poj3368（RMQ——ST）

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16543		Accepted: 5985

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

---

分析

因为题上说是有序的，就可以求一样的数的数量，再来使用RMQ

开始我也不会做，看了别人的写的，思路还放不开

还记得以前求众数只会用桶排序，现在想起来实在可笑，但是路还很长

AC代码

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #define MAX 100010
 using namespace std;
 int f[MAX][];
 int num[MAX];
 int sum[MAX];
 int n,Q,ans,a,b;
 int _max(int a,int b)
 {
     return a>b?a:b;
 }
 void ST()
 {
     for(int i=;i<=n;i++)
         f[i][]=sum[i];
     int k=log((double)(n+))/log(2.0);
     for(int i=;i<=k;i++)
       for(int j=;j+(<<i)<=n+;j++)//n+1
         f[j][i]=_max(f[j][i-],f[j+(<<(i-))][i-]);
 }
 int RMQ_Query(int l,int r)
 {
     if(l>r)return ;
     int k=log((double)(r-l+))/log(2.0);
     return _max(f[l][k],f[r-(<<k)+][k]);//r-(1<<k)一定要+1
 }
 int main()
 {
     while(scanf("%d",&n)!=EOF,n)
     {
         scanf("%d",&Q);
         for(int i=;i<=n;i++)
         {
             scanf("%d",num+i);
             if(i==)
             {
                 sum[i]=;
                 continue;
             }
             if(num[i]==num[i-])
             sum[i]=sum[i-]+;
             else sum[i]=;
         }
         ST();
         while(Q--)
         {
             scanf("%d%d",&a,&b);
             if(a==b||num[a]==num[b])
             {
                 cout<<b-a+<<endl;
                 continue;
             }
             int t=a;
             while(num[t]==num[t-]&&t<=b)//t 必须小于 b
                 t++;
             int cnt=RMQ_Query(t,b);
             ans=_max(cnt,t-a);
             printf("%d\n",ans);
         }
     }
     return ;
 }