【CUGBACM15级BC第37场 A】hdu 5202 Rikka with string

Rikka with string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1577    Accepted Submission(s): 587

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:


One day, Yuta got a string which contains n letters but Rikka lost it in accident. Now they want to recover the string. Yuta remembers that the string only contains lowercase letters and it is not a palindrome string. Unfortunately he cannot remember some letters. Can you help him recover the string?


It is too difficult for Rikka. Can you help her?

 

Input

This problem has multi test cases (no more than20 ). For each test case, The first line contains a number n (1≤n≤1000). The next line contains an n-length string which only contains lowercase letters and ‘?’ – the place which Yuta is not sure.

 

Output

For each test cases print a n-length string – the string you come up with. In the case where more than one string exists, print the lexicographically first one. In the case where no such string exists, output “QwQ”.

 

Sample Input

5a?bb?3aaa

 

Sample Output

aabbaQwQ

 

题目大意：给出一个字符串，字符串中有n个问号，要求你用小写字母填充哪些问号，使得填充完后，字符串的字典序最小且不形成回文，如果无论怎么填充都是回文的话，就输出QwQ

题目大意：处理的话，先将所有问号用填充，然后再考虑是否回文，如果是回文的话，优先修改的应该是最后一个填充位置的字符，如果还是回文的话，就只需修改倒数第二个填充位置的字符，最后一个不变。因为问号大于等于2的话，那么这个字符串必定不会回文了，这里几个陷阱样例
1 ？，输出应该是QwQ
5 aa？aa，输出应该是QwQ
5 a？？aa，输出abaaa
注意一下特殊情况，即最后一个问号在中间的情况

#include<cstdio>#include<cstring>#define maxn 1010int n;char str1[maxn], str2[maxn];int mark[maxn];bool check(char *s){    for (int i = 0; i < n / 2; i++)        if (s[i] != s[n - 1 - i])        {            return false;        }    return true;}int main(){    while (scanf("%d", &n) == 1)    {        scanf("%s", str1);        int cnt = 0;        for (int i = 0; i < n; i++)        {            str2[i] = str1[i];            if (str1[i] == '?')            {                str2[i] = 'a';                mark[cnt++] = i;            }        }        str2[n] = '\0';                if (n == 1 && str1[0] == '?')        {            printf("QwQ\n");            continue;        }        if (cnt == 0 && check(str1))        {            printf("QwQ\n") ;            continue;        }        if (!check(str2))        {            printf("%s\n", str2);            continue;        }        str2[mark[cnt - 1]] = 'b';        if (!check(str2))        {            printf("%s\n", str2);            continue;        }        if (cnt == 1)        {            printf("QwQ\n");            continue;        }                str2[mark[cnt - 2]] = 'b';        str2[mark[cnt - 1]] = 'a';        printf("%s\n", str2);    }    return 0;}