【Codeforces Round #372 (Div. 2)】Codeforces 716B Complete the Word

ZS the Coder loves to read the dictionary. He thinks that a word is

nice if there exists a substring (contiguous segment of letters) of it

of length 26 where each letter of English alphabet appears exactly

once. In particular, if the string has length strictly less than 26,

no such substring exists and thus it is not nice.

Now, ZS the Coder tells you a word, where some of its letters are

missing as he forgot them. He wants to determine if it is possible to

fill in the missing letters so that the resulting word is nice. If it

is possible, he needs you to find an example of such a word as well.

Can you help him? Input

The first and only line of the input contains a single string s

(1 ≤ |s| ≤ 50 000), the word that ZS the Coder remembers. Each

character of the string is the uppercase letter of English alphabet

(‘A’-‘Z’) or is a question mark (‘?’), where the question marks

denotes the letters that ZS the Coder can’t remember. Output

If there is no way to replace all the question marks with uppercase

letters such that the resulting word is nice, then print  - 1 in the

only line.

Otherwise, print a string which denotes a possible nice word that ZS

the Coder learned. This string should match the string from the input,

except for the question marks replaced with uppercase English letters.

If there are multiple solutions, you may print any of them.

滑动窗口，O(n)扫描。

#include<cstring>
#include<cstdio>
char s[];
int cnt[],l;
void make(int p)
{
    int i,j,k,x,y,z;
    for (i=p;i<=p+-;i++)
      if (s[i]=='?')
        for (j=;j<=;j++)
          if (!cnt[j])
          {
            s[i]=j+'A'-;
            cnt[j]++;
            break;
          }
    for (i=;i<=l;i++)
      if (s[i]=='?') s[i]='A';
    for (i=;i<=l;i++)
      printf("%c",s[i]);
    printf("\n");
}
int main()
{
    int i,j,k,m,n,p,q,x,y,z,cnt_emp,now;
    scanf("%s",s+);
    l=strlen(s+);
    if (l<)
    {
        printf("-1\n");
        return ;
    }
    cnt_emp=now=;
    for (i=;i<=;i++)
      if (s[i]=='?') cnt_emp++;
      else
      {
        x=s[i]-'A'+;
        if (!cnt[x]) now++;
        cnt[x]++;
      }
    if (now+cnt_emp==)
    {
        make();
        return ;
    }
    for (i=;i<=l-+;i++)
    {
        if (s[i-]=='?') cnt_emp--;
        else
        {
            x=s[i-]-'A'+;
            cnt[x]--;
            if (!cnt[x]) now--;
        }
        if (s[i+-]=='?') cnt_emp++;
        else
        {
            x=s[i+-]-'A'+;
            if (!cnt[x]) now++;
            cnt[x]++;
        }
        if (now+cnt_emp==)
        {
            make(i);
            return ;
        }
    }
    printf("-1\n");
}