bzoj2527

首先环可以变成链来处理，对于l>r的情况就是修改区间[1,r],[l,mx]
然后不难想到整体二分，二分答案k,然后算1~k场流星雨对国家的贡献
然后判定将国家划分变成子问题解决，没什么难的
终于不是tle，poi良心了一把

 type way=record
        po,next:longint;
      end;
      que=record
        p,n:longint;
      end;
      an=record
        l,r,v:longint;
      end;
 var a:array[..] of an;
     qq,q:array[..] of que;
     e:array[..] of way;
     c:array[..] of int64;
     p,ans,h:array[..] of longint;
     v:array[..] of boolean;
     tot,t,j,n,m,x,i:longint;
     s:int64;

 function lowbit(x:longint):longint;
   begin
     exit(x and (-x));
   end;

 procedure add(x,y:longint);
   begin
     e[i].po:=i;
     e[i].next:=p[x];
     p[x]:=i;
   end;

 procedure ins(x:longint;w:int64);
   begin
     while x<=n do
     begin
       if not v[x] then  //清理标记
       begin
         inc(tot);
         h[tot]:=x;
         v[x]:=true;
       end;
       c[x]:=c[x]+w;
       x:=x+lowbit(x);
     end;
   end;

 function ask(x:longint):int64;
   begin
     ask:=;
     while x> do
     begin
       ask:=ask+c[x];
       x:=x-lowbit(x);
     end;
   end;

 procedure work(f,t,l,r:longint);
   var mid,l1,l2:longint;
   begin
     if f>t then exit;
     if l>r then exit;
     mid:=(l+r) shr ;
     tot:=;
     for i:=l to mid do
       if a[i].l<=a[i].r then
       begin
         ins(a[i].l,a[i].v);
         ins(a[i].r+,-a[i].v);
       end
       else begin
         ins(,a[i].v);
         ins(a[i].r+,-a[i].v);
         ins(a[i].l,a[i].v);
       end;

     l1:=f;
     l2:=t;
     for i:=f to t do
     begin
       j:=p[q[i].p];
       s:=;
       while j<> do
       begin
         s:=s+ask(e[j].po);
         if s>=q[i].n then
         begin
           qq[l1]:=q[i];
           inc(l1);
           ans[q[i].p]:=mid;
           break;
         end;
         j:=e[j].next;
       end;
       if s<q[i].n then
       begin
         q[i].n:=q[i].n-s;  //对于还不够的国家，直接把这部分贡献减去即可，下次直接处理mid之后的流星雨的贡献
         qq[l2]:=q[i];
         dec(l2);
       end;
     end;
     for i:= to tot do
     begin
       c[h[i]]:=;
       v[h[i]]:=false;
     end;
     for i:=f to t do
       q[i]:=qq[i];
     work(f,l1-,l,mid-);
     work(l2+,t,mid+,r);
   end;

 begin
   readln(m,n);
   for i:= to n do
   begin
     read(x);
     add(x,i);
   end;
   for i:= to m do
   begin
     read(q[i].n);
     q[i].p:=i;
   end;
   readln(t);
   for i:= to t do
     readln(a[i].l,a[i].r,a[i].v);
   work(,m,,t);
   for i:= to m do
     if ans[i]= then writeln('NIE')
     else writeln(ans[i]);
 end.