poj3368 Frequent values（线段树）

Description

You are given a sequence of n integers a₁ , a₂ , ... , a_n in non-decreasing order. In addition to that, you are given several queries consisting of indices i and j (1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integers a_i , ... , a_j.

Input

The input consists of several test cases. Each test case starts with a line containing two integers n and q (1 ≤ n, q ≤ 100000). The next line contains n integers a₁ , ... , a_n (-100000 ≤ a_i ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: a_i ≤ a_i+1. The following q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), which indicate the boundary indices for the 
query.

The last test case is followed by a line containing a single 0.

Output

For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

Sample Input

10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0

Sample Output

1
4
3

Source

Ulm Local 2007

 

 

无脑的线段树。写着玩玩。

program rrr(input,output);
type
  treetype=record
     l,r,m,lm,rm:longint;
  end;
var
  a:array[0..400040]of treetype;
  c:array[0..100010]of longint;
  n,i,q,x,y:longint;
function max(a,b:longint):longint;
begin
   if a>b then exit(a) else exit(b);
end;
function min(a,b:longint):longint;
begin
   if a<b then exit(a) else exit(b);
end;
procedure build(k,l,r:longint);
var
  mid,i:longint;
begin
   a[k].l:=l;a[k].r:=r;
   if l=r then begin a[k].m:=1;a[k].lm:=1;a[k].rm:=1;exit; end;
   mid:=(l+r)>>1;i:=k+k;
   build(i,l,mid);build(i+1,mid+1,r);
   a[k].m:=max(a[i].m,a[i+1].m);if c[mid]=c[mid+1] then a[k].m:=max(a[k].m,a[i].rm+a[i+1].lm);
   if c[l]=c[mid+1] then a[k].lm:=a[i].lm+a[i+1].lm else a[k].lm:=a[i].lm;
   if c[mid]=c[r] then a[k].rm:=a[i].rm+a[i+1].rm else a[k].rm:=a[i+1].rm;
end;
function query(k:longint):longint;
var
  mid,i,ans:longint;
begin
    if (x<=a[k].l) and (a[k].r<=y) then exit(a[k].m);
    mid:=(a[k].l+a[k].r)>>1;i:=k+k;
    ans:=0;
    if x<=mid then ans:=query(i);
    if mid<y then ans:=max(ans,query(i+1));
    if (x<=mid) and (mid<y) and (c[mid]=c[mid+1]) then ans:=max(ans,min(mid-x+1,a[i].rm)+min(y-mid,a[i+1].lm));
    exit(ans);
end;
begin
   assign(input,'r.in');assign(output,'r.out');reset(input);rewrite(output);
   while true do
      begin
         read(n);if n=0 then break;
         readln(q);
         for i:=1 to n do read(c[i]);
         build(1,1,n);
         for i:=1 to q do begin readln(x,y);writeln(query(1)); end;
      end;
   close(input);close(output);
end.