题目链接：

(https://codeforces.com/contest/1512/problem/C)

思路：

1.遍历字符串，将可以确定的‘？’给确定了，处理完之后剩下的‘？’必然对称
2.遍历字符串，每当遇到‘？’，可以用的1比较多那就换成1，否则换成0，时刻需要注意是否不符合题意，若不符合题意，直接输出-1

1A代码（还可以精简）

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
int cnt[2];
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	int t;
	cin >> t;
	part:
	while (t--) {
		memset(cnt, 0, sizeof(cnt));
		int w, e;
		cin >> w >> e;
		string a;
		cin >> a;
		a = "." + a;
		int sz = a.size()-1;
		for (int i = 1; i <= sz; i++)if (a[i] == '1')cnt[1]++; else if (a[i] == '0')cnt[0]++;
		for (int i = 1; i <= sz / 2; i++) {
			int pos = sz - i + 1;
			if (a[i] != a[pos] ) {
				if (a[i] != '?' && a[pos] != '?') { cout << -1 << '\n'; goto part; }
				else if (a[i] == '?' && a[pos] != '?') { a[i] = a[pos]; cnt[a[pos] - '0']++; }
				else if (a[i] != '?' && a[pos] == '?') { a[pos] = a[i]; cnt[a[i] - '0']++; }
			}
		}
		int zero = w - cnt[0];
		int one = e - cnt[1];
		for (int i = 1; i <= sz; i++) {
			if (zero < 0 || one < 0) { cout << -1 << '\n'; goto part; }
			int pos = sz - i + 1;
			if (a[i] == '?') {
				if (zero > one) {
					if (zero - 2 >= 0) {
						a[i] = a[pos] = '0';
						zero -= 2;
					}
					else {
						if (pos == i&&zero>=1) {
								a[i] = '0';
								zero--;
						}
						else {
							cout << -1 << '\n';
							goto part;
						}
					}
				}
				else {
					if (one - 2 >= 0) {
						a[i] = a[pos] = '1';
						one -= 2;
					}
					else {
						if (pos == i && one >= 1) {
							a[i] = '1';
							one--;
						}
						else {
							cout << -1 << '\n';
							goto part;
						}
					}
				}
			}
		}
		for (int i = 1; i <= sz; i++)cout << a[i];
		cout << '\n';
	}
}