状态不佳，rating--，blog质量--，有点抱歉
A. Submission is All You Need
思路：mex操作优于max操作只限定于{0}，{0，1}，其他都是sum更优秀.所以直接判断0，1数量即可
吐槽：没判多余的0，wa了发
#include<iostream>
#include<queue>
#include<map>
#include<iomanip>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#include<functional>
#define ll long long
#define ull unsigned long long
#define lowbit(x) (x & -x)
#define endl "\n"// 交互题记得删除
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
const ll maxn = 2e5 + 5;
ll ksm(ll x, ll y)
{
ll ans = 1;
while (y)
{
if (y & 1)
{
ans = ans * x % mod;
}
x = x * x % mod;
y >>= 1;
}
return ans % mod;
}
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int main() {
fio();
ll t=1;
cin>>t;
while(t--){
ll n;
cin>>n;
vector<ll>a(n+2);
ll c0=0;
ll c1=0;
ll sum=0;
for(ll i=1;i<=n;i++){
cin>>a[i];
c0+=(a[i]==0);
c1+=(a[i]==1);
if(a[i]>=2)sum+=a[i];
}
sum+=min(c0,c1)*2;
ll u=min(c0,c1);
c0-=u;
c1-=u;
sum+=c1+c0;
cout<<sum<<endl;
}
return 0;
}
B. Pathless
思路：设sum为数组总和，如果s<sum随便顺序输出，如果s=sum则-1，如果s>sum，且s-sum的差不等于x * 2+y * 3(x>=0&&y>=0)，那么直接0021构造即可，否则怎么构造都无解
#include<iostream>
#include<queue>
#include<map>
#include<iomanip>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#include<functional>
#define ll long long
#define ull unsigned long long
#define lowbit(x) (x & -x)
#define endl "\n"// 交互题记得删除
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
const ll maxn = 2e5 + 5;
ll ksm(ll x, ll y)
{
ll ans = 1;
while (y)
{
if (y & 1)
{
ans = ans * x % mod;
}
x = x * x % mod;
y >>= 1;
}
return ans % mod;
}
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int main() {
fio();
ll t=1;
cin>>t;
while(t--){
ll n,s;
cin>>n>>s;
vector<ll>a(n+2);
ll sum=0;
for(ll i=1;i<=n;i++)cin>>a[i],sum+=a[i];
if(sum==s)cout<<-1<<endl;
else if(s<sum){
for(ll i=1;i<=n;i++)cout<<a[i]<<" ";
cout<<endl;
}
else {
ll dd=s-sum;
ll c0=0,c2=0;
for(ll i=1;i<=n;i++){
c0+=(a[i]==0);
c2+=(a[i]==2);
// c3+=(a[i]==3);
}
bool flag=0;
for(ll j=0;j<=1000;j++){
if(j*2<=dd){
if((dd-j*2)%3==0){
flag=1;
break;
}
}
}
if(flag)cout<<-1<<endl;
else {
for(ll i=1;i<=n;i++){
if(c0>0){
c0--;
cout<<0<<" ";
}
else if(c2>0){
c2--;
cout<<2<<" ";
}
else {
cout<<1<<" ";
}
}
cout<<endl;
}
}
}
return 0;
}
C. Double Perspective
思路：显然如果构成了环，那么其的并集将会重复一次，所以考虑直接连边，然后直接dfs得出答案即可，可能是多个连通分量，每个点都去dfs一下即可（使用vis标记），不走已被vis标记过的点（防止成环）
吐槽：这个n * n<=9e6看得真难受
#include<iostream>
#include<queue>
#include<map>
#include<iomanip>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#include<functional>
#define ll long long
#define ull unsigned long long
#define lowbit(x) (x & -x)
#define endl "\n"// 交互题记得删除
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
const ll maxn = 2e5 + 5;
ll ksm(ll x, ll y)
{
ll ans = 1;
while (y)
{
if (y & 1)
{
ans = ans * x % mod;
}
x = x * x % mod;
y >>= 1;
}
return ans % mod;
}
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int main() {
fio();
ll t=1;
cin>>t;
while(t--){
ll n;
cin>>n;
vector<vector<pair<ll,ll>>>g(n*2+4);
for(ll i=1;i<=n;i++){
ll l,r;
cin>>l>>r;
g[l].push_back({r,i<<1});
g[r].push_back({l,i<<1|1});
// b[i]={l,r};
}
vector<bool>vi(n*2+2,0);
vector<ll>d;
function<void(ll)>dfs=[&](ll x){
vi[x]=1;
for(auto [to,j]:g[x]){
if(vi[to])continue;
else {
d.push_back(j/2);
dfs(to);
}
}
};
for(ll i=1;i<=2*n;i++){
if(vi[i]==0){
dfs(i);
}
}
cout<<d.size()<<endl;
sort(d.begin(),d.end());
for(auto j:d)cout<<j<<" ";
cout<<endl;
}
return 0;
}
D. Stay or Mirror
思路：这道题分类讨论下，讨论x和2 * n-x产生的必定贡献，谁更小选谁，相等选x(不知道选n * 2-x是否一样)。这个思路没有严谨证明，赛时想了下试了下就过了。
#include<iostream>
#include<queue>
#include<map>
#include<iomanip>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#include<functional>
#define ll long long
#define ull unsigned long long
#define lowbit(x) (x & -x)
#define endl "\n"// 交互题记得删除
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
const ll maxn = 2e5 + 5;
ll ksm(ll x, ll y)
{
ll ans = 1;
while (y)
{
if (y & 1)
{
ans = ans * x % mod;
}
x = x * x % mod;
y >>= 1;
}
return ans % mod;
}
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int main() {
fio();
ll t = 1;
cin >> t;
while (t--) {
ll n;
cin >> n;
vector<ll>a(n + 2);
for (ll i = 1; i <= n; i++)cin >> a[i];
vector<ll>vis(n * 2 + 5, 0);
for (ll i = 1; i <= n; i++) {
ll l = min(a[i], 2 * n - a[i]);
ll r = max(a[i], 2 * n - a[i]);
ll c1 = 0, c2 = 0;
vector<ll>now(n * 2 + 5, 0);
for (ll j = r - 1; j >= 1; j--) {
if (vis[j] == 0 && now[j] == 0 && 2 * n - j <= r - 1) {
now[j] = now[2 * n - j] = 1;
c1++;
}
}
ll c3 = 0;
for (ll j = r + 1; j <= 2*n; j++) {
if (vis[j] == 1)c3++;
}
for (ll j = l + 1; j <= 2 * n; j++) {
if (vis[j] == 1)c2++;
}
if (c2 <= c3 + c1) {
a[i] = l;
vis[r] = -1, vis[l] = 1;
}
else a[i] = r, vis[l] = -1, vis[r] = 1;
}
ll ans = 0;
vector<bool>vi(n * 2 + 2, 0);
for (ll i = 1; i <= n; i++) {
for (ll j = a[i] + 1; j <= 2 * n; j++)ans += (vi[j] == 1);
vi[a[i]] = 1;
}
cout << ans << endl;
}
return 0;
}
E1. Interactive RBS (Easy Version)
思路：询问次数很多，先考虑二分出左括号位置和右括号位置（a1,a2），那么构造如下格式((()()a[i-1]a[i]即可根据值知道答案,最大询问次数为500+log2(1000) * 2。
#include<iostream>
#include<queue>
#include<map>
#include<iomanip>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#include<functional>
#define ll long long
#define ull unsigned long long
#define lowbit(x) (x & -x)
// #define endl "\n"// 交互题记得删除
using namespace std;
mt19937 rnd(time(0));
const ll mod = 1e9 + 7;
const ll maxn = 2e5 + 5;
ll ksm(ll x, ll y)
{
ll ans = 1;
while (y)
{
if (y & 1)
{
ans = ans * x % mod;
}
x = x * x % mod;
y >>= 1;
}
return ans % mod;
}
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int main() {
fio();
ll t = 1;
cin >> t;
while (t--) {
ll n;
cin >> n;
ll a1, a2;
ll l = 2, r = n;
function<ll(ll)>q1 = [&](ll x) {
cout << "? " << x;
for (ll i = 1; i <= x; i++)cout << " " << i;
cout << endl;
cout.flush();
cin >> x;
return x;
};
function<ll(ll)>q2 = [&](ll x) {
cout << "? " << n - x+1;
for (ll i = x; i <= n; i++)cout << " " << i;
cout << endl;
cout.flush();
cin >> x;
return x;
};
while (l < r) {
ll mid = l + r >> 1;
if (q1(mid))r = mid;
else l = mid + 1;
}
if (q1(r))a1 = r;
else a1 = 1;
l = 1, r = n - 1;
while (l <= r) {
ll mid = l + r >> 1;
if (q2(mid))l = mid + 1;
else r = mid - 1;
}
if (r == 0)r = 1;
if (q2(r))a2 = r;
else a2 = n;
swap(a1, a2);
vector<char>a(n + 2);
for (ll i = 2; i <= n; i += 2) {
cout << "? " << 8 << " " << a1 << " " << a1 << " " <<a1<<" "<<a2 << " " << a1 << " " << a2 << " " << i - 1 << " " << i << endl;
cout.flush();
ll x;
cin >> x;
if (x == 6)a[i - 1] = '(', a[i] = ')';
else if (x == 4)a[i - 1] = ')', a[i] = '(';
else if (x == 3)a[i - 1] = '(', a[i] = '(';
else a[i - 1] = ')', a[i] = ')';
}
if (n & 1) {
cout << "? " << 2 << " " << n << " " << a2 << endl;
cout.flush();
ll x;
cin >> x;
if (x > 0)a[n] = '(';
else a[n] = ')';
}
cout << "! ";
for (ll i = 1; i <= n; i++)cout << a[i];
cout << endl;
//((()()() 6
//((()())( 4
//((()()(( 3
//((()())) 5
}
return 0;
}
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