hdu1806——Frequent values

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 indicesi and j (1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integersa_i , ... , a_j.

Input

The input consists of several test cases. Each test case starts with a line containing two integersn and q (1 ≤ n, q ≤ 100000). The next line containsn integers a₁ , ... , a_n (-100000 ≤ a_i ≤ 100000, for eachi ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: a_i ≤ a_i+1. The followingq 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

大意：n个非递减数m条询问，输出询问区间内相同频率最高的次数

思路：创建a2数组记录出现次数：1,2,1,2,3,4,1,1,2,3，注意区间左边处理情况

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#define N 100003using namespace std;int dp[N][30],a1[N],a2[N],n;void RMQ(){    int i,j;    for(j=1; 1<<j<=n; j++)        for(i=1; i+(1<<j)-1<=n; i++)            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}int main(){    int m,i,a,b;    while(scanf("%d",&n),n)    {        scanf("%d",&m);        scanf("%d",&a1[1]);        a2[1]=1;        dp[1][0]=1;        for(i=2; i<=n; i++)        {            scanf("%d",&a1[i]);            if(a1[i]==a1[i-1])            {                a2[i]=a2[i-1]+1;                dp[i][0]=a2[i];            }            else            {                a2[i]=1;                dp[i][0]=1;            }        }        RMQ();        while(m--)        {            scanf("%d%d",&a,&b);            if(a>b)                swap(a,b);            int sum=0;            for(i=a; i<=b; i++)            {                if(a2[i]==1)                    break;                sum++;            }            if(i==b+1)                printf("%d\n",sum);            else            {                int k=log(b-i+1.0)/log(2.0);                printf("%d\n",max(sum,max(dp[i][k],dp[b-(1<<k)+1][k])));            }        }    }    return 0;}