PAT甲级——A1085 Perfect Sequence

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 (≤) is the number of integers in the sequence, and p (≤) is the parameter. In the second line there are N positive integers, each is no greater than 1.

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 8

2 3 20 4 5 1 6 7 8 9

Sample Output:

8

又是没看清题，这道题的子序列不需要是原来的连续子序列，只要求是原来里面的值就行，搞得又浪费了很多时间！！！！

 //靠，不需要是子排序，就是找数字就行

 #include <iostream>

 #include <deque>

 #include <vector>

 #include <algorithm>

 using namespace std;

 int N;

 long long P;

 int main()

 {

     cin >> N >> P;

     vector<int>num(N);

     for (int i = ; i < N; ++i)

         cin >> num[i];

     sort(num.begin(), num.end());

     int res = ;

     for(int L=,R=;L<=R && R<N;++R)

     {

         while (L <= R && num[R] > P * num[L])

             L++;

         res = res > R - L +  ? res : R - L + ;

     }

     cout << res << endl;

     return ;

 }

巴特西

PAT甲级——A1085 Perfect Sequence

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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