Codeforces Round ＃620 (Div. 2)ABC

A. Two Rabbits

两点之间的距离每次减少 a+b，初始时距离为 y-x
如果两个人恰好遇到，那么 (y-x)%(a+b)==0
时间为 (y-x)/(a+b)

#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
#define x first
#define y second
#define LL long long 
#define int LL
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define debug(x) cerr << #x << ": " << x << endl
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
LL Mod(LL a,LL mod){
    return (a%mod+mod)%mod;}
LL lowbit(LL x){
    return x&-x;}//最低位1及其后面的0构成的数值
LL qmi(LL a,LL b,LL mod) {
    LL ans = 1; while(b){
     if(b & 1) ans = ans * (a % mod) % mod; a = a % mod * (a % mod) % mod; b >>= 1;} return ans; }
int _;
int n;
int gcd(int a,int b)
{
    
	return b?gcd(b,a%b):a;
}
void solve()
{
    
	int x,y,a,b;
	cin>>x>>y>>a>>b;
	if((y-x)%(a+b))cout<<-1<<endl;
	else cout<<(y-x)/(a+b)<<endl;
}
signed main()
{
    
    io;
  	cin>>_; 
 	while(_--)
    solve();
    return 0;
}

B. Longest Palindrome

思路：暴力枚举，首先先找出自身回文得放在中间，只能有一个
然后存储互相回文得字符出输出即可

#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
#define x first
#define y second
#define LL long long 
#define int LL
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define debug(x) cerr << #x << ": " << x << endl
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
LL Mod(LL a,LL mod){
    return (a%mod+mod)%mod;}
LL lowbit(LL x){
    return x&-x;}//最低位1及其后面的0构成的数值
LL qmi(LL a,LL b,LL mod) {
    LL ans = 1; while(b){
     if(b & 1) ans = ans * (a % mod) % mod; a = a % mod * (a % mod) % mod; b >>= 1;} return ans; }
int _;
int n,m;
const int N=110;
vector<string>v;
int gcd(int a,int b)
{
    
	return b?gcd(b,a%b):a;
}
bool check(string s)
{
    
	string t=s;
	reverse(all(t));
	return s==t;
}
void solve()
{
    
	cin>>n>>m;
	string s;
	map<string,bool>mp;
	vector<string>v1,v2;
	string mid="";
	for(int i=0;i<n;i++)
	{
    
		cin>>s;
		v.pb(s);
	}
	for(int i=0;i<n;i++)
	{
    
		if(check(v[i]))
		{
    
			mid=v[i];
			continue;
		}
		for(int j=0;j<n;j++)
		{
    
			if(j==i)continue;
			string t=v[i]+v[j];
			if(check(t)&&mp[v[i]]==false&&mp[v[j]]==false)
			{
    
				v1.pb(v[i]);
				v2.pb(v[j]);
				mp[v[i]]=mp[v[j]]=true;
			}
		}
	}
	int len=v1.size()+v2.size();
	if(mid!="")len++;
	cout<<len*m<<endl;
	for(int i=0;i<v1.size();i++)cout<<v1[i];
	if(mid!="")cout<<mid;
	for(int i=v2.size()-1;i>=0;i--)cout<<v2[i];
	cout<<endl;
}
signed main()
{
    
    io;
  // cin>>_; 
 // while(_--)
    solve();
    return 0;
}

C. Air Conditioner

思路：贪心，维护区间

#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
#define x first
#define y second
#define LL long long 
#define int LL
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define debug(x) cerr << #x << ": " << x << endl
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
LL Mod(LL a,LL mod){
    return (a%mod+mod)%mod;}
LL lowbit(LL x){
    return x&-x;}//最低位1及其后面的0构成的数值
LL qmi(LL a,LL b,LL mod) {
    LL ans = 1; while(b){
     if(b & 1) ans = ans * (a % mod) % mod; a = a % mod * (a % mod) % mod; b >>= 1;} return ans; }
int _;
int n,m;
const int N=110;
struct node
{
    
	int t,l,r;
}c[N];
void solve()
{
    
	cin>>n>>m;
	for(int i=1;i<=n;i++)cin>>c[i].t>>c[i].l>>c[i].r;
	bool f=false;
	int l=m,r=m;
	for(int i=1;i<=n;i++)
	{
    
		int d=c[i].t-c[i-1].t;
		l-=d,r+=d;
		if(l>c[i].r||r<c[i].l)
		{
    
			f=true;
			break;
		}
		l=max(l,c[i].l),r=min(r,c[i].r);
	}
	if(f)cout<<"NO"<<endl;
	else cout<<"YES"<<endl;
}
signed main()
{
    
    io;
  	cin>>_; 
 	while(_--)
    solve();
    return 0;
}