POJ 3368 Frequent values(RMQ 求区间出现最多次数的数字的次数)
id=3368
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
Source
PS:
RMQ介绍+模板：http://blog.csdn.net/u012860063/article/details/40752197
代码例如以下：
#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 10
2 3
1 10
5 10
*/
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