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.
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
题目大意：
给出n个数和m个询问（l,r)，对于每个询问求出(l,r)之间连续出现次数最多的次数。
思路：
RMQ算法
预处理一个数组：
if(num[i]==num[i-1])
a[i]=a[i-1]+1;
else
a[i]=1;
对于每个询问(l,r)，分为两个部分，前半部分求与l之前相同的数的个数直到t，后半部分从t开始直接用RMQ求解最大值就行了。
最后结果为max(前半部分,后半部分)。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=100010;
int n,m,num[maxn],a[maxn],f[maxn][20];
int init()
{
int f=1,p=0;char c=getchar();
while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){p=p*10+c-'0';c=getchar();}
return f*p;
}
void prepare()
{
memset(f,-1,sizeof(-1));
for(int i=1;i<=n;i++)
f[i][0]=a[i];
int k=log((double)(n+1))/log(2.0);
for(int j=1;j<=k;j++)
for(int i=1;i+(1<<j)-1<=n;i++)
f[i][j]=max(f[i][j-1],f[i+(1<<j-1)][j-1]);
}
int rmq(int l,int r)
{
if(l>r) return 0;
int k=log((double)(r-l+1))/log(2.0);
return max(f[l][k],f[r-(1<<k)+1][k]);
}
int main()
{
int l,r;
while(cin>>n&&n)
{
m=init();
for(int i=1;i<=n;i++)
{
num[i]=init();
if(i==1)
{
a[i]=1;
continue;
}
if(num[i]==num[i-1])
a[i]=a[i-1]+1;
else a[i]=1;
}
prepare();
for(int i=1;i<=m;i++)
{
l=init();r=init();
int t=l;
while(t<=r&&num[t]==num[t-1])
t++;
int ans=rmq(t,r);
ans=max(ans,t-l);
printf("%d
",ans);
}
}
return 0;
}