Frequent values
You are given a sequence of n integers a 1 , a 2 , ... , 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 .
InputThe 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 a 1 , ... , 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.
OutputFor 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
Hint
A naive algorithm may not run in time! 题意:查询区间[l,r]中众数的个数. 因为数列是非递减的，所以可以转化为ST表.
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define N 100005
using namespace std;
int maxn[N][64],a[N],pre[N];
int n,q,i,j;
int main()
{
while(scanf("%d",&n)!=EOF&&n)
{
scanf("%d",&q);
memset(maxn,0,sizeof maxn);
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(i==1)
{
pre[i]=1;
continue;
}
if(a[i]==a[i-1])
pre[i]=pre[i-1]+1;
else
pre[i]=1;
}
for(i=1;i<=n;i++)
maxn[i][0]=pre[i];
int t=log2(n);
for(j=1;j<=t;j++)
{
for(i=1;i+(1<<j)-1<=n;i++)
{
maxn[i][j]=max(maxn[i][j-1],maxn[i+(1<<(j-1))][j-1]);
}
}
while(q--)
{
int MAX;
int l,r;
scanf("%d%d",&l,&r);
int m=l;
while(m<=r&&a[m]==a[m-1])//区间第一个数字要特判.
m++;
if(r<m)
MAX=0;
else
{
int k=log(r-m+1)/log(2);
MAX=max(maxn[m][k],maxn[r-(1<<k)+1][k]);
}
MAX=max(MAX,m-l);
printf("%d\n",MAX);
}
}
}
原文地址：https://www.cnblogs.com/switch-waht/p/11396907.html