codeforces 1512 C. A-B Palindrome（1200，回文）

链接：https://codeforces.com/problemset/problem/1512/C

题意：构造满足条件的字符串

题解：没啥解，就硬凑；具体看代码；

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		int a, b, n;
		cin >> a >> b;
		n = a + b;
		string s;
		cin >> s;
		for (int i = 0; i < n; i++)
		{
			if (s[i] == '?')s[i] = s[n - i - 1];
		}
		a -= count(s.begin(), s.end(), '0');//返回字符串中所有0的数量
		b -= count(s.begin(), s.end(), '1');//返回字符串中所有1的数量
		for (int i = 0; i <= n / 2; i++) 
		{
			if (i != n - i - 1 && s[i] == '?')
			{
				if (a > 1)
				{
					s[i] = s[n - i - 1] = '0';
					a -= 2;
				}
				else if (b > 1)
				{
					s[i] = s[n - i - 1] = '1'; 
					b -= 2;
				}
			}
			else if (s[i] == '?')//中间位置特判
			{
				if (a)
				{
					s[i] = '0';
					a--;
				}
				else
				{
					s[i] = '1';
					b--;
				}
			}
		}
		string t = s;
		reverse(t.begin(), t.end());//反转字符串
		if (t == s && a == 0 && b == 0)
			cout << s << '\n';
		else 
			cout << "-1\n";

	}
	return 0;
}

标签：begin,Palindrome,end,cin,int,1200,codeforces,--,else
来源： https://blog.csdn.net/m0_55858144/article/details/119395816