Educational Codeforces Round 3

609A - USB Flash Drives    20171108

609B - The Best Gift    20171108

前两题较为简单，故略过_(:з」∠)_

609C - Load Balancing    20171108

排一次序，求出mi的和之后就可以得出每个服务器的最终状态了，O(n)扫一遍求解即可

我的代码在求解时只考虑了减小负载的数目之和，如果求的是变换量之和记得要除以2

#include<bits/stdc++.h>
using namespace std;
int n,m[100001],s,ans;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&m[i]),s+=m[i];sort(m+1,m+n+1);
    for(int i=n;i>=1;i--)ans+=max(m[i]-((i<=(n-s%n))?s/n:s/n+1),0);
    return printf("%d\n",ans),0;
}

 

609D - Gadgets for dollars and pounds    20180904

预处理a[i]，b[i]的前缀最小值及对应天数编号，之后二分答案并进行O(n)的检查是否可以满足条件即可。开始二分前应先判断是否有解，即对天数为n时进行判断。

#include<bits/stdc++.h>
using namespace std;
#define N 200001
#define LL long long
struct rua{LL w;int d,id;}f[N];
int n,m,k,s,a[N],b[N],t[N],c[N],ma[N],mb[N];
priority_queue<LL,vector<LL>,greater<LL> >q;
bool cmp(rua x,rua y){return x.w<y.w;}
bool check(int x)
{
    while(!q.empty())q.pop();
    for(int i=1;i<=m;i++)
      if(t[i]==1)q.push(1ll*c[i]*a[x]);
      else q.push(1ll*c[i]*b[x]);
    LL res=0;
    for(int i=1;i<=k;i++)
      {
      res+=q.top(),q.pop();
      if(res>1ll*s)return false;
      }
    return true;
}
void print(int d)
{
    printf("%d\n",d);
    for(int i=1;i<=m;i++)
      if(t[i]==1)f[i]={1ll*c[i]*a[d],ma[d],i};
      else f[i]={1ll*c[i]*b[d],mb[d],i};
    sort(f+1,f+m+1,cmp);
    for(int i=1;i<=k;i++)
      printf("%d %d\n",f[i].id,f[i].d);
}
int main()
{
    a[0]=b[0]=19260817;
    scanf("%d%d%d%d",&n,&m,&k,&s);
    for(int i=1;i<=n;i++)
      {
      scanf("%d",&a[i]);
      if(a[i]<a[i-1])ma[i]=i;
      else ma[i]=ma[i-1],a[i]=a[i-1];
      }
    for(int i=1;i<=n;i++)
      {
      scanf("%d",&b[i]);
      if(b[i]<b[i-1])mb[i]=i;
      else mb[i]=mb[i-1],b[i]=b[i-1];
      }
    for(int i=1;i<=m;i++)
      scanf("%d%d",&t[i],&c[i]);
    if(!check(n))return printf("-1\n"),0;
    int l=1,r=n;
    while(l<r)
      {
      int mid=l+r>>1;
      if(check(mid))r=mid;
      else l=mid+1;
      }
    return print(l),0;
}

 

609E - Minimum spanning tree for each edge    20180905

[Educational Round 3][Codeforces 609E. Minimum spanning tree for each edge]

 

609F - Frogs and mosquitoes    20180910

[Educational Round 3][Codeforces 609F. Frogs and mosquitoes]