| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 15880 | Accepted: 5778 |
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：求区间内重复次数最多的元素所重复的次数。
思路：由于数据的范围很大，所以是一道很典型的RMQ，由于数组保证非降序，所以相同元素一定排列在一起，这样我们就可以将相同元素划分在一个区间内，用一个num数组储存第i个元素在新划分后的第一个小区间，left数组储存第i个小区间的左端点，right数组储存右端点，res存第i个区间有多少个元素，这样在求解时如果发现输入的l和r值在一个区间内，答案即为r-l+1，若不在一个区间内，答案即为right[num[l]]-l+1,r-left[num[r]]+1,RMQ(num[l]+1,num[r]-1))三者中的最大值。
代码：
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#define N 100010
using namespace std;
int num[N],lef[N],rig[N],res[N],dp[N][20],cnt,n;
int maxn(int a,int b,int c)
{
return max(max(a,b),c);
}
void RMQ_init()
{
memset(dp,0,sizeof(dp));
for(int i=1; i<=cnt; i++) dp[i][0]=res[i];
for(int j=1; (1<<j)<=cnt; j++)
for(int i=1; i+(1<<j-1)<=cnt; i++)
dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int RMQ(int L,int R)
{
if(L>R)
return 0;
int k=0;
while((1<<(k+1))<=R-L+1) k++;
return max(dp[L][k],dp[R-(1<<k)+1][k]);
}
int main()
{
int q,temp,before;
while(~scanf("%d",&n)&&n)
{
scanf("%d",&q);
cnt=0;
memset(num,0,sizeof(num));
memset(lef,0,sizeof(lef));
memset(rig,0,sizeof(rig));
memset(res,0,sizeof(res));
for(int i=1; i<=n; i++)
{
scanf("%d",&temp);
if(i==1)
{
++cnt;
lef[cnt]=1;
before=temp;
}
if(temp==before)
{
num[i]=cnt;
res[cnt]++;
rig[cnt]++;
}
else
{
num[i]=++cnt;
res[cnt]++;
lef[cnt]=rig[cnt]=i;
before=temp;
}
}
RMQ_init();
while(q--)
{
int l,r;
scanf("%d%d",&l,&r);
if(num[l]==num[r])
{
printf("%d
",r-l+1);
continue;
}
printf("%d
",maxn(rig[num[l]]-l+1,r-lef[num[r]]+1,RMQ(num[l]+1,num[r]-1)));
}
}
}