Frequent values
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 15134 | Accepted: 5519 |
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
题意:给出一个长度为n的不下降序列,然后m次查询,每次输入两个数l,r,输出区间[l,r]中连续的数中最长的数目是多少。
思路:将输入的不下降序列进行预处理,找出其中的相邻的数字的相邻的数目,存储在p数组中,然后作为RMQ算法中dp的初值,每次输入l,r时,输出该区间的最大值就可以了。
ps:不过之中有一个需要注意的地方,就是每次询问[l,r]时,要先将这个区间分成两个部分,前半部分求出与num[pt] == num[l]的个数,直到num[pt]!=num[l],当然pt<=r然后对[pt,r]进行RMQ求解,这样做的原因是排除在l之前还有数与num[l]相等。这样的话可能会影响结果。然后将RMQ求解得到的结果与pt-l的值进行取最大值就可以了。
#include<iostream> #include<algorithm> #include<stdio.h> #include<stdlib.h> #include<string.h> #include<math.h> using namespace std; const int N = 200010; int maxx[N][20]; int num[N]; int p[N]; int n,m; void play() { int l = floor(log10(double(n))/log10(double(2))); for(int j=1; j<=l; j++) { for(int i=1; i<=n+1-(1<<j); i++) { maxx[i][j] = max(maxx[i][j-1],maxx[i+(1<<(j-1))][j-1]); } } } int RMQ(int xx,int yy) { if(xx>yy) { return 0; } int pp = floor(log10(double(yy-xx))/log10(double(2))); return max(maxx[xx][pp],maxx[yy-(1<<pp)+1][pp]); } int main() { while(scanf("%d",&n)!=EOF) { if(n == 0) { break; } scanf("%d",&m); for(int i=1; i<=n; i++) { scanf("%d",&num[i]); if(i == 1) { p[i] = 1; } else { if(num[i] == num[i-1]) { p[i] = p[i-1] + 1; } else { p[i] = 1; } } } for(int i=1; i<=n; i++) { maxx[i][0] = p[i]; } play(); int x,y; while(m--) { scanf("%d%d",&x,&y); int pt = x; while(pt<=y && num[pt] == num[pt-1]) { pt++; } int ans = RMQ(pt,y); ans = max(ans,pt-x); printf("%d\n",ans); } } return 0; }
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