POJ 3368：Frequent values

Frequent values

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 14764 Accepted: 5361

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

题意是给出n个数，q次询问。每次询问一个区间中出现次数最多的那个数。

这题自己没有做出来，一直TLE，像当年做扩展欧几里德题目似的。看网上代码，才知道RMQ这种算法，总之是要记住了这种有动态规划思想的东西，还有这种询问这么多次的题目要离线处理啊啊啊啊，这都不懂你还搞毛啊光速小子。。。

我自己都不好意思舔这个大脸上这个代码，为了以后自己常常看看，记住它吧。

觉得这个代码写得真的好，很多细节做的真的不错(。。。。。。)：

#include <iostream>#include <cmath>#include <algorithm>#pragma warning(disable:4996) using namespace std;int num[100005];int fre[100005];int n,q;int max_v[100005][20];void RMQ(){int i,j;for(i=1;i<=n;i++){max_v[i][0]=fre[i];}int temp = log((double)n)/log(2.0)+1;for(j=1;j<=temp;j++){for(i=1;i+(1<<j)-1<=n;i++){max_v[i][j]=max(max_v[i][j-1],max_v[i+(1<<(j-1))][j-1]);}}}int cal(int h,int k){if(h>k)return 0;int temp=k-h+1;int temp2=log((double)temp)/log(2.0);return max(max_v[h][temp2],max_v[k-(1<<temp2)+1][temp2]);}int main(){//freopen("input.txt","r",stdin);//freopen("out.txt","w",stdout);while(scanf("%d",&n)){int i,h,k;if(n==0)break;scanf("%d",&q);memset(fre,0,sizeof(fre));memset(max_v,0,sizeof(max_v));for(i=1;i<=n;i++){scanf("%d",&num[i]);if(i==1){fre[1]=1;}else{if(num[i]==num[i-1])fre[i]=fre[i-1]+1;elsefre[i]=1;}}RMQ();for(i=1;i<=q;i++){scanf("%d%d",&h,&k);int temp;int t=0;while((h+t<=k)&&(num[h+t]==num[h])){t++;}temp=h+t;int temp2=cal(temp,k);cout<<max(t,temp2)<<endl;}}    return 0;}