Frequent values
Description
You are given a sequence of n integers a1 , a2 , ... , an 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 ai , ... , aj.
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 , ... , an (-100000 ≤ ai ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: ai ≤ ai+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.(多组数据)
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
解题思路：
可以求出相同数的的个数（可以离散），但是如果所求区间切断了连续区间怎么办，分成三段，左边一段，右边一段，中间的就可以用线段树（不用更新，RMQ要简单一点）或者RMQ算最大值。
具体代码：
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #define N 100005
5 using namespace std;
6 int a[N],lo[N],pos[N];
7 int n,m,x,y;
8 struct ii
9 {
10 int sta,end;
11 int num;
12 }aa[N];
13 int f[N][20];
14 int main()
15 {
16 freopen("poj3246.in","r",stdin);
17 freopen("poj3246.out","w",stdout);
18 //RQM 初始化，把log2(i),先打表出来
19 //f[][]是以连续区间个数，不是n个
20 while((scanf("%d%d",&n,&m))==2)
21 {
22 int i=1,q=2,p=0;
23 lo[i]=0;
24 while(i<=n)
25 {
26 i++;
27 if(i==q)
28 {
29 q*=2;
30 p++;
31 }
32 lo[i]=p;
33 }
34 int pre=N;
35 int k=0;
36 for(int i=1;i<=n;i++)//aa记录连续区间的起点中点个数
37 {
38 scanf("%d",&a[i]);
39 if(a[i]!=pre)
40 {
41 k++;
42 aa[k].sta=i;
43 aa[k].end=i;//记录k的始末
44 pre=a[i];
45 }
46 else aa[k].end=i;
47 pos[i]=k;//记录下标所对应的k
48 }
49 for(int i=1;i<=k;i++)
50 aa[i].num=aa[i].end-aa[i].sta+1;
51 for(int i=1;i<=k;i++)
52 f[i][0]=aa[i].num;
53 for(int j=1;j<=lo[n]+1;j++)
54 for(int i=1;i<=k;i++)
55 if(i+(1<<j)-1<=k)
56 f[i][j]=max(f[i][j-1],f[i+(1<<j-1)][j-1]);//分成等量两段长度为1<<j-1
57 for(int i=1;i<=m;i++)
58 {
59 scanf("%d%d",&x,&y);
60 if(pos[x]==pos[y])cout<<y-x+1<<endl;
61 else
62 {
63 int a1=aa[pos[x]].end;
64 int b1=aa[pos[y]].sta;
65 int n1=a1-x+1;
66 int n2=0;
67 int n3=y-b1+1;
68 int k1=pos[x],k2=pos[y];
69 int p=lo[k2-k1+1-2];//都是连续区间个数
70 if(k2-k1-1==0)cout<<max(n1,n3)<<endl;
71 else
72 {
73 n2=max(f[k1+1][p],f[k2-1-(1<<p)+1][p]);
74 cout<<max(max(n1,n2),n3)<<endl;
75 }
76 }
77 }
78 }
79 return 0;
80 }
总结：
1.RMQ求最值的用法
2，ＲＭＱ（线段树）的应用，不完全用，先分成三部分，中间一部分用ＲＭＱ
３,ＲＭＱ：ｆ［ｉ］［ｊ］初始化ｆ［ｉ］［０］＝ａ［ｉ］；
ｆ［ｉ］［ｊ］＝ｍａｘ（ｆ［ｉ］［ｊ－１］，ｆ［ｉ＋（１＜＜ｊ－１）］［ｊ－１］），递推，ｆｏｒ（ｊ：１－ｌｏｇ２（ｎ））ｆｏｒ（ｉ：１－ｎ）［ｊ是外层循环，ｉ是内层，先推两个的最大值，然后再是四个，八个］
４．看清题目，是多组数据