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  • 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 Mm×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!!!!

     1 //靠,不需要是子排序,就是找数字就行
     2 #include <iostream>
     3 #include <deque>
     4 #include <vector>
     5 #include <algorithm>
     6 using namespace std;
     7 int N;
     8 long long P;
     9 int main()
    10 {
    11     cin >> N >> P;
    12     vector<int>num(N);
    13     for (int i = 0; i < N; ++i)
    14         cin >> num[i];
    15     sort(num.begin(), num.end());
    16     int res = 0;
    17     for(int L=0,R=0;L<=R && R<N;++R)
    18     {
    19         while (L <= R && num[R] > P * num[L])
    20             L++;
    21         res = res > R - L + 1 ? res : R - L + 1;        
    22     }
    23     cout << res << endl;
    24     return 0;
    25 }
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  • 原文地址:https://www.cnblogs.com/zzw1024/p/11337278.html
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