B. Circus(枚举)

对于每个角色有四种可能属性，a:10 b:01 c:11 d:00。
思路：构造的情况要满足a1+c1==b2+c2，然后可以枚举第一层用到几个1，第二层枚举其中来自a的个数，然后移项推公式判断。。。。

int main()//a:10 b:01 c:11 d:00  枚举：外层枚举第一次有几个人生效，内层枚举其中来自a的个数 然后推变量。。
{
	int n;
	string s1, s2;
	while (cin >> n >> s1 >> s2)
	{
		vector<int> t1, t2, t3, t4;
		int n1 = 0, n2 = 0, n3 = 0, n4 = 0;
		f(i, 0, n - 1)
		{
			if (s1[i] == '1'&&s2[i] == '0')n1++, t1.emplace_back(i);
			else if (s1[i] == '0'&&s2[i] == '1')n2++, t2.emplace_back(i);
			else if (s1[i] == '1'&&s2[i] == '1')n3++, t3.emplace_back(i);
			else n4++, t4.emplace_back(i);
		}
		int fg = 0;
		int a = 0, c = 0, b = 0;
		f(i, 0, min(n / 2, n1 + n3))
		{
			for (int j = 0;j <= i && j <= n1;j++)
			{
				int a1 = j, c1 = i - j, c2 = n3 - c1;
				int imgb2 = a1 + c1 - c2;
				if (imgb2 >= 0 && imgb2 <= n2)
				{
					a = a1, c = c1, b = n2 - imgb2;
					if (n / 2 - a - c - b >= 0&&n/2-a-b-c<=n4)
					{
						fg = 1;
						goto he;
					}
				}
			}
		}
		he:
		if (!fg) { puts("-1");continue; }
		int re = n / 2 - a - c - b;
		f(i, 0, t1.size() - 1)
		{
			if (a == 0)break;
			printf("%d ", t1[i] + 1), a--;
		}
		f(i, 0, t3.size() - 1)
		{
			if (c == 0)break;
			printf("%d ", t3[i] + 1), c--;
		}

		f(i, 0, t2.size() - 1)
		{
			if (b == 0)break;
			printf("%d ", t2[i] + 1), b--;
		}
		f(i, 0, t4.size() - 1)
		{
			if (re == 0)break;
			printf("%d ", t4[i] + 1), re--;
		}puts("");
	}
	return 0;
}