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Codeforces Round 950 (Div. 3)

news 来源：原创 2024/7/1 16:53:04

好久没写题解了，今天来写个题解。

A - 问题 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 都是简单的模拟题，按照题意来写就行，可以参考代码。

C - Sofia and the Lost Operations

思路：给出的m个元素可以分为三类来看，一类是需要改成b的，一类是和b相等的元素，还有一类是ab 中都没有的元素。而这第三类元素必须要被第一类和第二类覆盖掉。所以只需要倒叙找最后的元素是不是第一第二类元素。（可以使用map 来存第一第二类元素）。

#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：通过贪心来遍历没所以最多只能处理一个递减的情况，可以先开个数组记录一下递减的位置和递减的对数数量。一点要特判的是在边界的话，是可以直接删掉最外面的数的。然后可以直接遍历一次，每次都对删去中间那个a。看操作后，是不是可以消去所有的不递增。

可以看代码理解。

#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：其实就是看两个矩阵的每行和每列的元素是不是一样的。

所以用两个map 分别存每个元素的x和 y坐标然后最后看两个矩阵的每一个元素的x和坐标是不是对应的。

#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：从底下往上算，每次都修改最左边的值和最下面的值，并加上左边的面积，就是总面积了。

其中x轴的排序时从大到小，y轴的排序是从小到大。如果想移除这个台灯后面积变大，这个台灯必须得位于边界，并且两个相邻的边为边界才行，这个点也是边界改变的点。所以每次改变边界的时候都把这个点标记。

#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();}
}