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 <bits/stdc++.h>
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<<i] = i;
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<<j)<=a[i];j++){
if ((1<<j) & a[i]) {
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<bit_pos;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<bit_pos;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<<j)<=a[i]; j++) {
if ((1<<j) & a[i]) {
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