Codeforces Round #590 (Div. 3)（e、f待补

https://codeforces.com/contest/1234/problem/A

A. Equalize Prices Again

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 int main(){
     int n,a;
     int t;
     cin>>t;
     ll sum = ,ans;
     while(t--){
         cin>>n;sum = ;
         for(int i = ;i < n;++i){
             cin>>a;sum+=a;
         }
         ans = sum/n;
         if(sum%n)ans+=;
         cout<<ans<<endl;
     }
 }

AC代码

https://codeforces.com/contest/1234/problem/B1

B1. Social Network (easy version)

https://codeforces.com/contest/1234/problem/B2

B2. Social Network (hard version)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a;
int n ,k ,now=,fi=;
vector<ll>s;
map<ll,int>mp;
int main(){
    cin>>n>>k;
    for(int i = ;i < n;++i){
        cin>>a;
        if(mp[a]==){
                mp[a]=;s.push_back(a);
            if(now<k){
                now++;
            }
            else if(now==k){
                mp[s[fi]]=;fi++;
            }
        }
    }
    cout<<now<<endl;
    int l =s.size()-;
    int cnt=;
    while(cnt<now&&l>=){
        if(mp[s[l]])cnt++,cout<<s[l]<<" ";
        l--;
    }
    cout<<endl;

    return ;
}

AC代码

https://codeforces.com/contest/1234/problem/C

C. Pipes

旋转一遍发现前两种其实是不同方向摆放的一种管道，后四个同理，也就是只有两个管道，一个是直流另一个会变向，然后问题就很简单了。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;

 int main(){
     int t,n;
     cin>>t;
     while(t--){
         cin>>n;
         string a[];cin>>a[]>>a[];
         int now=,flag =;
         for(int i = ;i <n;++i){
             if(a[now][i]==''||a[now][i]=='')continue;
             else{
                now=-now;
                if(a[now][i]==''||a[now][i]==''){
                 flag=;break;
                }
             }
         }
         if(flag==||now==)cout<<"no"<<endl;
         else cout<<"yes"<<endl;
     }
     return ;
 }

AC代码

https://codeforces.com/contest/1234/problem/D

D. Distinct Characters Queries

用线段树维护不同字母的个数orz学到了新东西，待会再看看set的做法？

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5+;
int a[N],ans[];
int tree[*N][];
void build(int l,int r,int rt){
    if(l==r){tree[rt][a[l]]++;return ;}
    int mid=(l+r)/;
    build(l,mid,rt*);
    build(mid+,r,rt*+);
    for(int i = ;i < ;++i)tree[rt][i]=tree[rt*][i]+tree[rt*+][i];
}
void f5(int l,int r,int rt,int x,int p,int f){
    if(l==r){tree[rt][p]--;tree[rt][f]++;return ;}
    int mid=(l+r)>>;
    if(x<=mid)f5(l,mid,rt<<,x,p,f);
    else f5(mid+,r,rt<<|,x,p,f);
    for(int i = ;i < ;++i)tree[rt][i]=tree[rt<<][i]+tree[rt<<|][i];
}
void query(int l,int r,int rt,int ll,int rr){
    if(r<=rr&&l>=ll){
        for(int i = ;i < ;++i)ans[i]+=tree[rt][i]; return ;
    }
    int mid=(l+r)>>;
    if(ll<=mid)query(l,mid,rt<<,ll,rr);
    if(rr>mid)query(mid+,r,rt<<|,ll,rr);
}
int main()
{
    ios::sync_with_stdio();
    string s;cin>>s;int n = s.size();
    for(int i = ;i < n;++i)a[i+]=s[i]-'a';
    build(,n,);
    int m;cin>>m;
    while(m--){
        int flag,x,l,r;char c;cin>>flag;
        if(flag==){
            cin>>x>>c;
            f5(,n,,x,s[x-]-'a',c-'a');
            s[x-]=c;
        }
        else{
            cin>>l>>r;memset(ans,,sizeof(ans));
            query(,n,,l,r);
            int tot=;
            for(int i = ;i < ;++i)if(ans[i])tot++;
            cout<<tot<<endl;
        }
    }
    return ;
}

AC代码