AtCoder Beginner Contest 350

本文主要是介绍AtCoder Beginner Contest 350，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

A - Past ABCs

简单的枚举判断即可

#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;
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; }
void solve()
{string s;cin>>s;int sum=0;for (int i=3;i<6;i++){sum=sum*10+(s[i]-'0');}string s1=s.substr(0,3);if(s1=="ABC" && sum>=1 && sum<=349 && sum!=316){YES}else {NO}}
signed main()
{IOSint t;t=1;//cin>>t;while(t--){solve();}
}

B - Dentist Aoki

如果一个洞奇数次进，则总数加一偶数次进总数减一。

#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;
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; }
void solve()
{int n,q;cin>>n>>q;int sum=n;vector<int> a(n+1,1);for (int i=1;i<=q;i++){int x;cin>>x;if(a[x]==1){a[x]=0;sum--;}else {a[x]=1;sum++;}}cout<<sum;
}
signed main()
{IOSint t;t=1;//cin>>t;while(t--){solve();}
}

C - Sort

先用一个数组记录第几个输入的数字，然后一个数组记录这个数字的位置，

然后再按照排列从小到大的顺序遍历，如果这个数不在相对应的位置上就和其需要到的位置上的数交换。时间复杂度(n).

#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;
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; }
void solve()
{int n;cin>>n;vi a(n+1);vi b(n+1);for (int i=1;i<=n;i++){cin>>a[i];b[a[i]]=i;}int cnt=0;vector<pi> v;for (int i=1;i<=n;i++){if(b[i]!=i){int j=b[i];swap(a[i],a[j]);swap(b[a[i]],b[a[j]]);v.push_back({i,j});}}cout<<(int)v.size()<<endl;for (int i=0;i<(int)v.size();i++){auto [x,y]=v[i];cout<<x<<" "<<y<<endl;}}
signed main()
{IOSint t;t=1;//cin>>t;while(t--){solve();}
}

D - New Friends

这题考察联通块的知识，每个联通块内都可以有（cnt)*(cnt-1)/2条边。

这题可以用两种做法：

1.图论

因为存在一个点被联通块内多个点连接的情况，所以每次先加上边，最后除以2.

#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;
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; }
int vis[200010];
void solve()
{int n,m;cin>>n>>m;vector<vector<int>> v(n+1);for (int i=1;i<=m;i++){int x,y;cin>>x>>y;v[x].push_back(y);v[y].push_back(x);}int ans=0;for (int i=1;i<=n;i++){if(vis[i]){continue;}queue<int> q;q.push(i);vis[i]=1;int cnt1=1,cnt2=0;while(q.size()){int u=q.front();q.pop();for (auto x : v[u]){cnt2++;if(vis[x]==0){vis[x]=1;cnt1++;q.push(x);}}}ans+=cnt1*(cnt1-1)-cnt2;}ans/=2;cout<<ans;
}
signed main()
{IOSint t;t=1;//cin>>t;while(t--){solve();}
}

2.并查集

#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;
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; }
int p[200010],cnt[200010];
int find(int x)
{if(p[x]!=x) p[x]=find(p[x]);return p[x];
}
void solve()
{int n,m;cin>>n>>m;iota(p+1,p+n+1,1);for (int i=1;i<=m;i++){int x,y;cin>>x>>y;p[find(x)]=find(y);}int ans=-m;for (int i=1;i<=n;i++){cnt[find(i)]++;}for (int i=1;i<=n;i++){ans+=cnt[i]*(cnt[i]-1)/2;}cout<<ans;}
signed main()
{IOSint t;t=1;//cin>>t;while(t--){solve();}
}

E - Toward 0

mp[n]是n的期望花费。

cost1=mp[n/a]+x。

cost2= $\sum$ 1~6 mp[n/i] / 6 + y 当i=1时两边有相同的式子，把它移到左边。

cost2= $\sum_{2}^{6}i mp[n/i]/5+\frac{6}{5}*y$

#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;
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; }void solve()
{int n,a,x,y;cin>>n>>a>>x>>y;map<int,double> mp;auto dfs=[&](auto dfs,int u)-> double{if(u==0){return 0;}if(mp.find(u)!=mp.end()){return mp[u];}double cost1=dfs(dfs,u/a)+x;double cost2=0;for (int i=2;i<=6;i++){cost2+=dfs(dfs,u/i);}cost2=cost2/5+1.0*y*6/5;return mp[u]=min(cost1,cost2);};dfs(dfs,n);cout<<fixed<<setprecision(10)<<mp[n];	}
signed main()
{IOSint t;t=1;//cin>>t;while(t--){solve();}
}

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