A. Lame King

没啥好说的，先穿插着走，在直走

#include <bits/stdc++.h>
#define int long long
#define pb push_back
#define fer(i,a,b) for(int i=a;i<=b;++i)
#define der(i,a,b) for(int i=a;i>=b;--i)
#define all(x) (x).begin(),(x).end()
#define pll pair<int,int>
#define et  cout<<'\n'
#define xx first
#define yy second
using namespace std;
template <typename _Tp>void input(_Tp &x){char ch(getchar());bool f(false);while(!isdigit(ch))f|=ch==45,ch=getchar();x=ch&15,ch=getchar();while(isdigit(ch))x=x*10+(ch&15),ch=getchar();if(f)x=-x;
}
template <typename _Tp,typename... Args>void input(_Tp &t,Args &...args){input(t);input(args...);}
const int N=1e6+10;
signed main()
{int T;cin>>T;while(T--){int a,b;cin>>a>>b;a=abs(a);b=abs(b);if(a<b){swap(a,b);}cout<<b*2+max(0ll,(a-b)*2-1)<<endl;}
}

B. Vaccination

用一个队列存病人

原则就是，拖到最晚开一瓶药，开了之后尽量用没

#include <bits/stdc++.h>
#define int long long
#define pb push_back
#define fer(i,a,b) for(int i=a;i<=b;++i)
#define der(i,a,b) for(int i=a;i>=b;--i)
#define all(x) (x).begin(),(x).end()
#define pll pair<int,int>
#define et  cout<<'\n'
#define xx first
#define yy second
using namespace std;
template <typename _Tp>void input(_Tp &x){char ch(getchar());bool f(false);while(!isdigit(ch))f|=ch==45,ch=getchar();x=ch&15,ch=getchar();while(isdigit(ch))x=x*10+(ch&15),ch=getchar();if(f)x=-x;
}
template <typename _Tp,typename... Args>void input(_Tp &t,Args &...args){input(t);input(args...);}
const int N=1e6+10;
int a[N];
struct node{int canuse;int time;
};
signed main()
{int T;cin>>T;while(T--){int n,w,t1,tn;queue<int> br;cin>>n>>w>>t1>>tn;fer(i,1,n){cin>>a[i];br.push(a[i]);}int res=0;while(br.size()){auto noww=w;auto nowt=br.front()+t1+tn;while(noww&&nowt>=br.front()&&br.size()){noww--;br.pop();}res++;}cout<<res<<'\n';}
}

C. Pull Your Luck

等差数列求和，如果超出循环节还没有解就是No

#include <bits/stdc++.h>
#define int long long
#define pb push_back
#define fer(i,a,b) for(int i=a;i<=b;++i)
#define der(i,a,b) for(int i=a;i>=b;--i)
#define all(x) (x).begin(),(x).end()
#define pll pair<int,int>
#define et  cout<<'\n'
#define xx first
#define yy second
using namespace std;
template <typename _Tp>void input(_Tp &x){char ch(getchar());bool f(false);while(!isdigit(ch))f|=ch==45,ch=getchar();x=ch&15,ch=getchar();while(isdigit(ch))x=x*10+(ch&15),ch=getchar();if(f)x=-x;
}
template <typename _Tp,typename... Args>void input(_Tp &t,Args &...args){input(t);input(args...);}
const int N=1e6+10;
const int MX=1e6;
using namespace std;
const int mod=998244353;
int T;
int n,x,p;
signed main()
{ios::sync_with_stdio(0);cin.tie(0);int T;cin>>T;while(T--){cin>>n>>x>>p;int now=1;int fl=0;while(1){if(now>p){fl=0;break;}if(now>3*n){fl=0;break;}if(!((x+now*(now+1)/2%n)%n)){cout<<"Yes\n";fl=1;break;}now++;}if(!fl) cout<<"No\n";}return 0;
}

D. Accommodation

最少就是把所有连续的1都贪心成两室一厅

最多就是把把所有连续的0都贪心成两室一厅

注意不要超过限制（m/4，m/2）

（赛后被fst了，循环应该循环到m-1，没注意写成m了，出题人用*造数据）

#include <bits/stdc++.h>
#define int long long
#define pb push_back
#define fer(i,a,b) for(int i=a;i<=b;++i)
#define der(i,a,b) for(int i=a;i>=b;--i)
#define all(x) (x).begin(),(x).end()
#define pll pair<int,int>
#define et  cout<<'\n'
#define xx first
#define yy second
using namespace std;
template <typename _Tp>void input(_Tp &x){char ch(getchar());bool f(false);while(!isdigit(ch))f|=ch==45,ch=getchar();x=ch&15,ch=getchar();while(isdigit(ch))x=x*10+(ch&15),ch=getchar();if(f)x=-x;
}
template <typename _Tp,typename... Args>void input(_Tp &t,Args &...args){input(t);input(args...);}
using namespace std;
const int N=1e6+10;
string s;
int n,m,tmp;
int vis[N];
int calc1()
{int res=0;int cnt=0;fer(i,1,m){if(s[i]=='1') cnt++;else{res+=cnt/2;cnt=0;}}res+=cnt/2;if(res>m/4){res=m/4;}res=tmp-res;return res;
}
int calc2()
{int res=0;int cnt=0;cnt=tmp;fer(i,1,m-1){if(s[i]=='0'||s[i+1]=='0'){if(!vis[i]&&!vis[i+1]){vis[i]=1;vis[i+1]=1;res+=1;}}}int dis=min(m/4,res);res=m/4-dis;return cnt-res;
}signed main()
{ios::sync_with_stdio(0);cin.tie(0);cin>>n>>m;int ans1=0;int ans2=0;fer(i,1,n){cin>>s;s='0'+s;tmp=0;for(auto t:s){if(t=='1'){tmp++;}}fer(j,1,m){vis[j]=0;}ans1+=calc1();ans2+=calc2();}cout<<ans1<<" "<<ans2<<'\n';
}

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Nebius Welcome Round (Div. 1 + Div. 2) 题解

A. Lame King

B. Vaccination

C. Pull Your Luck

D. Accommodation

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