bzoj1789 AHOI 维护数列（线段树）

首先想到线段树，然后刚开始写忽然想到树状数组求和岂不是更快，而且编程复杂度又小，于是把之前写的删掉，写树状数组，写完模版之后忽然发现这题竟然是区间修改！

于是又删掉重写，忽然发现不会处理又加又乘的，果断看题解……

经过几乎两个小时的调试，终于1A。

需要注意的是，一定要让线段树的每一个区间保存的值时刻为正确的，这样才能在调用时直接返回。比如说这道题的change和query操作的最后一句话：

sum:=f(g[k<<1]+g[k<<1+1])

而不是

sum:=f(t[k<<1].sum+t[k<<1+1].sum)

时刻记住这点就ok了。一开始我还以为我的模版记错了呢……

代码：

 type node=record
      l,r,ti,ad,sum:int64;
      end;
 var i,n,m,tagtime,tagadd,ch,x,y,c,p:longint;
     t:array[..] of node;
 function f(x:int64):int64;
  begin
  f:=x mod p;
  end;
 function g(k:longint):longint;
  begin
  with t[k] do
   begin
   g:=f(f(sum*ti)+f(ad*(r-l+)));
   end;
  end;
 procedure build(x,y,k:longint);
  var mid:longint;
  begin
  with t[k] do
   begin
   l:=x;r:=y;ad:=;ti:=;
   if l=r then begin read(sum);sum:=f(sum);exit;end;
   mid:=(l+r)>>;
   build(l,mid,k<<);
   build(mid+,r,k<<+);
   sum:=f(t[k<<].sum+t[k<<+].sum);
   end;
  end;
 procedure pushdown(k:longint);
  begin
  with t[k] do
   begin
   if ti<> then
    begin
    sum:=f(sum*ti);
    t[k<<].ti:=f(t[k<<].ti*ti);
    t[k<<].ad:=f(t[k<<].ad*ti);
    t[k<<+].ti:=f(t[k<<+].ti*ti);
    t[k<<+].ad:=f(t[k<<+].ad*ti);
    ti:=;
    end;
   if ad<> then
    begin
    sum:=f(sum+ad*(r-l+));
    t[k<<].ad:=f(t[k<<].ad+ad);
    t[k<<+].ad:=f(t[k<<+].ad+ad);
    ad:=;
    end;
   end;
  end;
 procedure change(x,y,k:longint);
  var mid:longint;
  begin
  with t[k] do
   begin
   if (l=x) and (r=y) then
    begin
     ti:=(ti*tagtime) mod p;
     ad:=(ad*tagtime+tagadd) mod p;
     exit;
    end;
   pushdown(k);
   mid:=(l+r)>>;
   if y<=mid then change(x,y,k<<)
   else if x>mid then change(x,y,k<<+)
   else
    begin
    change(x,mid,k<<);
    change(mid+,y,k<<+);
    end;
   sum:=f(g(k<<)+g(k<<+));
   end;
  end;
 function query(x,y,k:longint):longint;
  var mid:longint;
  begin
  with t[k] do
   begin
   pushdown(k);
   if (l=x) and (r=y) then exit(f(sum));
   mid:=(l+r)>>;
   if y<=mid then query:=f(query(x,y,k<<))
   else if x>mid then query:=f(query(x,y,k<<+))
   else query:=f(f(query(x,mid,k<<))+f(query(mid+,y,k<<+)));
   sum:=f(g(k<<)+g(k<<+));
   end;
  end;
 procedure init;
  begin
  readln(n,p);
  build(,n,);
  end;
 procedure main;
  begin
  readln(m);
  for i:= to m do
   begin
    read(ch);
    if ch= then
     begin
     readln(x,y,tagtime);
     tagadd:=;
     change(x,y,);
     end
    else if ch= then
     begin
     readln(x,y,tagadd);
     tagtime:=;
     change(x,y,);
     end
    else
     begin
     readln(x,y);
     writeln(query(x,y,));
     end;
   end;
  end;
 begin
  init;
  main;
 end.      

虽然现在已经12：30了，但我感觉，很开心。做了这道题，值！