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。
Code
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。
Code
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
Code#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。