1.油滴扩展（box）

搜索，注意精度。

 

#include<fstream>
#include<cmath>
#define pi 3.1415926575897932
using namespace std;
ifstream cin("box.in");
ofstream cout("box.out");
int n;
int x,y,xx,yy;
double tot=0,ans=0;
bool v[8];
struct list
{
       int x,y;
       double r;
}d[8];

void dfs(int a,double sum)
{
     if (a==(n+1))
     {
                  if (sum>ans) ans=sum;
                  return;
     }
     for (int j=1;j<=n;j++)
     if (!v[j])
     {
     d[j].r=min(min(min(abs((double)(d[j].x-x)),abs((double)(d[j].x-xx))),abs((double)(d[j].y-y))),abs((double)(d[j].y-yy)));
     for (int i=1;i<=n;i++)
     if (v[i])
     {
         double t=sqrt(abs((double)(d[j].x-d[i].x))*abs((double)(d[j].x-d[i].x))+abs((double)(d[j].y-d[i].y))*abs((double)(d[j].y-d[i].y)));
         t-=d[i].r;
         if (t<0) t=0;
         d[j].r<?=t;
     }
     v[j]=true;
     dfs(a+1,sum+pi*d[j].r*d[j].r);
     v[j]=false;
     }
}

int main()
{
    cin>>n;
    cin>>x>>y>>xx>>yy;
    for (int i=1;i<=n;i++) cin>>d[i].x>>d[i].y;
    dfs(1,0);
    tot=abs((double)(x-xx))*abs((double)(y-yy));
    int rest=(int)(tot-ans+0.5);
    cout<<rest<<endl;
    return 0;
}

2.数列（sequence）

 

f[i,j]=max{f[i-1,j-1],f[i-1,j]+(a[i]=i-j)}

 

#include<fstream>
using namespace std;
ifstream cin("sequence.in");
ofstream cout("sequence.out");
int n,a[1100];
int f[1100][1100]={0};
int main()
{
    int i,j;
    cin>>n;
    for (i=1;i<=n;i++) cin>>a[i];
    for (i=1;i<=n;i++)
      for (j=1;j<=i;j++)
      {
          f[i][j]=max(f[i-1][j],f[i-1][j-1]);
          if (a[i]==(i-j+1)) f[i][j]=max(f[i][j],f[i-1][j]+1);
      }
    int ans=0;
    for (i=1;i<=n;i++)
      ans>?=f[n][i];
    cout<<ans<<endl;
    return 0;
}

3.SOFTWARE

 

二分答案，判断条件f[i,j+k]=max{f[i-1,j]+(mid-k*a[i])/b[i]}  (k+j<=m&&mid>=a[i]*k)

 

#include<fstream>
#include<cstdlib>
using namespace std;
ifstream cin("software.in");
ofstream cout("software.out");
int n,m,maxtime;
struct list
{
       int a,b;
}d[110];
int f[110][110];
bool check(int mid)
{
     memset(f,-1,sizeof(f));
     f[0][0]=0;
     for (int i=1;i<=n;i++)
       for (int j=0;j<=m;j++)
       {
         if (f[i-1][j]<0)  continue;
         for (int k=0;(j+k<=m) && (mid>=d[i].a*k);k++)
             f[i][j+k]>?=f[i-1][j]+((mid-k*d[i].a)/d[i].b);
       }
     return (f[n][m]>=m);
}

int main()
{
    cin>>n>>m;
    for (int i=1;i<=n;i++) 
    {
        cin>>d[i].a>>d[i].b;
        maxtime>?=d[i].a;
        maxtime>?=d[i].b;
    }
    int l=0,r=m*maxtime,mid;
    while (l<r)
    {
          mid=(l+r)>>1;
          if (!check(mid)) l=mid+1; 
          else r=mid;
    }
    cout<<l<<endl;
    return 0;
}

4.黑匣子（blackbox）

 

二分查找树

 

代码 
program blackbox;
type rec=^node;
     node=record
     num,tot:longint;
     l,r:rec;
     end;
var a,b:array[0..200000]of longint;
    i,j,k,m,n,last:longint;
    root:rec;
procedure add(var p:rec;x:longint);
  begin
    if p=nil then
               begin
                 new(p);
                 p^.num:=x;
                 p^.tot:=1;
                 p^.l:=nil;
                 p^.r:=nil;
               end
               else
                 begin
                   inc(p^.tot);
                   if x<p^.num then add(p^.l,x)
                   else add(p^.r,x);
                 end;
  end;
procedure dfs(p:rec;k:longint);
  var tt:longint;
  begin
    if p^.l=nil then tt:=1
      else tt:=p^.l^.tot+1;
    if k=tt then writeln(p^.num)
      else if k<tt then dfs(p^.l,k) else dfs(p^.r,k-tt);
  end;

begin
  assign(input,'blackbox.in');reset(input);
  assign(output,'blackbox.out');rewrite(output);
  readln(m,n);
  for i:=1 to m do read(a[i]);
  last:=1;
  k:=0;
  root:=nil;
  for i:=1 to n do
    begin
      read(b[i]);
      for j:=last to b[i] do add(root,a[j]);
      inc(k);
      dfs(root,k);
      last:=b[i]+1;
    end;
  close(input);close(output);
end.