AtCoder Beginner Contest 371

A - Jiro

思路

创建变量a,b,c;根据输入进行增加，等于1的那个即为答案

AC代码

	#include<bits/stdc++.h>
	#define endl '\n'
	#define int int long long
	#define pb push_back
	#define bs bitset
	using namespace std;
	typedef pair<char,int> PCI;
	typedef pair<int,int> PII;
	typedef priority_queue<int> PQ;

	const int N = 2e5+10, MAX = 1e9, INF = -1e9;

	int a,b,c;
	string s;

	signed main()
	{
		ios::sync_with_stdio(false);
		cin.tie(0);
		a=b=c=0;
		cin>>s;
		if(s==">")a++;else b++;
		cin>>s;
		if(s==">")a++;else c++;
		cin>>s;
		if(s==">")b++;else c++;

		if(a==1)cout<<"A"<<endl;
		else if(b==1)cout<<"B"<<endl;
		else cout<<"C"<<endl;
		
		return 0;
	}

B - Taro

思路

创建st数组记录是否出现过长子，当读入为“M”的时候忽略即可

AC代码

	#include<bits/stdc++.h>
	#define endl '\n'
	#define int int long long
	#define pb push_back
	#define bs bitset
	using namespace std;
	typedef pair<char,int> PCI;
	typedef pair<int,int> PII;
	typedef priority_queue<int> PQ;

	const int N = 2e5+10, MAX = 1e9, INF = -1e9;

	bool st[N];
	int n,m;
	int e;
	string s;

	signed main()
	{
		ios::sync_with_stdio(false);
		cin.tie(0);
		cin>>n>>m;
		while(m--){
			cin>>e>>s;
			if(!st[e]&&s=="M"){
				cout<<"Yes"<<endl;
				st[e]=true;
			}
			else cout<<"No"<<endl;
		}

		return 0;
	}

C - Make Isomorphic

思路

由于\(N\)较小使用邻接表来表示图，方便比较和修改，同样由此原因，我们直接采取枚举\(H\)图的排列来挨个比对找到最小代价即可；

AC代码

	#include<bits/stdc++.h>
	#define endl '\n'
	#define int int long long
	#define pb push_back
	#define bs bitset
	using namespace std;
	typedef pair<char,int> PCI;
	typedef pair<int,int> PII;
	typedef priority_queue<int> PQ;

	const int N = 10, MAX = 1e9, INF = -1e9;

	int n;
	int mg,mh;
	int u,v;
	int a[N][N];
	int ans=MAX+10;
	int g[N][N];
	int h[N][N];
	vector<int> w;

	int counts(){
		int num=0;
		for(int i=1;i<=n;i++){
			for(int j=i+1;j<=n;j++){
				if(g[i][j]!=h[w[i]][w[j]]){
					num+=a[w[i]][w[j]];
				}
			}
		}
		return num;
	}

	signed main()
	{
		ios::sync_with_stdio(false);
		cin.tie(0);
		cin>>n;
		cin>>mg;
		for(int i=1;i<=mg;i++){
			cin>>u>>v;
			g[u][v]=1;
			g[v][u]=1;
		}
		cin>>mh;
		for(int i=1;i<=mh;i++){
			cin>>u>>v;
			h[u][v]=1;
			h[v][u]=1;
		}
		for(int i=1;i<=n-1;i++){
			for(int j=i+1;j<=n;j++){
				cin>>a[i][j];
				a[j][i]=a[i][j];
			}
		}
		w.pb(0);
		for(int i=1;i<=n;i++)w.pb(i);

		do{
			ans=min(ans,counts());
		}while(next_permutation(w.begin()+1,w.end()));
		cout<<ans<<endl;

		return 0;
	}

D - 1D Country

思路

离散化+前缀和+二分，首先对村庄的位置进行离散化处理，对村民数量进行前缀和处理，查找时采取二分查找，时间复杂度为\(O(qlogn)\)

AC代码

	#include<bits/stdc++.h>
	#define endl '\n'
	#define int int long long
	#define pb push_back
	#define bs bitset
	using namespace std;
	typedef pair<char,int> PCI;
	typedef pair<int,int> PII;
	typedef priority_queue<int> PQ;

	const int N = 2e5+10, MAX = 1e9, INF = -1e9;

	int n,q;
	vector<int> v;
	int s[N];
	int e;
	int l,r;

	signed main()
	{
		ios::sync_with_stdio(false);
		cin.tie(0);
		cin>>n;
		for(int i=1;i<=n;i++){
			cin>>e;
			v.pb(e);
		}
		sort(v.begin(),v.end());
		for(int i=1;i<=n;i++)cin>>s[i];
		s[0]=0;
		for(int i=1;i<=n;i++)s[i]+=s[i-1];
		cin>>q;
		while(q--){
			cin>>l>>r;
			int st=lower_bound(v.begin(),v.end(),l)-v.begin()+1;
			int end=upper_bound(v.begin(),v.end(),r)-v.begin();
			cout<<s[end]-s[st-1]<<endl;
		}

		return 0;
	}

t.b.c.