'+i;
if(i==0)r[i]='A';
if(i==4) r[i]='3';
if(i==7||i==8||i==12||i==14||(i>=19&&i<=25))
r[i]=a[i];
if(i==9) r[i]='L';
if(i==11)r[i]='J';
if(i==18) r[i]='2';
if(i==25) r[i]='5';
}
for(int i=1;i<=9;i++){
a[25+i]='0'+i;
if(i==1||i==8) r[25+i]=a[25+i];
if(i==2) r[25+i]='S';
if(i==3) r[25+i]='E';
if(i==5) r[25+i]='Z';
}
}
int is_same1(char ch1,char ch2){
for(int i=0;i<35;i++){
if(a[i]==ch1&&ch2==r[i]){
return 1;
}
}
return 0;
}
int is_mirror(char *ch){
int len=strlen(ch);
int mid=len/2;
for(int l=0,h=len-1;l<=mid;l++,h--){
if(!is_same1(ch[l],ch[h])){
if(!(ch[l]=='0'&&(ch[h]=='0'||ch[h]=='O')||(ch[l]=='O'&&(ch[h]=='0'||ch[h]=='O'))))///o和0没差别
return 0;
}
}
return 1;
}
int is_pali(char*ch){
int len=strlen(ch);
int mid=len/2;
for(int l=0,h=len-1;l<=mid;l++,h--){
if(ch[l]!=ch[h]){
if(!(ch[l]=='0'&&(ch[h]=='0'||ch[h]=='O')||(ch[l]=='O'&&(ch[h]=='0'||ch[h]=='O'))))
return 0;
}
}
return 1;
}
char res[N];
int main(){
pre();
while(scanf("%s",res)!=EOF){
if(is_pali(res)){
if(is_mirror(res)){
printf("%s -- is a mirrored palindrome.\n\n",res);
}
else{
printf("%s -- is a regular palindrome.\n\n",res);
}
}
else if(is_mirror(res)){
printf("%s -- is a mirrored string.\n\n",res);
}
else{
printf("%s -- is not a palindrome.\n\n",res);
}
}
return 0;
}