2821: 作诗(Poetize)

2821: 作诗(Poetize)

Time Limit: 50 Sec  Memory Limit: 128 MB
Submit: 1078  Solved: 348
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Description

神犇SJY虐完HEOI之后给傻×LYD出了一题：
SHY是T国的公主，平时的一大爱好是作诗。
由于时间紧迫，SHY作完诗之后还要虐OI，于是SHY找来一篇长度为N的文章，阅读M次，每次只阅读其中连续的一段[l,r]，从这一段中选出一些汉字构成诗。因为SHY喜欢对偶，所以SHY规定最后选出的每个汉字都必须在[l,r]里出现了正偶数次。而且SHY认为选出的汉字的种类数（两个一样的汉字称为同一种）越多越好（为了拿到更多的素材！）。于是SHY请LYD安排选法。
LYD这种傻×当然不会了，于是向你请教……
问题简述：N个数，M组询问，每次问[l,r]中有多少个数出现正偶数次。

Input

输入第一行三个整数n、c以及m。表示文章字数、汉字的种类数、要选择M次。
第二行有n个整数，每个数Ai在[1, c]间，代表一个编码为Ai的汉字。
接下来m行每行两个整数l和r，设上一个询问的答案为ans(第一个询问时ans=0)，令L=(l+ans)mod n+1, R=(r+ans)mod n+1，若L>R，交换L和R，则本次询问为[L,R]。

Output

输出共m行，每行一个整数，第i个数表示SHY第i次能选出的汉字的最多种类数。

Sample Input

5 3 5
1 2 2 3 1
0 4
1 2
2 2
2 3
3 5

Sample Output

2
0
0
0
1

HINT

对于100%的数据，1<=n,c,m<=10^5

Source

By lydrainbowcat

挖个坑，以后来填。。。

本来用这题来练分块的，结果完全被二分坑到了QAQ。。。TLE...哪天有空A了这题

我的代码：

1.二分自己yy的

{$inline on}
const maxn=100000+100;
type node=record
     p,v:longint;
     end;
var b:array[0..maxn] of node;
    a,belong,v,head,tail:array[0..maxn] of longint;
    mark:array[0..maxn] of boolean;
    f:array[0..400,0..400] of longint;
    l,r:array[0..400] of longint;
    i,j,n,m,tot,ans,ll,rr,block,cnt,x:longint;
function max(x,y:longint):longint;inline;
 begin
 if x>y then exit(x) else exit(y);
 end;
procedure swap(var x,y:longint);inline;
 var t:longint;
 begin
 t:=x;x:=y;y:=t;
 end;

function cmp(x,y:node):boolean;inline;
 begin
  if x.v=y.v then exit(x.p<y.p) else exit(x.v<y.v);
 end;
procedure sort(l,r:longint);inline;
 var i,j:longint;x,y:node;
 begin
   i:=l;j:=r;x:=b[(i+j)>>1];
   repeat
     while cmp(b[i],x) do inc(i);
     while cmp(x,b[j]) do dec(j);
     if i<=j then
      begin
        y:=b[i];b[i]:=b[j];b[j]:=y;
        inc(i);dec(j);
      end;
   until i>j;
   if i<r then sort(i,r);
   if j>l then sort(l,j);
 end;
procedure init;
 begin
   readln(n,m,m);
   for i:=1 to n do
    begin
      read(a[i]);
      b[i].p:=i;b[i].v:=a[i];
    end;
   readln;
   sort(1,n);
   block:=trunc(sqrt(n)*ln(n)/ln(2));
   cnt:=(n-1) div block+1;
   for i:=1 to n do belong[i]:=(i-1) div block+1;
   for i:=1 to cnt do
    begin
      l[i]:=(i-1)*block+1;r[i]:=i*block;
    end;
   r[cnt]:=n;
   for i:=1 to n do
    begin
     x:=b[i].v;
     if head[x]=0 then head[x]:=i;
     tail[x]:=i;
    end;
 end;
procedure prepare;
 begin
  for i:=1 to cnt do
   begin
     fillchar(v,sizeof(v),0);
     tot:=0;
     for j:=l[i] to r[cnt] do
      begin
       inc(v[a[j]]);
       if (v[a[j]] and 1=1) and (v[a[j]]<>1) then dec(tot)
       else if v[a[j]] and 1=0 then inc(tot);
       f[i,belong[j]]:=tot;
      end;
   end;
 end;
function findup(z,x:longint):longint;inline;
 var l,r,mid:longint;
 begin
  if x>=b[tail[z]].p then exit(tail[z]);
  l:=head[z];r:=tail[z];
  while l<r do
   begin
    mid:=(l+r)>>1;
    if b[mid].p<x then l:=mid+1
    else if b[mid].p>x then r:=mid
    else exit(mid);
   end;
  exit(l-1);
 end;
function finddown(z,x:longint):longint;inline;
 var l,r,mid:longint;
 begin
  l:=head[z];r:=tail[z];
  while l<r do
   begin
    mid:=(l+r)>>1;
    if b[mid].p<x then l:=mid+1
    else if b[mid].p>x then r:=mid
    else exit(mid);
   end;
  exit(l);
 end;

function find(z,x,y:longint):longint;inline;
 begin
  find:=max(findup(z,y)-finddown(z,x)+1,0);
 end;
procedure query(x,y:longint);inline;
 var i,bx,by,t1,t2,z:longint;
 begin
 ans:=0;
 bx:=belong[x];by:=belong[y];
 if by-bx<=1 then
   begin
     fillchar(v,sizeof(v),0);
     for i:=x to y do
      begin
       inc(v[a[i]]);
       if (v[a[i]] and 1=1) and (v[a[i]]<>1) then dec(ans)
       else if v[a[i]] and 1=0 then inc(ans);
      end;
   end
 else
   begin
     fillchar(mark,sizeof(mark),false);
     ans:=f[bx+1,by-1];
     for i:=x to r[bx] do
      begin
       z:=a[i];if mark[z] then continue;mark[z]:=true;
       t1:=find(z,x,y);t2:=find(z,l[bx+1],r[by-1]);
       if (t1 and 1=1) and (t2 and 1=0) and (t2<>0) then dec(ans)
       else if (t1 and 1=0) and ((t2 and 1=1) or (t2=0)) then inc(ans);
      end;
     for i:=l[by] to y do
      begin
       z:=a[i];if mark[z] then continue;mark[z]:=true;
       t1:=find(z,x,y);t2:=find(z,l[bx+1],r[by-1]);
       if (t1 and 1=1) and (t2 and 1=0) and (t2<>0) then dec(ans)
       else if (t1 and 1=0) and ((t2 and 1=1) or (t2=0)) then inc(ans);
      end;
   end;
 end;
procedure main;
 begin
   ans:=0;
   for i:=1 to m do
    begin
      readln(ll,rr);
      ll:=(ll+ans) mod n+1;rr:=(rr+ans) mod n+1;
      if ll>rr then swap(ll,rr);
      query(ll,rr);
      writeln(ans);
    end;
 end;

begin
  assign(input,'input.txt');assign(output,'output.txt');
  reset(input);rewrite(output);
  init;
  prepare;
  main;
  close(input);close(output);
end.    

2.二分摘抄自hzwer

{$inline on}
const maxn=100000+100;
type node=record
     p,v:longint;
     end;
var b:array[0..maxn] of node;
    a,belong,v,head,tail:array[0..maxn] of longint;
    mark:array[0..maxn] of boolean;
    f:array[0..400,0..400] of longint;
    l,r:array[0..400] of longint;
    i,j,n,m,tot,ans,ll,rr,block,cnt,x:longint;
function max(x,y:longint):longint;inline;
 begin
 if x>y then exit(x) else exit(y);
 end;
procedure swap(var x,y:longint);inline;
 var t:longint;
 begin
 t:=x;x:=y;y:=t;
 end;

function cmp(x,y:node):boolean;inline;
 begin
  if x.v=y.v then exit(x.p<y.p) else exit(x.v<y.v);
 end;
procedure sort(l,r:longint);inline;
 var i,j:longint;x,y:node;
 begin
   i:=l;j:=r;x:=b[(i+j)>>1];
   repeat
     while cmp(b[i],x) do inc(i);
     while cmp(x,b[j]) do dec(j);
     if i<=j then
      begin
        y:=b[i];b[i]:=b[j];b[j]:=y;
        inc(i);dec(j);
      end;
   until i>j;
   if i<r then sort(i,r);
   if j>l then sort(l,j);
 end;
procedure init;
 begin
   readln(n,m,m);
   for i:=1 to n do
    begin
      read(a[i]);
      b[i].p:=i;b[i].v:=a[i];
    end;
   readln;
   sort(1,n);
   block:=trunc(sqrt(n)*ln(n)/ln(2));
   cnt:=(n-1) div block+1;
   for i:=1 to n do belong[i]:=(i-1) div block+1;
   for i:=1 to cnt do
    begin
      l[i]:=(i-1)*block+1;r[i]:=i*block;
    end;
   r[cnt]:=n;
   for i:=1 to n do
    begin
     x:=b[i].v;
     if head[x]=0 then head[x]:=i;
     tail[x]:=i;
    end;
 end;
procedure prepare;
 begin
  for i:=1 to cnt do
   begin
     fillchar(v,sizeof(v),0);
     tot:=0;
     for j:=l[i] to r[cnt] do
      begin
       inc(v[a[j]]);
       if (v[a[j]] and 1=1) and (v[a[j]]<>1) then dec(tot)
       else if v[a[j]] and 1=0 then inc(tot);
       f[i,belong[j]]:=tot;
      end;
   end;
 end;
function findup(z,x:longint):longint;inline;
 var l,r,mid,tmp:longint;
 begin
  l:=head[z];r:=tail[z];tmp:=0;
  while l<=r do
   begin
    mid:=(l+r)>>1;
    if b[mid].p>x then l:=mid+1
    else begin tmp:=mid;r:=mid-1;end;
   end;
  exit(tmp);
 end;
function finddown(z,x:longint):longint;inline;
 var l,r,mid,tmp:longint;
 begin
  l:=head[z];r:=tail[z];tmp:=maxlongint;
  while l<=r do
   begin
    mid:=(l+r)>>1;
    if b[mid].p<x then l:=mid+1
    else begin tmp:=mid;r:=mid-1;end;
   end;
  exit(tmp);
 end;

function find(z,x,y:longint):longint;inline;
 begin
  find:=max(findup(z,y)-finddown(z,x)+1,0);
 end;
procedure query(x,y:longint);inline;
 var i,bx,by,t1,t2,z:longint;
 begin
 ans:=0;
 bx:=belong[x];by:=belong[y];
 if by-bx<=1 then
   begin
     fillchar(v,sizeof(v),0);
     for i:=x to y do
      begin
       inc(v[a[i]]);
       if (v[a[i]] and 1=1) and (v[a[i]]<>1) then dec(ans)
       else if v[a[i]] and 1=0 then inc(ans);
      end;
   end
 else
   begin
     fillchar(mark,sizeof(mark),false);
     ans:=f[bx+1,by-1];
     for i:=x to r[bx] do
      begin
       z:=a[i];if mark[z] then continue;mark[z]:=true;
       t1:=find(z,x,y);t2:=find(z,l[bx+1],r[by-1]);
       if (t1 and 1=1) and (t2 and 1=0) and (t2<>0) then dec(ans)
       else if (t1 and 1=0) and ((t2 and 1=1) or (t2=0)) then inc(ans);
      end;
     for i:=l[by] to y do
      begin
       z:=a[i];if mark[z] then continue;mark[z]:=true;
       t1:=find(z,x,y);t2:=find(z,l[bx+1],r[by-1]);
       if (t1 and 1=1) and (t2 and 1=0) and (t2<>0) then dec(ans)
       else if (t1 and 1=0) and ((t2 and 1=1) or (t2=0)) then inc(ans);
      end;
   end;
 end;
procedure main;
 begin
   ans:=0;
   for i:=1 to m do
    begin
      readln(ll,rr);
      ll:=(ll+ans) mod n+1;rr:=(rr+ans) mod n+1;
      if ll>rr then swap(ll,rr);
      query(ll,rr);
      writeln(ans);
    end;
 end;

begin
  assign(input,'input.txt');assign(output,'output.txt');
  reset(input);rewrite(output);
  init;
  prepare;
  main;
  close(input);close(output);
end.