Codeforces Round 920 (Div. 3)（A~F）

A

按题意模拟即可

#include  
#define int long long
#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 pii pair<int, int>
#define pll pair<long long, long long>
#define ll long long
#define db double
#define endl '\n'
#define x first
#define y second
#define pb push_back

using namespace std;

void solve()
{
	set<pii>s;
	rep(i,1,4){
		int x,y;cin>>x>>y;
		s.insert({x,y});
	}
	for(auto [x1,y1]:s){
		for(auto [x2,y2]:s){
			if(x1==x2&&y1==y2)	continue;
			if(x1==x2){
				cout<<abs(y1-y2)*abs(y1-y2)<<endl;
				return;
			}
		}
	}
}
signed main(){
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
//   	freopen("1.in", "r", stdin);
  	int _;
	cin>>_;
	while(_--)
	solve();
	return 0;
}

B

统计01和10的数量取最大值

#include  
#define int long long
#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 pii pair<int, int>
#define pll pair<long long, long long>
#define ll long long
#define db double
#define endl '\n'
#define x first
#define y second
#define pb push_back

using namespace std;

void solve()
{
	string s1,s2;
	int n;cin>>n;
	cin>>s1>>s2;
	int cnt1=0,cnt2=0;
	rep(i,0,s1.size()-1){
		if(s1[i]=='0'&&s2[i]=='1')	{
			cnt1++;
		}else if(s1[i]=='1'&&s2[i]=='0'){
			cnt2++;
		}
	}
	cout<<min(cnt1,cnt2)+abs(cnt1-cnt2)<<endl;
}
signed main(){
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
//   	freopen("1.in", "r", stdin);
  	int _;
	cin>>_;
	while(_--)
	solve();
	return 0;
}

C

考虑贪心，到达当前如果a操作的代价大选择b操作，b操作的代价大选择a操作

#include  
#define int long long
#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 pii pair<int, int>
#define pll pair<long long, long long>
#define ll long long
#define db double
#define endl '\n'
#define x first
#define y second
#define pb push_back

using namespace std;

void solve()
{
	int n,f,a,b;cin>>n>>f>>a>>b;
	vector<int>t(n+1);
	int ans=0;
	rep(i,1,n){
		cin>>t[i];
		int costa=(t[i]-t[i-1])*a;
		int costb=b;
		ans+=min(costa,costb);
	}
	cout<<(ans>=f?"no":"yes")<<endl;
}
signed main(){
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
//   	freopen("1.in", "r", stdin);
  	int _;
	cin>>_;
	while(_--)
	solve();
	return 0;
}

D

双指针+贪心
先将两个数组排序，a升序，b降序。
然后比较两个端点的对答案的贡献，选择贡献大的

#include  
#define int long long
#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 pii pair<int, int>
#define pll pair<long long, long long>
#define ll long long
#define db double
#define endl '\n'
#define x first
#define y second
#define pb push_back

using namespace std;

void solve()
{
	int n,m;cin>>n>>m;
	vector<int>a(n+1),b(m+1);
	rep(i,1,n){
		cin>>a[i];
	}
	rep(i,1,m){
		cin>>b[i];
	}
	sort(a.begin()+1,a.begin()+1+n);
	sort(b.begin()+1,b.begin()+1+m,greater<int>());
	int ans=0,la=1,ra=n,lb=1,rb=m;
	rep(i,1,n){
		int dl=abs(a[la]-b[lb]),dr=abs(a[ra]-b[rb]);
		if(dl>dr){
			ans+=dl;
			la++;lb++;
		}else{
			ans+=dr;
			ra--;rb--;
		}
	}
	cout<<ans<<endl;
}
signed main(){
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
//   	freopen("1.in", "r", stdin);
  	int _;
	cin>>_;
	while(_--)
	solve();
	return 0;
}

E

博弈论的题目
将行列分开考虑
先考虑行，最终两者一定会走到一行。
什么情况下alice可能会赢，一定是$yb-ya$为奇数时
当$yb-ya$为偶数时，bob可能赢
然后考虑列一定是一方经过turn伦将对方挤到边界处获胜

#include  
#define int long long
#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 pii pair<int, int>
#define pll pair<long long, long long>
#define ll long long
#define db double
#define endl '\n'
#define x first
#define y second
#define pb push_back

using namespace std;

void solve()
{
	int n,m;cin>>n>>m;
	int xa,ya,xb,yb;cin>>ya>>xa>>yb>>xb;
	int diff=yb-ya;
	if(diff<=0){
		cout<<"draw"<<endl;
		return;
	}
	int turn=diff>>1;
	if(diff&1){
		if(xb>xa){
			xa=min(xa+turn+1,m);
			xb=min(xb+turn,m);
			if(xa>=xb)	cout<<"alice"<<endl;
			else	cout<<"draw"<<endl;
		}else{
			xa=max(xa-turn-1,1*1ll);
			xb=max(xb-turn,1*1ll);
			if(xa<=xb)	cout<<"alice"<<endl;
			else	cout<<"draw"<<endl;
		}
	}else{
		if(xa>xb){
			xa=min(xa+turn,m);
			xb=min(xb+turn,m);
			if(xa<=xb)	cout<<"bob"<<endl;
			else	cout<<"draw"<<endl;
		}else{
			xa=max(xa-turn,1*1ll);
			xb=max(xb-turn,1*1ll);
			if(xa>=xb)	cout<<"bob"<<endl;
			else	cout<<"draw"<<endl;
		}
	}
}

signed main(){
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
//   	freopen("1.in", "r", stdin);
  	int _;
	cin>>_;
	while(_--)
	solve();
	return 0;
}

F

根号分治

根号分治

#include  
#define int long long
#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 pii pair<int, int>
#define pll pair<long long, long long>
#define ll long long
#define db double
#define endl '\n'
#define x first
#define y second
#define pb push_back

using namespace std;

int f[330][100010],g[330][100010];

void solve()
{
	int n,q;cin>>n>>q;
	vector<int>a(n+1);
	rep(i,1,n){
		cin>>a[i];
	}
	//预处理出来模数比较小的情况
	int sq=sqrt(n);

	rep(i,1,sq){
		rep(j,1,n){
			f[i][j]=(j-i>=0?f[i][j-i]:0)+j/i*a[j];
			g[i][j]=(j-i>=0?g[i][j-i]:0)+a[j];
		}
	}
	
	while(q--){
		int s,d,k;cin>>s>>d>>k;
		//模数较小直接查询预处理的结果
		if(d<=sq){
			int ans=0;
			ans+=f[d][s+(k-1)*d]-(s-d>=0?f[d][s-d]:0);
			ans-=(s/d-1)*(g[d][s+(k-1)*d]-(s-d>=0?g[d][s-d]:0));
			cout<<ans<<' ';
		}else{
			//较大的话暴力去做
			int ans=0;
			for(int i=s,j=1;j<=k;i+=d,j++){
				ans+=j*a[i];
			}
			cout<<ans<<' ';
		}
	}
	cout<<endl;		
}
signed main(){
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
//   	freopen("1.in", "r", stdin);
  	int _;
	cin>>_;
	while(_--)
	solve();
	return 0;
}