比赛链接：https://codeforces.com/contest/2026
A. Perpendicular Segments
题目说了必定有答案，直接对角线即可
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#define ll long long
#define lowbit(x) (x & -x)
using namespace std;
mt19937 rnd(time(0));
const ll mod=1e9+7;
const ll N=2e5+5;
ll ksm(ll x,ll y)
{ll ans=1;while(y){if(y&1)ans=(ans%mod*x%mod)%mod;x=x%mod*(x%mod)%mod;y>>=1;}return ans%mod%mod;
}
ll gcd(ll x,ll y)
{if(y==0)return x;else return gcd(y,x%y);
}
void fio()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
}
int main()
{fio();ll t;cin>>t;while(t--){ll x,y,k;cin>>x>>y>>k;ll j=min(x,y);cout<<0<<" "<<0<<" "<<j<<" "<<j<<endl;cout<<j<<" "<<0<<" "<<0<<" "<<j<<endl;}
}B. Black Cells
n为偶数答案是固定的，n为奇数可以二分答案
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#define ll long long
#define lowbit(x) (x & -x)
using namespace std;
mt19937 rnd(time(0));
const ll mod=1e9+7;
const ll N=2e5+5;
ll ksm(ll x,ll y)
{ll ans=1;while(y){if(y&1)ans=(ans%mod*x%mod)%mod;x=x%mod*(x%mod)%mod;y>>=1;}return ans%mod%mod;
}
ll gcd(ll x,ll y)
{if(y==0)return x;else return gcd(y,x%y);
}
void fio()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
}
ll a[250000];
int main()
{fio();ll t;cin>>t;while(t--){ll n;cin>>n;if(n==1){ll b;cin>>b;cout<<1<<endl;continue;}for(ll i=1;i<=n;i++)cin>>a[i];sort(a+1,a+1+n);//a[n+1]=0;if(n%2==0){ll ans=0;for(ll i=1;i<=n-1;i+=2){ans=max(ans,a[i+1]-a[i]);}cout<<ans<<endl;}else {ll l=1,r=1e18;while(l<r){ll mid=(l+r)>>1;ll gs=1;ll cnt=0;ll pd=0;while(gs<=n-1){if(a[gs+1]-a[gs]<=mid){gs+=2;continue;}else {if(cnt==1){pd=1;break;}gs++;cnt++;continue;}}if(pd)l=mid+1;else r=mid;}cout<<r<<endl;}}
}C. Action Figures
分析可得0点必选买，非0点优先用0点消除，其次用下标最小的1点消除
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#define ll long long
#define lowbit(x) (x & -x)
using namespace std;
mt19937 rnd(time(0));
const ll mod=1e9+7;
const ll N=2e5+5;
ll ksm(ll x,ll y)
{ll ans=1;while(y){if(y&1)ans=(ans%mod*x%mod)%mod;x=x%mod*(x%mod)%mod;y>>=1;}return ans%mod%mod;
}
ll gcd(ll x,ll y)
{if(y==0)return x;else return gcd(y,x%y);
}
void fio()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
}
ll a[550000];
int main()
{fio();ll t;cin>>t;while(t--){ll mo=1;ll n;cin>>n;string f;cin>>f;set<ll>q,k;ll op=0;ll ans=0;for(ll i=1;i<=n;i++){if(f[i-1]=='0')q.insert(i);else k.insert(i);}for(ll i=f.size()-1;i>=0;i--){if(f[i]=='1'){ll c=(i+1);auto j=k.lower_bound(c);if(j!=k.end()){if(q.size()>0){ll u=*q.rbegin();q.erase(u);ans+=u;k.erase(c);}else{if(k.size()==1){ans+=c;k.clear();}else {ll u=*k.begin();k.erase(c);k.erase(u);ans+=u;}}}}else {ll u=i+1;if(q.lower_bound(u)!=q.end()){q.erase(u);ans+=u;}}//cout<<ans<<endl;}cout<<ans<<endl;}
}
D. Sums of Segments
直接暴力统计优化，数据不会爆long long
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#define ll long long
#define lowbit(x) (x & -x)
using namespace std;
mt19937 rnd(time(0));
const ll mod=1e9+7;
const ll N=2e5+5;
ll ksm(ll x,ll y)
{ll ans=1;while(y){if(y&1)ans=(ans%mod*x%mod)%mod;x=x%mod*(x%mod)%mod;y>>=1;}return ans%mod%mod;
}
ll gcd(ll x,ll y)
{if(y==0)return x;else return gcd(y,x%y);
}
void fio()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
}
ll a[325000];
ll pre[325000];//一维前缀
ll pre1[325000];//类似二维
ll sub[320555];
ll pre3[325000];
//1
//1 2
//1 2 5
//1 2 5 10
ll n;
set<pair<ll,ll>>q;
ll qu(ll x)
{ll ans=0;auto j=q.lower_bound({x,0});ll zq=(*j).second;ans+=pre3[zq-1];j--;ll wz=x-(*j).first+zq-1;ans+=pre1[zq]-sub[wz+1]-(pre[wz]-pre[zq-1])*(n-zq-(wz-zq));return ans;
}
int main()
{fio();ll t;t=1;while(t--){q.clear();cin>>n;ll cnt=0;for(ll i=1;i<=n;i++)cin>>a[i],pre[i]=pre[i-1]+a[i],cnt+=pre[i];pre1[1]=cnt;for(ll i=2;i<=n;i++)pre1[i]=pre1[i-1]-(n-i+2)*a[i-1];for(ll i=1;i<=n;i++)pre3[i]=pre3[i-1]+pre1[i];sub[n+1]=0;for(ll i=n;i>=1;i--)sub[i]=sub[i+1]+a[i]*(n-i+1);q.insert({0,0});cnt=0;for(ll i=1;i<=n;i++){cnt+=(n-i+1);q.insert({cnt,i});}ll op;cin>>op;while(op--){ll l,r;cin>>l>>r;cout<<qu(r)-(l-1?qu(l-1):0)<<endl;}}
}
