A. Subset Mex
大意：把原数组分成俩数组A,B，求(最小的A里没出现的自然数+最小的B里没出现的自然数)的最大值
答案是最小的出现次数<2的自然数，加上最小的没出现的自然数
#include <bits/stdc++.h>
using namespace std;
#define repeat(i,a,b) for(int i=(a),_=(b);i<_;i++)
#define repeat_back(i,a,b) for(int i=(b)-1,_=(a);i>=_;i--)
#define mst(a,x) memset(a,x,sizeof(a))
#define fi first
#define se second
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
int cansel_sync=(ios::sync_with_stdio(0),cin.tie(0),0);
const int N=200010; typedef long long ll; const int inf=~0u>>2; const ll INF=~0ull>>2; ll read(){ll x; if(scanf("%lld",&x)==-1)exit(0); return x;} typedef double lf; const lf pi=acos(-1.0); lf readf(){lf x; if(scanf("%lf",&x)==-1)exit(0); return x;} typedef pair<ll,ll> pii; template<typename T> void operator<<(vector<T> &a,T b){a.push_back(b);}
const ll mod=(1?1000000007:998244353); ll mul(ll a,ll b,ll m=mod){return a*b%m;} ll qpow(ll a,ll b,ll m=mod){ll ans=1; for(;b;a=mul(a,a,m),b>>=1)if(b&1)ans=mul(ans,a,m); return ans;}
#define int ll
map<int,int> mp;
void Solve(){
int n=read(); mp.clear();
repeat(i,0,n)mp[read()]++;
int ans=0;
repeat(i,0,inf)if(mp[i]<2){
ans+=i;
break;
}
repeat(i,0,inf)if(mp[i]<1){
ans+=i;
break;
}
cout<<ans<<endl;
}
signed main(){
//freopen("data.txt","r",stdin);
int T=1; T=read();
repeat(ca,1,T+1){
Solve();
}
return 0;
}
B. Maximum Product
大意：求数组中的五个数之积的最大值
最小的x个数和最大的5-x个数之积(x=0,1,2,3,4,5)，这六个数取最大值
#include <bits/stdc++.h>
using namespace std;
#define repeat(i,a,b) for(int i=(a),_=(b);i<_;i++)
#define repeat_back(i,a,b) for(int i=(b)-1,_=(a);i>=_;i--)
#define mst(a,x) memset(a,x,sizeof(a))
#define fi first
#define se second
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
int cansel_sync=(ios::sync_with_stdio(0),cin.tie(0),0);
const int N=200010; typedef long long ll; const int inf=~0u>>2; const ll INF=~0ull>>2; ll read(){ll x; if(scanf("%lld",&x)==-1)exit(0); return x;} typedef double lf; const lf pi=acos(-1.0); lf readf(){lf x; if(scanf("%lf",&x)==-1)exit(0); return x;} typedef pair<ll,ll> pii; template<typename T> void operator<<(vector<T> &a,T b){a.push_back(b);}
const ll mod=(1?1000000007:998244353); ll mul(ll a,ll b,ll m=mod){return a*b%m;} ll qpow(ll a,ll b,ll m=mod){ll ans=1; for(;b;a=mul(a,a,m),b>>=1)if(b&1)ans=mul(ans,a,m); return ans;}
#define int ll
vector<int> a;
void Solve(){
int n=read(); a.clear();
repeat(i,0,n)a<<read();
#define b a.rbegin()
sort(a.begin(),a.end());
int ans=max({
a[0]*a[1]*a[2]*a[3]*a[4],
a[0]*a[1]*a[2]*a[3]*b[0],
a[0]*a[1]*a[2]*b[1]*b[0],
a[0]*a[1]*b[2]*b[1]*b[0],
a[0]*b[3]*b[2]*b[1]*b[0],
b[4]*b[3]*b[2]*b[1]*b[0]
});
cout<<ans<<endl;
}
signed main(){
//freopen("data.txt","r",stdin);
int T=1; T=read();
repeat(ca,1,T+1){
Solve();
}
return 0;
}
C. Link Cut Centroids
大意：给定一棵树，要求删掉一条边并加上一条边，满足得到的图仍然是一棵树并且重心的个数只有一个
首先找到其中一个重心a，然后找到这个重心的重儿子（子树最大的儿子）b，然后找到这个儿子的随便一个儿子c，去掉b-c添加a-c即可（特判b没有儿子的情况）
证明：一棵树最多只能有两个重心。如果原来只有一个重心，那么该操作只会增大b的最大子树大小，a显然还是是唯一重心；如果原来有两个重心，那么b就是原来另一个重心，该操作会增大b的最大子树大小并减小a最大子树大小，a就成为了唯一重心
#include <bits/stdc++.h>
using namespace std;
#define repeat(i,a,b) for(int i=(a),_=(b);i<_;i++)
#define repeat_back(i,a,b) for(int i=(b)-1,_=(a);i>=_;i--)
#define mst(a,x) memset(a,x,sizeof(a))
#define fi first
#define se second
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
int cansel_sync=(ios::sync_with_stdio(0),cin.tie(0),0);
const int N=200010; typedef long long ll; const int inf=~0u>>2; const ll INF=~0ull>>2; ll read(){ll x; if(scanf("%lld",&x)==-1)exit(0); return x;} typedef double lf; const lf pi=acos(-1.0); lf readf(){lf x; if(scanf("%lf",&x)==-1)exit(0); return x;} typedef pair<ll,ll> pii; template<typename T> void operator<<(vector<T> &a,T b){a.push_back(b);}
const ll mod=(1?1000000007:998244353); ll mul(ll a,ll b,ll m=mod){return a*b%m;} ll qpow(ll a,ll b,ll m=mod){ll ans=1; for(;b;a=mul(a,a,m),b>>=1)if(b&1)ans=mul(ans,a,m); return ans;}
#define int ll
vector<int> a[N];
namespace center{
vector<int> rec;
int sz[N],maxx[N];
void dfs(int x,int fa=-1){
rec<<x;
sz[x]=1; maxx[x]=0;
for(auto i:a[x]){
int p=i;
if(p!=fa){
dfs(p,x);
sz[x]+=sz[p];
maxx[x]=max(maxx[x],sz[p]);
}
}
}
int get(int x){ //get center
rec.clear(); dfs(x); int n=sz[x],ans=x;
for(auto x:rec){
maxx[x]=max(maxx[x],n-sz[x]);
if(maxx[x]<maxx[ans])ans=x;
}
return ans;
}
}using namespace center;
void Solve(){
int n=read(); repeat(i,1,n+1)a[i].clear();
repeat(i,0,n-1){
int x=read(),y=read();
a[x]<<y; a[y]<<x;
}
int c=get(1);
get(c);
if(maxx[c]==1){
cout<<c<<' '<<a[c][0]<<endl;
cout<<c<<' '<<a[c][0]<<endl;
return;
}
int p1,p2;
for(auto i:a[c])if(maxx[c]==sz[i])p1=i;
for(auto i:a[p1])if(c!=i)p2=i;
cout<<p1<<' '<<p2<<endl;
cout<<p2<<' '<<c<<endl;
}
signed main(){
//freopen("data.txt","r",stdin);
int T=1; T=read();
repeat(ca,1,T+1){
Solve();
}
return 0;
}
D. Three Sequences
大意：给定ai，令bi为单调不减序列，ci为单调不增序列，且ai=bi+ci。对a数组区间加操作，询问max(b1,b2,...,bn,c1,c2,...,cn)的最小值
设ai的差分数组为di。（具体思路写不了，太难讲了）发现答案只与di所有正数（或者负数）之和以及a1（或者an）有关。计算可得，令 (a_n=lst,sum_{d_i<0}d_i=dec)，答案就是 (lst+lfloor dfrac{lst+dec}2 floor)（形式不唯一）。维护lst和dec都简单的一批
#include <bits/stdc++.h>
using namespace std;
#define repeat(i,a,b) for(int i=(a),_=(b);i<_;i++)
#define repeat_back(i,a,b) for(int i=(b)-1,_=(a);i>=_;i--)
#define mst(a,x) memset(a,x,sizeof(a))
#define fi first
#define se second
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
int cansel_sync=(ios::sync_with_stdio(0),cin.tie(0),0);
const int N=200010; typedef long long ll; const int inf=~0u>>2; const ll INF=~0ull>>2; ll read(){ll x; if(scanf("%lld",&x)==-1)exit(0); return x;} typedef double lf; const lf pi=acos(-1.0); lf readf(){lf x; if(scanf("%lf",&x)==-1)exit(0); return x;} typedef pair<ll,ll> pii; template<typename T> void operator<<(vector<T> &a,T b){a.push_back(b);}
const ll mod=(1?1000000007:998244353); ll mul(ll a,ll b,ll m=mod){return a*b%m;} ll qpow(ll a,ll b,ll m=mod){ll ans=1; for(;b;a=mul(a,a,m),b>>=1)if(b&1)ans=mul(ans,a,m); return ans;}
#define int ll
int a[N],b[N];
void Solve(){
int n=read();
repeat(i,1,n+1)a[i]=read();
int lst=a[n],ans=0;
auto up=[&](int x,int d=1){
if(b[x]<0)ans+=d*b[x];
};
repeat(i,1,n){
b[i]=a[i+1]-a[i];
up(i);
}
int q=read();
cout<<lst-llround((lst+ans)/2.0-0.1)<<endl;
while(q--){
int l=read(),r=read(),k=read();
if(r==n)lst+=k;
if(l!=1){
up(l-1,-1);
b[l-1]+=k;
up(l-1,1);
}
if(r!=n){
up(r,-1);
b[r]-=k;
up(r,1);
}
cout<<lst-llround((lst+ans)/2.0-0.1)<<endl;
}
}
signed main(){
//freopen("data.txt","r",stdin);
int T=1; //T=read();
repeat(ca,1,T+1){
Solve();
}
return 0;
}