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。

Codeuses 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;BEGINreadln(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。

Codevar ch:char;    next,pre,num:array[0..5000000] of longint;    i,j,h,t,n,len:longint;BEGINreset(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。