好久沒寫題解了，今天來寫個題解。
A - 問題 Generator
#include "bits/stdc++.h"
using namespace std;#define int long long
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0);
#define all(x) x.begin(),x.end()
#define pi pair<int,int>
#define vi vector<int>
#define si set<int>
#define mi map<int,int>
#define mc map<char,int>
#define YES cout<<"Yes"<<endl;
#define NO cout<<"No"<<endl;
#define pb(x) push_back(x);
#define fi first
#define sc second
#define is insert
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
const int INF =1e18;
void solve()
{int n,m;cin>>n>>m;string s;cin>>s;map<char,int> mp;//int n=s.size();for (int i=0;i<n;i++){mp[s[i]]++;}int sum=0;for (char i='A';i<='G';i++){if(mp[i]<m){sum+=(m-mp[i]);}}cout<<sum<<endl;}signed main()
{IOSint t;cin>>t;while(t--){solve();}
}B - Choosing Cubes
#include "bits/stdc++.h"
using namespace std;#define int long long
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0);
#define all(x) x.begin(),x.end()
#define pi pair<int,int>
#define vi vector<int>
#define si set<int>
#define mi map<int,int>
#define mc map<char,int>
#define YES cout<<"Yes"<<endl;
#define NO cout<<"No"<<endl;
#define pb(x) push_back(x);
#define fi first
#define sc second
#define is insert
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
const int INF =1e18;
void solve()
{int n,f,k;cin>>n>>f>>k;int x;vi a(n+1);for (int i=1;i<=n;i++){cin>>a[i];if(i==f){x=a[i];}}int pos=0;sort(a.begin()+1,a.end());reverse(a.begin()+1,a.end());for (int i=1;i<=n;i++){if(a[i]==x){pos=i;break;}}int cnt=pos;for (int i=pos+1;i<=n;i++){if(a[i]==a[pos]){cnt++;}else {break;}}if(cnt<=k){cout<<"YES"<<endl;}else if (cnt>k && pos<=k){cout<<"MAYBE"<<endl;}else if(cnt>k){cout<<"NO"<<endl;}}signed main()
{IOSint t;cin>>t;while(t--){solve();}
}A和B 都是簡單的模擬題，按照題意來寫就行，可以參考代碼。
C - Sofia and the Lost Operations
思路：給出的m個元素可以分為三類來看，一類是需要改成b的，一類是和b相等的元素，還有一類是ab 中都沒有的元素。 而這第三類元素必須要被第一類和第二類覆蓋掉。所以只需要倒敘找最后的元素是不是第一第二類元素。? （可以使用map 來存第一第二類元素）。
#include "bits/stdc++.h"
using namespace std;#define int long long
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0);
#define all(x) x.begin(),x.end()
#define pi pair<int,int>
#define vi vector<int>
#define si set<int>
#define mi map<int,int>
#define mc map<char,int>
#define YES cout<<"Yes"<<endl;
#define NO cout<<"No"<<endl;
#define pb(x) push_back(x);
#define fi first
#define sc second
#define is insert
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
const int INF =1e18;
void solve()
{int n;cin>>n;vi a(n+1),b(n+1);for (int i=1;i<=n;i++){cin>>a[i];}for (int i=1;i<=n;i++){cin>>b[i];}map<int,int> mp,mp1;for (int i=1;i<=n;i++){if(a[i]!=b[i]){mp[b[i]]++;}else {mp1[b[i]]=1;}}int m;cin>>m;int fl=0;vi c;for (int i=1;i<=m;i++){int x;cin>>x;c.push_back(x);}int pos;for (int i=m-1;i>=0;i--){if(mp[c[i]]|| mp1[c[i]]){pos=i;break;}}for (int i=0;i<m;i++){if(mp[c[i]]){mp[c[i]]--;}else if(mp1[c[i]]){continue;}else {if(i>pos){fl=1;}}}if(fl==1){cout<<"NO"<<endl;return ;}for (int i=1;i<=n;i++){if(mp[b[i]]){cout<<"NO"<<endl;return ;}}cout<<"YES"<<endl;}signed main()
{IOSint t;cin>>t;while(t--){solve();}
}D - GCD-sequence
思路：通過貪心來遍歷沒所以最多只能處理一個遞減的情況，可以先開個數組記錄一下遞減的位置和遞減的對數數量。一點要特判的是在邊界的話，是可以直接刪掉最外面的數的。然后可以直接遍歷一次，每次都對刪去中間那個a。看操作后，是不是可以消去所有的不遞增。
可以看代碼理解。
#include "bits/stdc++.h"
using namespace std;#define int long long
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0);
#define all(x) x.begin(),x.end()
#define pi pair<int,int>
#define vi vector<int>
#define si set<int>
#define mi map<int,int>
#define mc map<char,int>
#define YES cout<<"Yes"<<endl;
#define NO cout<<"No"<<endl;
#define pb(x) push_back(x);
#define fi first
#define sc second
#define is insert
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
const int INF =1e18;
void solve()
{int n;cin>>n;vi a(n+1);for (int i=1;i<=n;i++){cin>>a[i];}int ans=0;vi b(n),c(n+1);for (int i=1;i<=n-1;i++){b[i]=__gcd(a[i],a[i+1]);}for (int i=1;i<n-1;i++){if(b[i+1]<b[i]){c[i]=1;ans++;}}if(ans==0){cout<<"YES"<<endl;return ;}if(ans==1 &&c[1]==1){cout<<"YES"<<endl;return ;}if(ans==1 &&c[n-2]==1){cout<<"YES"<<endl;return ;}int fl=0;for (int i=1;i<n-1;i++){int tmp1=0,tmp2=0,tmp3=1e9;tmp2=__gcd(a[i],a[i+2]);int cnt=ans;if(c[i]) cnt--;if(c[i-1]) cnt--;if(c[i+1] ) cnt--;if(i>1){tmp1=b[i-1];}if(i<n-2){tmp3=b[i+2];}if(tmp1>tmp2){cnt++;}if(tmp2>tmp3){cnt++;}if(cnt==0){fl=1;}}if(fl){cout<<"yes"<<endl;return ;}else {cout<<"NO"<<endl;}
}signed main()
{IOSint t;cin>>t;while(t--){solve();}
}E - Permutation of Rows and Columns
思路：其實就是看兩個矩陣的每行和每列的元素是不是一樣的。
所以用兩個map 分別存每個元素的x和 y坐標 然后最后看兩個矩陣的每一個元素的x和坐標是不是對應的。
#include "bits/stdc++.h"
using namespace std;#define int long long
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0);
#define all(x) x.begin(),x.end()
#define pi pair<int,int>
#define vi vector<int>
#define si set<int>
#define mi map<int,int>
#define mc map<char,int>
#define YES cout<<"Yes"<<endl;
#define NO cout<<"No"<<endl;
#define pb(x) push_back(x);
#define fi first
#define sc second
#define is insert
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
const int INF =1e18;map<int,int> mpx,mpy;
void solve()
{int n,m;cin>>n>>m;vector<vector<int>> a(n+1,vector<int>(m+1));vector<vector<int>> b(n+1,vector<int>(m+1));int fl=0;for (int i=1;i<=n;i++){for (int j=1;j<=m;j++){cin>>a[i][j];mpx[a[i][j]]=i;mpy[a[i][j]]=j;}}for (int i=1;i<=n;i++){for (int j=1;j<=m;j++){cin>>b[i][j];if(mpx[b[i][j]]!=mpx[b[i][1]] || mpy[b[i][j]]!=mpy[b[1][j]]){fl=1;}}}if(fl){cout<<"NO"<<endl;return ;}else {cout<<"yes"<<endl;}}signed main()
{IOSint t;cin>>t;while(t--){solve();}
}Field Division (easy version)
思路：從底下往上算，每次都修改最左邊的值和最下面的值，并加上左邊的面積，就是總面積了。
其中x軸的排序時從大到小，y軸的排序是從小到大。 如果想移除這個臺燈后面積變大，這個臺燈必須得位于邊界，并且兩個相鄰的邊為邊界才行，?這個點也是邊界改變的點。 所以每次改變邊界的時候都把這個點標記。
#include "bits/stdc++.h"
using namespace std;#define int long long
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0);
#define all(x) x.begin(),x.end()
#define pi pair<int,int>
#define vi vector<int>
#define si set<int>
#define mi map<int,int>
#define mc map<char,int>
#define YES cout<<"Yes"<<endl;
#define NO cout<<"No"<<endl;
#define pb(x) push_back(x);
#define fi first
#define sc second
#define is insert
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
const int INF =1e18;struct node{int x,y;int id;
}a[200010];bool cmp(node x,node y){if(x.x==y.x){return x.y<y.y;}else {return x.x>y.x;}
}void solve()
{int n,m,k;cin>>n>>m>>k;for (int i=1;i<=k;i++){cin>>a[i].x>>a[i].y;a[i].id=i;}sort(a+1,a+k+1,cmp);vi ans(k+1);int l=m+1,d=n;int sum=0;for (int i=1;i<=k;i++){if(l>a[i].y){ans[a[i].id]=1;sum+=(d-a[i].x)*(l-1);l=a[i].y;d=a[i].x;}}sum+=d*(l-1);cout<<sum<<endl;for (int i=1;i<=k;i++){cout<<ans[i]<<" ";}cout<<endl;}signed main()
{IOSint t;cin>>t;while(t--){solve();}
}
