Codeforces Round 891 (Div. 3)
A - Array Coloring
思路：需要两部分的奇偶相同，判断奇数的个数是否为偶数即可
#include<bits/stdc++.h> using namespace std; #define int long long //#define int __int128 #define double long double typedef pair<int,int>PII; typedef pair<string,int>PSI; typedef pair<string,string>PSS; const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353; const double eps=1e-6; int ksm(int x,int y){ int res=1; while(y){ if(y&1)res=res*x%mod; x=x*x%mod; y>>=1; } return res; } int c(int a,int b){ if(b>a)return 0; int res=1; for(int i=1,j=a;i<=b;++i,--j){ res=res*j%mod; res=res*ksm(i,mod-2)%mod; } return res; } int Lucas(int a,int b){ if(a<mod&&b<mod)return c(a,b); return c(a%mod,b%mod)*Lucas(a/mod,b/mod)%mod; } void init(){ } void solve(){ int n;cin>>n; vector<int>ve(n); int cnt=0; for(int i=0;i<n;++i){ cin>>ve[i]; if(ve[i]%2)cnt++; } if(cnt%2==0)cout<<"YES\n"; else cout<<"NO\n"; return; } signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int t=1; init(); cin>>t; while(t--){ solve(); } return 0; }
B - Maximum Rounding
思路：找到第一个大于4的数，给前面最后一个小于4的数加一，之后的数全为0
#include<bits/stdc++.h> using namespace std; #define int long long //#define int __int128 #define double long double typedef pair<int,int>PII; typedef pair<string,int>PSI; typedef pair<string,string>PSS; const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353; const double eps=1e-6; int ksm(int x,int y){ int res=1; while(y){ if(y&1)res=res*x%mod; x=x*x%mod; y>>=1; } return res; } int c(int a,int b){ if(b>a)return 0; int res=1; for(int i=1,j=a;i<=b;++i,--j){ res=res*j%mod; res=res*ksm(i,mod-2)%mod; } return res; } int Lucas(int a,int b){ if(a<mod&&b<mod)return c(a,b); return c(a%mod,b%mod)*Lucas(a/mod,b/mod)%mod; } void init(){ } void solve(){ string s; cin>>s; for(int i=0;i<s.size();++i){ if(s[i]>='5'){ while(i>0&&s[i-1]>='4')i--; for(int j=i;j<s.size();++j)s[j]='0'; if(i)s[i-1]++; else s.insert(s.begin(),'1'); break; } } cout<<s<<'\n'; return; } signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int t=1; init(); cin>>t; while(t--){ solve(); } return 0; }
C - Assembly via Minimums
思路1：对于原数组a，将其从小到大排序后，可以发现b数组中出现a的个数为 （n-1）~1；那么将b从小到大排序后，按个数取依次放到a中去
思路2：按从大到小统计所有的b及其出现的个数，枚举去重后的所有b，cnti为bi的个数，pre表示a中大于bi的个数，令a中bi的个数为x，那么会有pre*x+x*（x-1）/2个bi在b中，可以二分找到满足条件的x。
#include<bits/stdc++.h> using namespace std; #define int long long //#define int __int128 typedef pair<int,int>PII; typedef pair<string,int>PSI; typedef pair<string,string>PSS; const int N=5e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353,M=44750; const int inf=0x3f3f3f3f3f3f; const double eps=1e-6; void solve(){ int n,m;cin>>n; m=n*(n-1)/2; vector<int>ve(m); for(int i=0;i<m;++i)cin>>ve[i]; sort(ve.begin(),ve.end()); for(int i=1,j=0;i<=n;++i){ cout<<ve[j]<<' '; j+=n-i; j=min(j,m-1); } cout<<'\n'; } signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int T=1; cin>>T; while(T--){ solve(); } return 0; }
#include<bits/stdc++.h> using namespace std; #define int long long //#define int __int128 #define double long double typedef pair<int,int>PII; typedef pair<string,int>PSI; typedef pair<string,string>PSS; const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353; const double eps=1e-6; int ksm(int x,int y){ int res=1; while(y){ if(y&1)res=res*x%mod; x=x*x%mod; y>>=1; } return res; } int c(int a,int b){ if(b>a)return 0; int res=1; for(int i=1,j=a;i<=b;++i,--j){ res=res*j%mod; res=res*ksm(i,mod-2)%mod; } return res; } int Lucas(int a,int b){ if(a<mod&&b<mod)return c(a,b); return c(a%mod,b%mod)*Lucas(a/mod,b/mod)%mod; } void init(){ } map<int,int>mp; void solve(){ int n;cin>>n; mp.clear(); for(int i=0,u;i<(n*(n-1)/2);++i){ cin>>u; mp[u]++; } vector<PII>g(mp.size()); int idx=mp.size()-1; for(auto v:mp)g[idx--]={v.first,v.second}; vector<int>ans(mp.size(),0); for(int i=0,pre=0;i<mp.size();++i){ int c=g[i].second; int l=1,r=1e3,x; while(l<=r){ int mid=l+r>>1; if((pre*mid+mid*(mid-1)/2)>=c)r=mid-1,x=mid; else l=mid+1; } ans[i]=x; pre+=x; } for(int i=0;i<mp.size();++i){ while(ans[i]--)cout<<g[i].first<<' '; }cout<<'\n'; return; } signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int t=1; init(); cin>>t; while(t--){ solve(); } return 0; }
D - Strong Vertices
思路：找出等于最大值的
E - Power of Points
思路：对于每个s，求的是所有x到s的距离和，二分找到s的位置，前缀和求大于和小于等于s的值，对s求距离即可
#include<bits/stdc++.h> using namespace std; #define int long long //#define int __int128 #define double long double typedef pair<int,int>PII; typedef pair<string,int>PSI; typedef pair<string,string>PSS; const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353; const double eps=1e-6; int ksm(int x,int y){ int res=1; while(y){ if(y&1)res=res*x%mod; x=x*x%mod; y>>=1; } return res; } int c(int a,int b){ if(b>a)return 0; int res=1; for(int i=1,j=a;i<=b;++i,--j){ res=res*j%mod; res=res*ksm(i,mod-2)%mod; } return res; } int Lucas(int a,int b){ if(a<mod&&b<mod)return c(a,b); return c(a%mod,b%mod)*Lucas(a/mod,b/mod)%mod; } void init(){ } void solve(){ int n;cin>>n; vector<PII>ve(n+1); vector<int>sum(n+1,0),ans(n+1); for(int i=1;i<=n;++i)cin>>ve[i].first,ve[i].second=i; sort(ve.begin()+1,ve.end()); for(int i=1;i<=n;++i)sum[i]=sum[i-1]+ve[i].first; for(int i=1;i<=n;++i){ int p= lower_bound(ve.begin()+1,ve.end(),ve[i])-ve.begin(); int l=sum[p],r=sum[n]-sum[p],cl=p,cr=n-p; int s=0; if(cl>0)s+=abs(ve[i].first*cl-l)+cl; if(cr>0)s+=abs(r-ve[i].first*cr)+cr; ans[ve[i].second]=s; } for(int i=1;i<=n;++i)cout<<ans[i]<<' ';cout<<'\n'; return; } signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int t=1; init(); cin>>t; while(t--){ solve(); } return 0; }
F - Sum and Product
思路：由a+b=x，ab=y，可以得出b2-xb+y=0，解方程即可