题目分析：
1.直接枚举回文串的起始位置和重点位置会超时，O(n^3)，会超时，最长的回文串为2000；
2. 直接枚举中间位置，O(n^2)，
3.先对字符文本进行预处理，对非字母符号全部去掉，s1[20100],开一个p[20100]记录s1[i]在原字符串中的位置，
方便输出原字符串.
4.注意它的输入，是以文本的形式输入，可能有好几行，
5.FILE *fin=fopen("calfflac.in","r");
FILE *fout=fopen("calfflac.out","w");
int k=0;
while(!feof(fin))
{
s[k++]=fgetc(fin);
}
fprintf(fout,"%d\n",ans);
for(int i=p[start];i<=p[end];i++)
fprintf(fout,"%c",s[i]);
fprintf(fout,"\n");
的用法
/*
ID:wconvey
PROG:calfflac
LANG:C++
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctype.h>
#include<algorithm>
using namespace std;
char s[20100],s1[20100];
int p[20100];
int main()
{
//freopen("calfflac.in","r",stdin);
//freopen("calfflac.out","w",stdout);
FILE *fin=fopen("calfflac.in","r");
FILE *fout=fopen("calfflac.out","w");
int k=0;
while(!feof(fin))
{
s[k++]=fgetc(fin);
}
s[k]='\0';
//gets(s);
{
int len=strlen(s);
int c=0;
for(int i=0;i<len;i++)
{
if(isalpha(s[i]))
{
p[c]=i;
s1[c++]=toupper(s[i]);
}
}
c--;
//printf("%d****\n",c);
//puts(s);
//puts(s1);
/*for(int i=0;i<=c;i++)
printf("p[%d]=%d %c ",i,p[i],s[p[i]]);
printf("*****\n");*/
int ans=0,start=0,end=0;
for(int i=1;i<=c-1;i++)
{
int temp=0,j;
if(s1[i]==s1[i+1])
{
temp+=2;
for(j=1;j<=i;j++)
{
if(i+j<=c && s1[i-j]==s1[i+1+j])
temp+=2;
else
break;
}
if(temp>ans)
{
ans=temp;
start=i-(j-1);
end=i+(j-1)+1;
//printf("temp=%d start=%d end=%d\n",temp,start,end);
}
}
else
{
temp+=1;
for(j=1;j<=i;j++)
{
if(i+j<=c&&s1[i-j]==s1[i+j] )//写错了,s1[i-j]==s1[i-j]
temp+=2;
else
break;
}
if(temp>ans)
{
ans=temp;
start=i-(j-1);
end=i+(j-1);
//printf("temp=%d start=%d end=%d\n",temp,start,end);
}
}
}//******for
/*printf("%d\n",ans);
for(int i=p[start];i<=p[end];i++)
printf("%c",s[i]);
printf("\n");*/
fprintf(fout,"%d\n",ans);
for(int i=p[start];i<=p[end];i++)
fprintf(fout,"%c",s[i]);
fprintf(fout,"\n");
}
//system("pasue");
return 0;
}