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 10 102 31 105 100

Sample Output

143

Source

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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

1 #include
      
        2 #include
       
         3 #include
        
          4 #define MAX 100010 5 using namespace std; 6 int f[MAX][20]; 7 int num[MAX]; 8 int sum[MAX]; 9 int n,Q,ans,a,b;10 int _max(int a,int b)11 {12     return a>b?a:b;13 }14 void ST()15 {16     for(int i=1;i<=n;i++)17         f[i][0]=sum[i];18     int k=log((double)(n+1))/log(2.0);19     for(int i=1;i<=k;i++)20       for(int j=1;j+(1<
         
          <=n+1;j++)//n+121         f[j][i]=_max(f[j][i-1],f[j+(1<<(i-1))][i-1]);    22 }23 int RMQ_Query(int l,int r)24 {25     if(l>r)return 0;26     int k=log((double)(r-l+1))/log(2.0);27     return _max(f[l][k],f[r-(1<