POJ 3368 Frequent values(区间频繁次数 RMQ)

题目链接：http://poj.org/problem?id=3368

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

Ulm Local 2007

题意：

求区间内出现次数最多的数字出现的次数！

PS:

 先离散化处理！

代码如下：

#include <cstdio>#include <cmath>#include <algorithm>using namespace std;const int maxn = 100017;int num[maxn], f[maxn], MAX[maxn][20];int n;int max(int a,int b){    return a>b ? a:b;}int rmq_max(int l,int r){    if(l > r)        return 0;    int k = log((double)(r-l+1))/log(2.0);    return max(MAX[l][k],MAX[r-(1<<k)+1][k]);}void init(){    for(int i = 1; i <= n; i++)    {        MAX[i][0] = f[i];    }    int k = log((double)(n+1))/log(2.0);    for(int i = 1; i <= k; i++)    {        for(int j = 1; j+(1<<i)-1 <= n; j++)        {            MAX[j][i] = max(MAX[j][i-1],MAX[j+(1<<(i-1))][i-1]);        }    }}int main(){    int a, b, q;    while(scanf("%d",&n) && n)    {        scanf("%d",&q);        for(int i = 1; i <= n; i++)        {            scanf("%d",&num[i]);        }        sort(num+1,num+n+1);        for(int i = 1; i <= n; i++)        {            if(i == 1)            {                f[i] = 1;                continue;            }            if(num[i] == num[i-1])            {                f[i] = f[i-1]+1;            }            else            {                f[i] = 1;            }        }        init();        for(int i = 1; i <= q; i++)        {            scanf("%d%d",&a,&b);            int t = a;            while(t<=b && num[t]==num[t-1])            {                t++;            }            int cnt = rmq_max(t,b);            int ans = max(t-a,cnt);            printf("%d\n",ans);        }    }    return 0;}/*10 3-1 -1 1 2 1 1 1 10 10 102 31 105 10*/