Codeforces Round #798 (Div. 2) - E. ANDfinity

题目链接：https://codeforces.com/contest/1689/problem/E

解题思路：

        根据位运算可以简单的推算出，其实至多只需要2次操作就可以了。也就是操作最小bit位最大的那个，极端情况下需要对两个最小bit位最大的进行操作，一个+1，一个-1。

#include 
using namespace std;
#define lson l, mid, rt<<1
#define rson mid+1, r, rt<<1|1
typedef long long ll;
typedef unsigned long long ull;
const int mx = 2e3 + 10;
const int mod = 998244353;
typedef pair <int, int> Pa;
 
int a[mx];
int fa[mx];
set <int> st, vec[30], val[mx];
map <int, int> mp;
int ans;
 
int findfa(int x) {
	return x == fa[x]? x : fa[x] = findfa(fa[x]);
}
 
int cal(int n)
{
	st.clear();
	for (int i=1;i<=n;i++)
		fa[i] = i;
	for (int i=0; i<30; i++) {
		if (vec[i].size() > 1) {
			auto it1 = vec[i].begin();
			auto it2 = it1;
			it2++;
			while (it2 != vec[i].end()) {
				int fu = findfa(*it1);
				int fv = findfa(*it2);
				fa[fv] = fu;
				it2++, it1++;
			}
		}
	}
	for (int i=1;i<=n;i++) {
		int fu = findfa(i);
		st.insert(fu);
	}
	return st.size();
}
 
void print(int n)
{
	printf("%d\n", ans);
	for (int i=1;i<=n;i++)
		printf("%d%c", a[i], i==n?'\n':' ');
}
 
int main()
{
	for (int i=0;i<30;i++)
		mp[1<
	int t;
	scanf("%d", &t);
	while (t--) {
		int n;
		scanf("%d", &n);
		for (int i=0;i<30;i++)
			vec[i].clear();
		ans = 0;
		for (int i=1;i<=n;i++) {
			val[i].clear();
			fa[i] = i;
			scanf("%d", a+i);
			if (a[i] == 0)
				ans++, a[i]++;
			for (int j=0; (1<
				if ((1<
					vec[j].insert(i);
					val[i].insert(j);
				}
			}
		}
		if (cal(n) == 1) {
			print(n);
			continue;
		}
 
		for (int i=1;i<=n;i++) {
			if (a[i] != 1) {
				int bit_val = a[i] - (a[i] & (a[i] - 1));
				int bit_pos = mp[bit_val];
				a[i]--;
				vec[bit_pos].erase(i);
				for (int j=0;j
					vec[j].insert(i);
				//printf("########### %d %d\n", i, bit_pos);
				if (cal(n) == 1) {
					ans++;
					print(n);
					break;
				}
				a[i]++;
				vec[bit_pos].insert(i);
				for (int j=0;j
					vec[j].erase(i);
			}
			for (int v: val[i])
				vec[v].erase(i);
			set <int> temp_st;
			a[i]++;
			for (int j=0; (1<
				if ((1<
					vec[j].insert(i);
					temp_st.insert(j);
				}
			}
			if (cal(n) == 1) {
				ans++;
				print(n);
				break;
			}
			a[i]--;
			for (int v: temp_st)
				vec[v].erase(i);
			for (int v: val[i])
				vec[v].insert(i);
		}
		if (st.size() != 1) {
			int max_val = 0, pos;
			for (int i=1;i<=n;i++) {
				int bit_val = a[i] - (a[i] & (a[i] - 1));
				if (bit_val > max_val) {
					max_val = bit_val;
					pos = i;
				}
			}
			a[pos]--;
			for (int i=1;i<=n;i++) {
				if (a[i] & max_val) {
					a[i]++;
					break;
				}
			}
			ans += 2;
			print(n);
		}
 
	}
	return 0;
}
// 8
// 8
// 4 4 8 8 16 16 32 32