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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 43
也许看完题你可能想到了线段树做法，但是这道题的确可以用RMQ来写。
具体就是查询区间中重复最多的次数，但是RMQ只能查询区间最大值，我们
初始化的时候也只能初始化连续的那一部分，如果区间中不连续了了。
所以我们要在查询上下功夫。从查询左端点起，找到不连续的点pos,pos
以后的就可以RMQ查询，具体答案就是max(pos-l,RMQ(POS,l))
#include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<string> #include<iostream> #include<queue> #include<cmath> #include<map> #include<stack> #include<bitset> using namespace std; #define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i ) #define REP( i , n ) for ( int i = 0 ; i < n ; ++ i ) #define CLEAR( a , x ) memset ( a , x , sizeof a ) typedef long long LL; typedef pair<int,int>pil; const int INF = 0x3f3f3f3f; const int maxn=1e5+100; int num[maxn],sum[maxn]; int n,q; int dp[maxn][20]; void init() { REPF(i,1,n) dp[i][0]=sum[i]; for(int j=1;(1<<j)<=n;j++) for(int i=1;i+(1<<j)-1<=n;i++) dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]); } int RMQ(int l,int r) { int k=(int)(log(r-l+1)/log(2.0)); return max(dp[l][k],dp[r-(1<<k)+1][k]); } void work() { int x,y,ans; while(q--) { scanf("%d%d",&x,&y); int pos=x+1; while(num[pos]==num[pos-1]&&pos<=y) pos++; if(pos>y) ans=pos-x; else ans=max(pos-x,RMQ(pos,y)); printf("%d\n",ans); } } int main() { int x; while(~scanf("%d",&n)&&n) { scanf("%d",&q); CLEAR(sum,0);num[0]=INF; REPF(i,1,n) { scanf("%d",&num[i]); if(num[i]==num[i-1]) sum[i]=sum[i-1]+1; else sum[i]=1; } init(); work(); } return 0; }
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原文地址：http://blog.csdn.net/u013582254/article/details/44186691
