Educational Codeforces Round 61 (Div.2)

A.(c1=0&&c3>0)||(c1!=c4)

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 typedef long long ll;
 using namespace std;

 int c1,c2,c3,c4;

 int main(){
     freopen("a.in","r",stdin);
     freopen("a.out","w",stdout);
     scanf("%d%d%d%d",&c1,&c2,&c3,&c4);
     if ((!c1 && c3) || c1!=c4) puts(""); else puts("");
     return ;
 }

A

B.每次免费的显然应该是第n-q+1大的那个。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 typedef long long ll;
 using namespace std;

 const int N=;
 int n,m,x,a[N];
 ll sm;

 int main(){
     scanf("%d",&n);
     rep(i,,n) scanf("%d",&a[i]),sm+=a[i];
     sort(a+,a+n+); scanf("%d",&m);
     rep(i,,m) scanf("%d",&x),cout<<sm-a[n-x+]<<endl;
     return ;
 }

B

C.正难则反，先计算所有区间的并集sm，再计算最少要使多长区间不被覆盖ans，答案即sm-ans。

考虑枚举区间i,j，那么不涂这两个区间会使b[i]+b[j]+c[i][j]的区间不被覆盖。其中b[i]为只能由第i个区间覆盖的区间长度，c[i][j]为只能被i,j两个区间覆盖的区间长度。先O(nq)预处理出b,c，再O(q^2)计算答案即可。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 typedef long long ll;
 using namespace std;

 const int N=;
 int n,m,l,r,sm,ans,x[N],y[N],b[N],mp[N][N];

 int main(){
     scanf("%d%d",&n,&m); ans=n+;
     rep(i,,m){
         scanf("%d%d",&l,&r);
         rep(j,l,r){
             if (y[j]) x[j]=y[j]=-;
             if (x[j] && !y[j]) y[j]=i;
             if (!x[j]) x[j]=i;
         }
     }
     rep(i,,n){
         if (x[i]) sm++;
         if (x[i]> && !y[i]) b[x[i]]++;
         if (x[i]> && y[i]>) mp[x[i]][y[i]]++,mp[y[i]][x[i]]++;
     }
     rep(i,,m) rep(j,,m) if (i!=j) ans=min(ans,b[i]+b[j]+mp[i][j]);
     printf("%d\n",sm-ans);
     return ;
 }

C

D.二分答案，每次找到最先降到0以下的那个同学充电。

 #include<queue>
 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 typedef long long ll;
 using namespace std;

 const int N=;
 int n,k;
 ll mx,a[N],b[N];
 struct P{ ll a,b,r; };
 bool operator <(const P &a,const P &b){ return a.r>b.r; }
 priority_queue<P>Q;

 bool chk(ll mid){
     while (!Q.empty()) Q.pop();
     rep(i,,n) Q.push((P){a[i],b[i],a[i]/b[i]});
     rep(t,,k){
         P x=Q.top(); Q.pop();
         if (x.a/x.b+<t) return ;
         if (x.a/x.b+>=k) return ;
         Q.push((P){x.a+mid,x.b,(x.a+mid)/x.b});
     }
     return ;
 }

 int main(){
     ios::sync_with_stdio();
     cin>>n>>k;
     rep(i,,n) cin>>a[i];
     rep(i,,n) cin>>b[i],mx=max(mx,b[i]);
     ll L=,R=(k-)*mx+;
     while (L<R){
         ll mid=(L+R)>>;
         if (chk(mid)) R=mid; else L=mid+;
     }
     if (L>=(k-)*mx+) puts("-1"); else cout<<L<<endl;
     return ;
 }

D