感覺這場的題面都好長，看的腦殼子疼（英語不好，啊我掛了

A. The King's Race

考慮對角線為分界

#include <iostream>
using namespace std;
int main() {
	long long n,x,y;
	cin>>n>>x>>y;
	if((x+y)<=n+1)cout<<"White";
	else cout<<"Black";
	return 0;
}

B. Taxi drivers and Lyft

每個0都會選擇離他近的1，預處理一下每個0的前後最近的1的位置，然後跑一邊就能算出每個0應該會歸屬哪個1了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+5;
ll n,m;
ll res[maxn];
struct T{
	ll pre;
	ll last;
}tt[maxn];
ll x[maxn];
ll type[maxn];
int main(){
	scanf("%lld%lld",&n,&m);
	for(int i=0;i<n+m;++i){
		scanf("%lld",x+i);
	}
	for(int i=0;i<n+m;++i){
		scanf("%lld",type+i);
	}
	ll pp=-1e9;
	for(int i=0;i<n+m;++i){//更新前驅 
		if(type[i]==0){
			tt[i].pre=pp;
		}
		else{
			pp=i;
		} 
	}
	ll la=1e9+5;
	for(int i=n+m-1;i>=0;--i){//更新後繼 
		if(type[i]==0){
			tt[i].last=la;
		}
		else{
			la=i;
		} 
	}
	
	for(int i=0;i<m+n;++i){
		if(type[i]==0){
			if(tt[i].last==1e9+5){
				res[tt[i].pre]++;
			}
			else if(tt[i].pre==(-1e9)){
				res[tt[i].last]++;
			}
			else{
				ll xx=x[i]-x[tt[i].pre];
				ll yy=x[tt[i].last]-x[i];
				if(xx<=yy){
					res[tt[i].pre]++;
				}
				else{
					res[tt[i].last]++;
				}
			}
		}
	}
	for(int i=0;i<n+m;++i){
		if(type[i]==1){
			cout<<res[i]<<' ';
		}
	}
	return 0;
}




 

C. The Tower is Going Home