Vijos复活邀请赛

滑动输入法

　 　 乱搞即可，不得吐槽题目数据，明明说单词长度小于等于10，结果有很多11的。DFS生成所有可能字符串还有50分，判断矩形与圆相交时半径忘记平方居然有90分，彻底无语。考试的时候还犹豫会不会太暴力，其实看到数据范围小就不应该害怕了。

var x1,x2,y1,y2,count:array[1..20] of longint;
    c:array[1..20] of char;
    ch:char;
    word:array[1..20] of string;
    x,y:array[1..20] of longint;
    w,tot,R,m,n,i,j,ans,minj:longint;
    find:array[1..20] of boolean;
    Q:array[1..20,1..20] of char;
function  ok(i,j:integer):boolean;
          begin
          ok:=((x1[j]<=x[i])and(x[i]<=x2[j])and
               (y1[j]-R<=y[i])and(y[i]<=y2[j]+R))
          or ((y1[j]<=y[i])and(y[i]<=y2[j])and
              (x1[j]-R<=x[i])and(x[i]<=x2[j]+R))
          or((x[i]-x1[j])*(x[i]-x1[j])+(y[i]-y1[j])*(y[i]-y1[j])<=R*R)
          or((x[i]-x1[j])*(x[i]-x1[j])+(y[i]-y2[j])*(y[i]-y2[j])<=R*R)
          or((x[i]-x2[j])*(x[i]-x2[j])+(y[i]-y1[j])*(y[i]-y1[j])<=R*R)
          or((x[i]-x2[j])*(x[i]-x2[j])+(y[i]-y2[j])*(y[i]-y2[j])<=R*R);
          end;
function  check(str:string):boolean;
          var i,j:longint;
              ok:boolean;
          begin
          check:=true;
          for i:=1 to length(str) do
            begin
            ok:=false;
            for j:=1 to count[i] do
              if Q[i][j]=str[i]
                then begin
                     ok:=true;
                     break;
                     end;
            if not ok then exit(false);
            end;
          end;
procedure Sort;
          var i,j:longint;
              temp:string;
          begin
          for i:=1 to tot-1 do
            begin
            temp:=word[i];minj:=i;
            for j:=i+1 to tot do
              if word[j]<temp
                then begin
                     temp:=word[j];
                     minj:=j;
                     end;
            word[minj]:=word[i];
            word[i]:=temp;
            end;
          end;
BEGIN
readln(n,m,w);readln(r);
for i:=1 to n do
  read(x1[i],y1[i],x2[i],y2[i],ch,c[i]);
for i:=1 to m+1 do readln(x[i],y[i]);
for i:=1 to w do
  begin
  inc(tot);
  readln(word[tot]);
  if length(word[tot])<>m+1
    then dec(tot);
  end;
if tot=0 then begin writeln(0);halt;end;
for i:=1 to m+1 do
  for j:=1 to n do
    if ok(i,j)
      then begin
           inc(count[i]);
           Q[i][count[i]]:=c[j];
           end;
Sort;
for i:=1 to tot do
  begin
  find[i]:=check(word[i]);
  if find[i] then inc(ans);
  end;
writeln(ans);
for i:=1 to tot do
  if find[i] then writeln(word[i]);
END.

Edward的可乐旅行

  一个大坑
Mrw的工资计划

　 　 又一个大坑

hzzstep

　 　 找规律，在excel里面填一下，单纯凑规律。有些地方要特判。

uses math;
var x,y,t,x0,y0:int64;
    dis:qword;
    count:int64;
BEGIN
readln(x,y);
t:=max(abs(x),abs(y));
dis:=(t<<1)*(t<<1);
x0:=-t;y0:=t;
if(x=x0) then inc(dis,y0-y)else
if(y=y0) then dec(dis,x-x0)else
if(x=-x0)and(y<>-y0)
  then begin
       dec(dis,t<<1);
       dec(dis,y0-y);
       end
  else inc(dis,t<<1+x-x0);
writeln(dis);       
END.            

　 　 std写的挺短

#include<math.h>
#include<stdio.h>
#include<iostream>
using namespace std;
int x,y,ans,s;
int main()
{
    scanf("%d%d",&x,&y);
    s=max(abs(x),abs(y));
    ans=2*s*(2*s-1);
    if(x==s)
      ans-=(s-y);
    if(y==s)
      ans+=(s-x);
    else
      if(x==-s)
        ans+=(2*s+(s-y));
      else
        if(y==-s)
          ans+=((4*s)+(s+x));
    if(x==s && y==-s)
      ans=2*s*(2*s+2);
    printf("%d",ans);
    system("pause");
    return 0;
}

　 　 751人参赛，500多人有分，我150分名次80多，唉，第一题过了的话是200分，那就是40多名了。DP啊，树啊都是漏洞。LCA只会一次查询的，离线做法还不会啊。