原题链接：洛谷
题目描述

You are given a string s consisting of the characters ‘0’, ‘1’,
and ‘?’. You need to replace all the characters with ‘?’ in the string
s by ‘0’ or ‘1’ so that the string becomes a palindrome and has
exactly a characters ‘0’ and exactly b characters ‘1’. Note that each
of the characters ‘?’ is replaced independently from the others.

A string t of length n is called a palindrome if the equality t[i] =
t[n-i+1]t[i]=t[n−i+1] is true for all i (1≤i≤n ).

For example, if s=s= “01???0”, a=4 and b=4 , then you can replace
the characters ‘?’ in the following ways:

“01011010”; “01100110”. For the given string s and the numbers a and b
, replace all the characters with ‘?’ in the string s by ‘0’ or ‘1’ so
that the string becomes a palindrome and has exactly a characters ‘0’
and exactly b characters ‘1’.

输入格式

The first line contains a single integer t ( 1≤t≤10^4 ). Then t test
cases follow.

The first line of each test case contains two integers a and b (
0≤a,b≤2⋅10 ^5 , a+b≥1 ).

The second line of each test case contains the string s of length a+b
, consisting of the characters ‘0’, ‘1’, and ‘?’.

It is guaranteed that the sum of the string lengths of s over all test
cases does not exceed 2⋅10^5

输出格式

For each test case, output: “-1”, if you can’t replace all the
characters ‘?’ in the string s by ‘0’ or ‘1’ so that the string
becomes a palindrome and that it contains exactly a characters ‘0’ and
exactly b characters ‘1’; the string that is obtained as a result of
the replacement, otherwise. If there are several suitable ways to
replace characters, you can output any.

题意翻译

一句话题意，给出 T 个由 0、1、? 组成的字符串，以及这个字符串由 a 个 0 和 b 个 1 组成，要求将字符串中的 ? 替换成 0
或 1 之后是一个回文串并且恰好有 a 个 0 和 b 个 1。

输入输出样例

输入 #1复制 9
4 4
01???0
3 3
??? 1 0 ?
2 2
0101
2 2
01?0
0 1
0
0 3
1?1
2 2
?00?
4 3
??010?0

输出 #1复制
01011010
-1
0
-1
0110
-1
111
1001
0101010

题解
暴力模拟

题解：先把能确定的先确定，不能确定的再根据0和1 确定，n分两种，n为偶数，n为奇数，当n为奇数的时候，会出现i和j相等的情况，所以要考虑0的个数是加1还是加2

代码样例：

#include<bits/stdc++.h>
using namespace std;
const int maxn=2*1e5+10;//数组范围
char s[maxn];//数组范围开的较大时 
int main()
{
    int t;
    scanf("%d",&t);//输入多组数据的组数
    int a,b,n;
    int s0,s1;//记录‘0’‘1’出现的次数 
    int flag;// 
    while(t--)
    {
        s0=0,s1=0;
        scanf("%d%d",&a,&b);//a的个数 b的个数
        n=a+b;//提供的字符的总个数
        getchar();//吸收换行
        for(int i=1; i<=n; i++)
            scanf("%c",&s[i]);//输入
        if(n==1)//总共一个字符
        {
            if(s[1]=='?')//第一个字符为‘？’
            {
                if(a==1)//一个字符为‘0’
                {
                    s[1]='0';
                    s0++;//记录‘0’出现的次数
                }
                else//一个字符为‘1’
                {
                    s[1]='1';
                    s1++;//记录‘1’出现的次数
                }
            }
            else//第一个字符不为‘？’
            {
                if(s[1]=='0')//第一个字符为‘0’
                    s0++;//记录‘0’出现的次数
                else
                    s1++;//记录‘1’出现的次数
            }
            if(s0==a&&s1==b)//'0'的总数为a,'1'的总数为b
                printf("%c\n",s[1]);
            else
                printf("-1\n");
            continue;
        }
        int flag=0;//标记初始状态 
        if(n%2==0)//当n为偶数
        {
            int k=n/2;
            for(int i=1,j=n; i<=k; i++,j--)
            {
                if(s[i]!='?'&&s[j]!='?')//首尾对称都不为‘？’ 
                {
                    if(s[i]!=s[j])//首尾改变后不相等 
                    {
                        flag=1;//状态改变 
                        break;
                    }
                    else//首尾改变后相等
                    {
                        if(s[i]=='0')//都为‘0’ 
                            s0+=2;
                        else//都为‘1’ 
                            s1+=2; 
                    }
                }
                else if(s[i]=='?'&&s[j]!='?')//首为‘？’尾不为‘？’ 
                {
                    s[i]=s[j];// 首尾相同 
                    if(s[j]=='0')
                        s0+=2;
                    else
                        s1+=2;
                }
                else if(s[i]!='?'&&s[j]=='?')//首不为‘？’尾为‘？’
                {
                    s[j]=s[i];// 首尾相同
                    if(s[i]=='0')//首位是‘0’ 
                        s0+=2;
                    else//首位不是‘0’ 
                        s1+=2;
                }
                else
                    continue;
            }
            for(int i=1,j=n; i<=k; i++,j--)
            {
                if(s[i]=='?'&&s[j]=='?')//首尾都为‘？’ 
                {
                    if((s0+2)<=a)//*‘0’的数量加2小于等于给定的‘0’的个数 
                    {
                        s[i]='0',s[j]='0';//首尾都是‘0’ 
                        s0+=2;
                    }
                    else
                    {
                        s[i]='1',s[j]='1';//首尾都是‘1’
                        s1+=2;
                    }
                }
                if(s1>b)//‘1’的个数大于给定的‘1’出现的次数 
                {
                    flag=1;//改变状态 
                    break;
                }

            }
        }
        else//当n为奇数 
        {
            int k=(n/2)+1;//中间值 
            for(int i=1,j=n; i<=k; i++,j--)
            {
                if(s[i]!='?'&&s[j]!='?')//首尾都不为‘？’ 
                {
                    if(s[i]!=s[j])//首尾不相同 
                    {
                        flag=1;//改变状态 
                        break;
                    }
                    else//首尾相同 
                    {
                        if(i==j) 
                        {
                            if(s[i]=='0')//首尾都为‘0’ 
                                s0++;
                            else//首尾都为‘1’ 
                                s1++;
                        }
                        else
                        {
                            if(s[i]=='0')
                                s0+=2;
                            else
                                s1+=2;
                        }
                    }
                }
                else if(s[i]=='?'&&s[j]!='?')//i和j肯定不相等
                {
                    s[i]=s[j];
                    if(s[j]=='0')//末位为‘0’ 
                        s0+=2;
                    else
                        s1+=2;
                }
                else if(s[i]!='?'&&s[j]=='?')//i和j肯定不相等
                {
                    s[j]=s[i];
                    if(s[i]=='0')
                        s0+=2;
                    else
                        s1+=2;
                }
                else
                    continue;
            }
            for(int i=1,j=n; i<=k; i++,j--)
            {
                if(s[i]=='?'&&s[j]=='?')//首尾相同 
                {
                    if(i==j)
                    {
                        if((s0+1)<=a)
                        {
                            s[i]='0';
                            s0++;
                        }
                        else
                        {
                            s[i]='1';
                            s1++;
                        }
                        if(s1>b)
                        {
                            flag=1;
                            break;
                        }
                    }
                    else
                    {
                        if((s0+2)<=a)
                        {
                            s[i]='0',s[j]='0';
                            s0+=2;
                        }
                        else
                        {
                            s[i]='1',s[j]='1';
                            s1+=2;
                        }
                        if(s1>b)
                        {
                            flag=1;
                            break;
                        }
                    }
                }
            }
        }
        if(flag==1)
            printf("-1\n");
        else if(s0==a&&s1==b)
        {
            for(int i=1; i<=n; i++)
                printf("%c",s[i]);
            printf("\n");
        }
        else
            printf("-1\n");
    }
    return 0;
}