[POJ3368 Frequent values]

[题目来源]：Ulm Local 2007

[关键字]：RMQ

[题目大意]：给定一个非递减的整数序列（n），然后给出m个查询区间，针对每一个查询，输出此区间内重复最多的数字重复的次数。

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[分析]：如果给定的区间恰好是连续的区间的两端则可直接同过RMQ问题求解。但如果不是这样则需要先求出两遍的最大值，再求中间就转换成了之前所说的问题。

[代码]：

View Code

 1 var
 2   n, q: longint;
 3   a, b, c: array[0..100100] of longint;
 4   f: array[0..100100,0..40] of longint;
 5 
 6 function max(x, y: longint):longint;
 7 begin
 8   if x > y then exit(x) else exit(y);
 9 end;
10 
11 procedure st;
12 var
13   t, i, j: longint;
14 begin
15   for i := 1 to n do f[i,0] := a[i];
16   t := trunc(ln(n)/ln(2));
17   for j := 1 to t do
18     for i := 1 to n+1-1 shl j do
19        f[i,j] := max(f[i,j-1],f[i+1 shl (j-1),j-1])
20 end;
21 
22 procedure init;
23 var
24   i: longint;
25 begin
26   read(n);
27   if n = 0 then halt;
28   readln(q);
29   for i := 1 to n do
30     read(c[i]);
31   a[1] := 1;
32   for i := 2 to n do
33     if c[i] = c[i-1] then a[i] := a[i-1]+1 else a[i] := 1;
34   b[n] := a[n];
35   for i := n-1 downto 1 do
36     if c[i] = c[i+1] then b[i] := b[i+1] else b[i] := a[i];
37   fillchar(f,sizeof(f),0);
38   st;
39 end;
40 
41 function find(x, y: longint):longint;
42 var
43   ans, t: longint;
44 begin
45   ans := b[x]-a[x]+1;
46   if ans < a[y] then ans := a[y];
47   if ans > y-x+1 then exit(y-x+1);
48   
49   x := x+b[x]-a[x]+1;
50   y := y-a[y];
51   if x >= y then exit(ans);
52   
53   t := trunc(ln(y-x+1)/ln(2));
54   ans := max(max(f[x,t],f[y-1 shl t+1,t]),ans);
55   
56   exit(ans);
57 end;
58 
59 procedure work;
60 var
61   i, ans, x ,y: longint;
62 begin
63   for i := 1 to q do
64     begin
65       readln(x,y);
66       ans := find(x,y);
67       writeln(ans);
68     end;
69 end;
70 
71 begin
72   while 1 = 1 do
73     begin
74       init;
75       work;
76     end;
77 end.