Clover杯系列模拟赛 I

援助方案

　 　 坐标和图形数比较小，每处理一个图形暴力枚举可能在图形里的整点然后判断即可。圆用距离公式，矩形。。。，三角形用向量。注意圆可能覆盖到二三四象限。

uses math;
var f:array[-60..110,-60..110] of boolean;
    ans:int64;
    n,i:longint;
    ch:char;
function cross(x1,y1,x2,y2:longint):longint;
         begin
         Cross:=x1*y2-x2*y1;
         end;
function ok(a,b,c:longint):boolean;
         begin
         ok:=false;
         if(a>=0)and(b>=0)and(c>=0)then ok:=true;
         if(a<=0)and(b<=0)and(c<=0)then ok:=true;
         end;
procedure Triangle;
          var i,j,x1,x2,x3,xa,xb,xc,y1,y2,y3,ya,yb,yc,up,down,left,right:longint;
          begin
          readln(x1,y1,x2,y2,x3,y3);
          xa:=x2-x1;ya:=y2-y1;
          xb:=x3-x2;yb:=y3-y2;
          xc:=x1-x3;yc:=y1-y3;
          up:=max(max(y1,y2),y3);
          down:=min(min(y1,y2),y3);
          right:=max(max(x1,x2),x3);
          left:=min(min(x1,x2),x3);
          for i:=left to right do
            for j:=down to up do
              if(not f[i][j])and
                (ok(Cross(i-x1,j-y1,xa,ya),
                    Cross(i-x2,j-y2,xb,yb),
                    Cross(i-x3,j-y3,xc,yc))
                )
                then begin
                     inc(ans);
                     f[i][j]:=true;
                     end;
          end;
procedure circle;
          var x,y,r,i,j,rr:longint;
          begin
          readln(x,y,r);rr:=r*r;
          for i:=x-r to x+r do
            for j:=y-r to y+r do
              if (not f[i][j])and
                 ((i-x)*(i-x)+(j-y)*(j-y)<=rr)
                then begin
                     inc(ans);
                     f[i][j]:=true;
                     end;
          end;
procedure Square;
          var x,y,l,i,j:longint;
          begin
          readln(x,y,l);
          for i:=x to x+l do
            for j:=y to y+l do
              if not f[i][j]
                then begin
                     inc(ans);
                     f[i][j]:=true;
                     end;
          end;
BEGIN
readln(n);
for i:=1 to n do
  begin
  read(ch);
  case ch of
    'T':Triangle;
    'C':circle;
    'S':Square;
    end;
  end;
writeln(ans);
END.

数字游戏
　 　 标程是单调队列。我用数组模拟双链表A掉的。实际上我是维护了一个单调不降的线性表。内存泄露神马的不管了。

var ch:char;
    next,pre,num:array[0..5000000] of longint;
    i,j,h,t,n,len:longint;
BEGIN
reset(input);rewrite(output);
while not eoln do
  begin
  read(ch);
  inc(len);
  num[len]:=ord(ch)-ord('0');
  end;
readln(n);
if n=len then begin writeln(0);halt;end;
for i:=1 to len do
  begin
  pre[i]:=i-1;
  next[i]:=i+1;
  end;
pre[1]:=0;next[len]:=0;h:=1;t:=len;
i:=1;
while n>0 do
  begin
  while(i<>0)and(num[i]<=num[next[i]])do
  i:=next[i];
  if i=0 then i:=t;
  dec(n);dec(len);
  next[pre[i]]:=next[i];
  pre[next[i]]:=pre[i];
  if pre[i]=0 then h:=next[i];
  if next[i]=0 then t:=pre[i];
  i:=pre[i];if i=0 then i:=h;
  end;
i:=h;
while(i<>0)and(num[i]=0)do i:=next[i];
  if i=0 then writeln(0) else begin
  while(i<>0)do begin
    write(num[i]);
    i:=next[i];
    end;
    writeln;
  end;
END.

哈密顿路

      NPC问题做不来。 

总结

　 　 610人参赛，298人有分，两个AK的，3个280以上，我150分，69名，如果第二题仔细一点就好了，200分的话是26名。写完程序必须自己造数据。