luogu 2827(noip2016) 蚯蚓

https://www.luogu.org/problem/show?pid=2827
就是贪心 从a[1],a[2],a[3]中取出最大的一个，切掉，切下来的放到a[1],a[2]中
因为被切的虫子在这一秒不增长 所以可以将其减去q 来保证虫子们的数量关系的正确性
最后归并三个数组得到答案数组
注意：要取模卡常
上代码

var a:array[1..3,1..7100000] of longint;
    d:array[1..10000000] of longint;
    b,c,h:array[1..3] of longint;
    i,j,k,l,m,n,q,w,t,v,u,ct,cut,h1,h2,h3,tot,hj:longint;
    p:double;
procedure sort(l,r: longint);
      var i,j,x,y: longint;
      begin
         i:=l; j:=r; x:=a[3,(l+r) div 2];
         repeat
           while a[3,i]>x do inc(i);
           while x>a[3,j] do dec(j);
           if not(i>j) then
             begin
                y:=a[3,i]; a[3,i]:=a[3,j]; a[3,j]:=y;
                inc(i); j:=j-1;
             end;
         until i>j;
         if lthen sort(l,j);
         if ithen sort(i,r);
      end;
begin
    for i:=1 to 3 do begin b[i]:=1;c[i]:=1; end;
    fillchar(a,sizeof(a),128);
    readln(n,m,q,u,v,t);
    p:=u/v;
    for i:=1 to n do read(a[3,i]); readln;
    sort(1,n);
    for i:=1 to m do begin
      if(a[1,b[1]]>=a[2,b[2]])and((a[1,b[1]])>=a[3,b[3]]) then
            begin ct:=1; cut:=a[1,b[1]]; end else
      if(a[2,b[2]]>=a[1,b[1]])and((a[2,b[2]])>=a[3,b[3]]) then
            begin ct:=2; cut:=a[2,b[2]]; end else
      if(a[3,b[3]]>=(a[2,b[2]]+k))and((a[1,b[1]])<=a[3,b[3]]) then
            begin ct:=3; cut:=a[3,b[3]];end;
      inc(b[ct]);
      cut:=cut+(i-1)*q;
      a[1,c[1]]:=trunc(cut*u/v)-q*i;
      a[2,c[2]]:=cut-trunc(cut*u/v)-i*q;
      inc(c[1]);inc(c[2]);
      inc(hj);
      if hj=t then begin write(cut,' '); hj:=0; end;
end;
    writeln;
    h[1]:=b[1];h[2]:=b[2];h[3]:=b[3];
    while tot < m+n do begin
        inc(tot);
        if(a[1,h[1]]>=a[2,h[2]])and(a[1,h[1]]>=a[3,h[3]]) then begin
           ct:=1; d[tot]:=a[1,h[1]] end;
        if(a[2,h[2]]>=a[1,h[1]])and(a[2,h[2]]>=a[3,h[3]]) then begin
           ct:=2; d[tot]:=a[2,h[2]] end;
        if(a[3,h[3]]>=a[2,h[2]])and(a[3,h[3]]>=a[1,h[1]]) then begin
           ct:=3; d[tot]:=a[3,h[3]] end;
        inc(h[ct]);
    end; k:=m*q;   i:=t;
    while i<=(m+n) do begin write(d[i]+k,' '); i:=i+t; end;
end.

END.