本文迁移自老博客，原始链接为 https://seven.blog.ustc.edu.cn/usacocalf-flac/
寻找最长的回文，寻找回文时忽略标点符号、空格(但应该保留下来以便做为答案输出),只用考虑字母’A’-‘Z’和’a’-‘z’。 解法：先处理一下字符串，使其变成全小写字母，然后枚举判断。
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
struct posi
{
int n;
char a;
}hi[20001];
int main()
{
FILE *fin = fopen ("calfflac.in", "r");
FILE *fout = fopen ("calfflac.out", "w");
char str[20001];
char str1[20001];
char tmp[81];
char c;
while( fgets (tmp, 80, fin) != NULL)
{
strcat(str, tmp);
}
int len = strlen(str);
str[len] = '\0';
int t =0 ;
for(int i = 0; i < len ;i++)
{
if((isupper(str[i])) || (islower(str[i])))
{
hi[t].a = tolower(str[i]);
hi[t].n = i;
str1[t++] = tolower(str[i]);
}
}
str1[t] = '\0';
int num = 1,tem;
int maxn = 0;
for(int i = 0; i < t - 1 ; i++)
{
int r;
if(str1[i] == str1[i + 1])
{
num = 2;
r = 1;
while((i - r >= 0) && (i + 1 + r < t))
{
if(str1[i - r] == str1[i + 1 + r])
{
num += 2;
r++;
}
else
{
break;
}
}
}
if(num > maxn)
{
maxn = num;
tem = i;
}
num = 1;
int r1 = 1;
while((i - r1 >= 0) && (i + r1 < t))
{
if(str1[i - r1] == str1[i + r1])
{
num += 2;
r1++;
}
else
{
break;
}
}
if(num > maxn)
{
maxn = num;
tem = i;
}
}
fprintf(fout,"%d\n" , maxn);
if(maxn % 2 == 0)
{
int s = hi[tem - (maxn / 2 ) + 1].n;
int e = hi[tem + (maxn / 2)].n;
for(int i = s;i <= e; i++)
{
fprintf(fout,"%c", str[i]);
}
}
else
{
int s = hi[tem - (maxn / 2 )].n;
int e = hi[tem + (maxn /2)].n;
for(int i = s;i <= e; i++)
{
fprintf(fout,"%c", str[i]);
}
}
fprintf(fout,"\n");
fclose(fin);
fclose(fout);
return 0;
}
