Codeforces Round 914 (Div. 2)A~C

A
反过来考虑，由皇后和国王的位置去寻找骑士的位置，当一个点既可以被皇后找到，也可以被国王找到时就说明这个点是满足条件的

#include <bits/stdc++.h> 
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define PII pair<int, int>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
#define x first
#define y second
using namespace std;

const int N=3e5+10;
int n,m;
int dx[]={0,1,1,-1,-1};
int dy[]={0,1,-1,1,-1};
void solve()
{
	map<PII,int>mp;
	int xk,yk,xq,yq,a,b;cin>>a>>b>>xk>>yk>>xq>>yq;
	rep(i,1,4)
	{
		mp[{xk+dx[i]*a,yk+dy[i]*b}]++;
		if(a!=b)	mp[{xk+dx[i]*b,yk+dy[i]*a}]++;
		mp[{xq+dx[i]*a,yq+dy[i]*b}]++;
		if(a!=b)	mp[{xq+dx[i]*b,yq+dy[i]*a}]++;
	}
	int ans=0;
	for(auto cnt:mp)	if(cnt.y>1)	ans++;
	cout<<ans<<endl;
}

int main()
{
	IOS	
//  	freopen("1.in", "r", stdin);
  	int t;
	cin>>t;
	while(t--)
	solve();
	return 0;
}

B
思路：排序+前缀和
先排序，对于$a[i],\mathrm{前}\mathrm{面}\mathrm{的}\mathrm{数}\mathrm{一}\mathrm{定}\mathrm{是}\mathrm{小}\mathrm{于}\mathrm{它}\mathrm{的}\mathrm{一}\mathrm{定}\mathrm{可}\mathrm{以}\mathrm{被}\mathrm{删}\mathrm{除}，\mathrm{需}\mathrm{要}\mathrm{找}\mathrm{到}\mathrm{后}\mathrm{面}\mathrm{最}\mathrm{远}\mathrm{能}\mathrm{到}\mathrm{达}\mathrm{哪}\mathrm{里}$
$\mathrm{假}\mathrm{设}\mathrm{最}\mathrm{远}\mathrm{能}\mathrm{到}\mathrm{达}r，\mathrm{那}\mathrm{么}[i,r]\mathrm{这}\mathrm{一}\mathrm{段}\mathrm{区}\mathrm{间}\mathrm{中}\mathrm{所}\mathrm{有}\mathrm{数}\mathrm{的}\mathrm{答}\mathrm{案}\mathrm{是}\mathrm{一}\mathrm{样}\mathrm{的}\mathrm{都}\mathrm{是}\mathrm{到}r，\mathrm{如}\mathrm{何}\mathrm{判}\mathrm{断}\mathrm{哪}\mathrm{里}\mathrm{是}\mathrm{最}\mathrm{远}？$
$\mathrm{当}s[r]<a[r+1]\mathrm{时}\mathrm{就}\mathrm{说}\mathrm{明}\mathrm{此}\mathrm{时}r\mathrm{到}\mathrm{达}\mathrm{了}\mathrm{最}\mathrm{远}$

#include <bits/stdc++.h> 
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define PII pair<int, int>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
#define x first
#define y second
using namespace std;

const int N=1e5+10;
ll n,m,a[N],s[N],ans[N];
struct node
{
	int val,pos;
}b[N];

bool cmp(node a, node b)
{
	return a.val<b.val;
}

void solve()
{
	cin>>n;
	rep(i,1,n)	cin>>a[i],b[i].val=a[i],b[i].pos=i;
	sort(b+1,b+1+n,cmp);
	rep(i,1,n)	s[i]=s[i-1]+b[i].val;
	int last=0;
	rep(i,1,n)
	{
		if(last<i)
		{
			last=i;
			while(s[last]>=b[last+1].val&&last+1<=n)	last++;
		}
		ans[b[i].pos]=last-1;
	}
	rep(i,1,n)	cout<<ans[i]<<' ';
	cout<<endl;
	rep(i,1,n)	s[i]=0,ans[i]=0;
}

int main()
{
	IOS	
//  	freopen("1.in", "r", stdin);
  	int t;
	cin>>t;
	while(t--)
	solve();
	return 0;
}

C
思路：
$\mathrm{当}k=1\mathrm{时}$显然可以直接暴力求解，$O(n^2)\mathrm{枚}\mathrm{举}$
$\mathrm{当}k>=3\mathrm{时}$结果显然是0，前两次都选择一样的i、j，后面选择新产生的两个数直接相减为0
$\mathrm{当}k=2\mathrm{时}，\mathrm{对}{n}^2\mathrm{个}\mathrm{数}\mathrm{每}\mathrm{个}\mathrm{数}\mathrm{在}\mathrm{排}\mathrm{好}\mathrm{序}\mathrm{的}a\mathrm{数}\mathrm{组}\mathrm{上}\mathrm{面}\mathrm{二}\mathrm{分}\mathrm{找}\mathrm{到}\mathrm{最}\mathrm{接}\mathrm{近}\mathrm{的}\mathrm{维}\mathrm{护}\mathrm{最}\mathrm{小}\mathrm{值}\mathrm{就}\mathrm{是}\mathrm{答}\mathrm{案}$
需要注意一定要和最开始的a取一下最小值，因为答案可能在最初的a中

#include <bits/stdc++.h> 
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define PII pair<int, int>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
#define x first
#define y second
using namespace std;

const int N=5e6+10;
ll n,k,a[N],b[N];


void solve()
{
	cin>>n>>k;
	ll ans=1e18;
	rep(i,1,n)	cin>>a[i],ans=min(ans,a[i]);
	if(k>=3)
	{
		cout<<0<<endl;
		return;
	}
	sort(a+1,a+1+n);
	rep(i,2,n)	ans=min(ans,a[i]-a[i-1]);
	if(k==1)
	{
		cout<<ans<<endl;
		return;
	}
	if(k==2)
	{
		rep(i,1,n)	rep(j,i+1,n)
		{
			ll xx=abs(a[j]-a[i]);
			int k=lower_bound(a+1,a+1+n,xx)-a;
			if(k<=n)	ans=min(ans,a[k]-xx);
			if(k>=2)	ans=min(ans,xx-a[k-1]);
		}
		cout<<ans<<endl;
	}
}

int main()
{
	IOS	
//  	freopen("1.in", "r", stdin);
  	int t;
	cin>>t;
	while(t--)
	solve();
	return 0;
}