Codeforces Round #250 D - The Child and Sequence/[TYVJ3838] DQS和序列（by 帝江&Darkfalmes）

题目大意

• 有三种操作:求和，区间取模，单点修改
• n,m<=105,a[i]<=109

官方题解

http://codeforces.com/blog/entry/12513
vfk大大的题目
简单来说就是解释了一下取模操作的复杂度可以不超过 nlog(n)
然而我不停地TTTTTTTT,好像的确比别人慢很多QAQAQ….
UPD:重写过了啦啦啦~

const
 maxn=700000;
var
 w:array[0..4*maxn,1..2]of int64;
 i,j,k:longint;
 n,m:longint;
 a,b,c,d:longint;
procedure build(a,l,r:longint);
var mid:longint;
begin
 if l=r then begin read(w[a,1]); w[a,2]:=w[a,1]; exit; end;
 mid:=(l+r)>>1;
 build(a*2,l,mid); build(a*2+1,mid+1,r);
 w[a,1]:=w[a*2,1]+w[a*2+1,1];
 if w[a*2,2]>w[a*2+1,2]
 then w[a,2]:=w[a*2,2] else w[a,2]:=w[a*2+1,2];
end;

function query(a,x,y,l,r:longint):int64;
var mid:longint;
begin
 if (x=l)and(y=r)
 then exit(w[a,1]);
 mid:=(x+y)>>1;
 if r<=mid then exit(query(a*2,x,mid,l,r)) else 
 if l>mid then exit(query(a*2+1,mid+1,y,l,r))
 else exit(query(a*2,x,mid,l,mid)+query(a*2+1,mid+1,y,mid+1,r));
end;

procedure update1(a,x,y,l,r,mmod:longint);
var mid:longint;
begin
 if w[a,2]<mmod then exit;
 if x=y then begin w[a,1]:=w[a,1] mod mmod; w[a,2]:=w[a,1]; exit; end;
 mid:=(x+y)>>1;
 if r<=mid then update1(a*2,x,mid,l,r,mmod) else 
 if l>mid then update1(a*2+1,mid+1,y,l,r,mmod)
 else begin update1(a*2,x,mid,l,mid,mmod); update1(a*2+1,mid+1,y,mid+1,r,mmod); end;
 w[a,1]:=w[a*2,1]+w[a*2+1,1];
 if w[a*2,2]>w[a*2+1,2]
 then w[a,2]:=w[a*2,2] else w[a,2]:=w[a*2+1,2];
end;

procedure update2(a,x,y,b,c:longint);
var mid:longint;
begin
 if x=y then begin w[a,1]:=c; w[a,2]:=c; exit; end;
 mid:=(x+y)>>1;
 if b<=mid then update2(a*2,x,mid,b,c)
 else update2(a*2+1,mid+1,y,b,c);
 w[a,1]:=w[a*2,1]+w[a*2+1,1];
 if w[a*2,2]>w[a*2+1,2]
 then w[a,2]:=w[a*2,2] else w[a,2]:=w[a*2+1,2];
end;

begin
 readln(n,m);
 build(1,1,n);
 for i:=1 to m do
  begin
   read(a);
   case a of
   1:begin readln(b,c); writeln(query(1,1,n,b,c)); end;
   2:begin readln(b,c,d); update1(1,1,n,b,c,d); end;
   3:begin readln(b,c); update2(1,1,n,b,c); end;
   end;
  end;
end.