Codeforces 1138B（列方程枚举）

构造模拟要分情况讨论感觉不是够本质，然后官解是因为只有四个量所以可以根据限制条件列两个方程，再枚举一下解就可以了。

const int maxn = 5000 + 5;
int n, c[maxn], a[maxn];
vector<int> zero, two, cl, ac, ans1, ans2;

int main() {
	cin >> n;
	getchar();
	rep(i, 1, n) {
		char ch = getchar();
		c[i] = ch - '0';
	}
	getchar();
	rep(i, 1, n) {
		char ch = getchar();
		a[i] = ch - '0';
	}
	rep(i, 1, n) {
		if (c[i] && a[i])	two.push_back(i);
		else if (c[i])	cl.push_back(i);
		else if (a[i])	ac.push_back(i);
		else	zero.push_back(i);
	}
	
	for (int a = 0; a <= zero.size(); a++) {
		int d = a - n / 2 + ac.size() + two.size();
		if (d < 0 || d > two.size() || a + d > n / 2)	continue;
		int tmp = n / 2 - a - d;
		for (int c = 0; c <= cl.size(); c++) {
			if (tmp - c <= ac.size()) {
				for (int i = 0; i < a; i++)	cout << zero[i] << " ";
				for (int i = 0; i < tmp - c; i++)	cout << ac[i] << " ";
				for (int i = 0; i < c; i++)	cout << cl[i] << " ";
				for (int i = 0; i < d; i++)	cout << two[i] << " ";
				return 0;
			}
		}
	}
	printf("-1
");
	return 0;
}