H - Frequent values


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Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u


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bjtu_lyc (2011-08-08) System Crawler (2013-08-20)



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.



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



思路：计算区间的相同个数，然后定个左右边界点，外加一个简单的RMQ。




#include<iostream>
#include<cstring>
#include<cstdio>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
#define ll(x) (1<<x)
using namespace std;
const int mm=100000+9;
int N,Q;
int f[mm],L[mm],R[mm];
int rmq[mm][33],bit[mm];
void initRMQ()
{
  bit[0]=-1;
  FOR(i,1,mm-1)bit[i]=(i&(i-1))==0?bit[i-1]+1:bit[i-1];
  FOR(i,1,N)rmq[i][0]=f[i];
  FOR(i,1,bit[N])
  for(int j=1;j+ll(i)-1<=N;++j)
    rmq[j][i]=max(rmq[j][i-1],rmq[j+ll(i-1)][i-1]);
}
int RMQ(int l,int r)
{ if(l>r)return 0;
  int t=bit[r-l+1];
  r-=ll(t)-1;
  return max(rmq[l][t],rmq[r][t]);
}
int main()
{ int a,b;
  while(scanf("%d",&N)&&N)
  { scanf("%d",&Q);
    f[0]=mm;f[N+1]=mm;
    FOR(i,1,N)
    {
      scanf("%d",&f[i]);
    }
    int z=0,zs=f[1],kai=1;
    FOR(i,1,N+1)
    {
      if(f[i]!=zs)
        {
          for(int j=i-1;f[j]==zs;--j)
            f[j]=z,L[j]=kai,R[j]=i-1;
          kai=i;z=1;zs=f[i];
        }
       else ++z;
    }
    initRMQ();
    while(Q--)
    {
     scanf("%d%d",&a,&b);
     int ans;
     if(R[a]>L[b])ans=b-a+1;
     else
     ans=max(R[a]-a+1,b-L[b]+1);
     ans=max(ans,RMQ(R[a]+1,L[b]-1));
     printf("%d\n",ans);
    }
  }
}