Codeforces Round 867 (Div. 3)
Codeforces Round 867 (Div. 3)
A - TubeTube Feed
#include<bits/stdc++.h> using namespace std; typedef pair<int,int>PII; typedef pair<string,int>PSI; const int N=50+5,INF=0x3f3f3f3f,Mod=1e6; const double eps=1e-6; typedef long long ll; int q,n,t,a[N],b[N]; int main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); cin>>q; while(q--){ cin>>n>>t; for(int i=1;i<=n;++i)cin>>a[i]; for(int i=1;i<=n;++i)cin>>b[i]; int s=0,ma=0,p=-1; for(int i=1;i<=n;++i){ if(s+a[i]<=t){ if(b[i]>ma){ ma=b[i];p=i; } } s++; } cout<<p<<'\n'; } return 0; }
B - Karina and Array
#include<bits/stdc++.h> using namespace std; typedef pair<int,int>PII; typedef pair<string,int>PSI; const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e6; const double eps=1e-6; typedef long long ll; ll t,n,a[N]; int main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); cin>>t; while(t--){ cin>>n; for(int i=0;i<n;++i)cin>>a[i]; sort(a,a+n); ll x=a[0]*a[1],y=a[n-1]*a[n-2]; cout<<max(x,y)<<'\n'; } return 0; }
C - Bun Lover
#include<bits/stdc++.h> using namespace std; typedef pair<int,int>PII; typedef pair<string,int>PSI; const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e6; const double eps=1e-6; typedef long long ll; ll t,n; int main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); cin>>t; while(t--){ cin>>n; ll s=n*n+2*n+2; cout<<s<<'\n'; } return 0; }
D - Super-Permutation
#include<bits/stdc++.h> using namespace std; typedef pair<int,int>PII; typedef pair<string,int>PSI; const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e6; const double eps=1e-6; typedef long long ll; int t,n; int main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); cin>>t; while(t--) { cin>>n; if(n==1)cout<<1<<'\n'; else if(n%2)cout<<-1<<'\n'; else{ int l=n,r=1,p=n/2; while(p--){ cout<<l<<' ';l-=2; cout<<r<<' ';r+=2; }cout<<"\n"; } } return 0; }
E - Making Anti-Palindromes
思路：分类讨论，奇数和某种字母个数超过n/2的不行，否则交换有两种：a.回文对与回文对，b.回文对与非回文对；显然优先进行a；
判断回文对中出现次数的最大值ma，若ma大于总回文对数的一半，则交换ma次，否则直接回文对两两互换，交换总回文对数/2（上取）
#include<bits/stdc++.h> using namespace std; typedef pair<int,int>PII; typedef pair<string,int>PSI; const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e6; const double eps=1e-6; typedef long long ll; int t,n,a[30],b[30]; string s; int main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); cin>>t; while(t--) { cin>>n; cin>>s; memset(a,0,sizeof a); memset(b,0,sizeof b); int ma=0,cnt=0,mma=0; if(n%2)cout<<-1<<'\n'; else{ for(int i=0;i<n;++i){ a[s[i]-'a']++; ma=max(ma,a[s[i]-'a']); if(i<n/2){ if(s[i]==s[n-i-1]){ cnt++; b[s[i]-'a']++; mma=max(mma,b[s[i]-'a']); } } } if(ma>n/2)cout<<-1<<'\n'; else{ int s=max(mma,(int)ceil(1.0*cnt/2)); cout<<s<<'\n'; } } } return 0; }
F - Gardening Friends
思路：找到直径的两端点p，q，根节点到树上任意一点的最远路径一定过p或q，求出1，p，q到每个点最短路即可算出最大收益
#include<bits/stdc++.h> using namespace std; typedef pair<int,int>PII; typedef pair<string,int>PSI; const int N=1e6+5,INF=0x3f3f3f3f,Mod=1e6,S=1e6; const double eps=1e-6; typedef long long ll; ll t,n,k,c; auto bfs(vector<vector<int>>ve,int u){ vector<ll>d(n+1,-1); queue<int>q; q.push(u),d[u]=0; while(q.size()){ int v=q.front();q.pop(); for(auto w:ve[v]){ if(d[w]!=-1)continue; d[w]=d[v]+1;q.push(w); } } return d; } int main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); cin>>t; while(t--){ cin>>n>>k>>c; vector<vector<int>>ve(n+1); for(int i=0,x,y;i<n-1;++i){ cin>>x>>y; ve[x].push_back(y),ve[y].push_back(x); } auto d1=bfs(ve,1); int p= max_element(d1.begin(),d1.end())-d1.begin(); auto d2=bfs(ve,p); int q=max_element(d2.begin(),d2.end())-d2.begin(); auto d3=bfs(ve,q); ll res=0; for(int i=1;i<=n;++i){ res=max(res,max(d2[i],d3[i])*k-d1[i]*c); } cout<<res<<'\n'; } return 0; }
G1 - Magic Triples (Easy Version)
思路：ai，aj=b*ai，ak=b*b*ai，枚举ai和b可以求出总个数，b的范围为1e3
#include<bits/stdc++.h> using namespace std; typedef pair<int,int>PII; typedef pair<string,int>PSI; const int N=1e6+5,INF=0x3f3f3f3f,Mod=1e6; const double eps=1e-6; typedef long long ll; ll t,n,cnt[N]; int main(){ cin>>t; while(t--){ set<ll>se; ll x; cin>>n; for(int i=0;i<n;++i){ cin>>x; se.insert(x);cnt[x]++; } ll res=0; for(auto v:se){ if(cnt[v]>=3)res+=(cnt[v]*(cnt[v]-1)*(cnt[v]-2)); for(int b=2;b*b*v<=*se.rbegin();++b){ res+=(cnt[v]*cnt[v*b]*cnt[v*b*b]); } } for(auto v:se)cnt[v]=0; cout<<res<<'\n'; } return 0; }
G2 - Magic Triples (Hard Version)
思路：可以枚举aj，那么 ai=aj/b，aj，ak=aj*b，ai和b为aj的因子，那就枚举aj所有因子，复杂度为aj开根号，暴力枚举会超；
方法：当aj<S，枚举因子，否则暴力枚举，可知当S取1e6时，复杂度为1e3
#include<bits/stdc++.h> using namespace std; typedef pair<int,int>PII; typedef pair<string,int>PSI; const int N=1e6+5,INF=0x3f3f3f3f,Mod=1e6,S=1e6; const double eps=1e-6; typedef long long ll; int t,n; int main(){ cin>>t; while(t--){ cin>>n; map<ll,ll>cnt; for(int i=0;i<n;++i){ ll x; cin>>x; cnt[x]++; } ll res=0; for(auto [x,y]:cnt){ res+=y*(y-1)*(y-2); if(x<S){ vector<ll>f; for(int i=1;i<=x/i;++i){ if(x%i)continue; f.push_back(i); if(x/i!=i)f.push_back(x/i); } for(auto v:f){ if(v==1)continue; if(v*x>cnt.rbegin()->first)continue; if(cnt.count(x/v)&&cnt.count(x*v)) res+=cnt[x/v]*cnt[x*v]*y; } } else{ for(int b=2;b*x<=cnt.rbegin()->first;++b){ if(x%b==0&&cnt.count(x/b)&&cnt.count(x*b)) res+=cnt[x/b]*cnt[x*b]*y; } } } cout<<res<<'\n'; } return 0; }
