1085. Perfect Sequence (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CAO, Peng

Given a sequence of positive integers and another positive integer p. The sequence is said to be a "perfect sequence" if M <= m * p where M and m are the maximum and minimum numbers in the sequence, respectively.

Now given a sequence and a parameter p, you are supposed to find from the sequence as many numbers as possible to form a perfect subsequence.

Input Specification:

Each input file contains one test case. For each case, the first line contains two positive integers N and p, where N (<= 10⁵) is the number of integers in the sequence, and p (<= 10⁹) is the parameter. In the second line there are N positive integers, each is no greater than 10⁹.

Output Specification:

For each test case, print in one line the maximum number of integers that can be chosen to form a perfect subsequence.

Sample Input:

10 82 3 20 4 5 1 6 7 8 9

Sample Output:

8

得先看明白题意，minimum number是可以变化的，maximum  number也可以随着minimum number的增大而增大，找到满足maximum number < minimum number * p所跨的线性表最大长度。

#include<iostream>#include<vector>#include<algorithm>using namespace std;int main(){int i,N,p;int num;vector<long int> s;//要考虑N的范围cin>>N>>p;for(i=0;i<N;i++){cin>>num;s.push_back(num);}sort(s.begin(),s.end());//排序int begin=0,last=0,maximun=0;while(last<N){while(last<N && s[last]<=s[begin]*p){last++;}if(maximun<last-begin)//最大长度记录及更新maximun=last-begin;begin++;//最小数的变化}cout<<maximun<<endl;return 0;}