PAT甲级1085

Posted 2022-08-30 wx630c98f24f6b8

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 8 2 3 20 4 5 1 6 7 8 9

Sample Output:

8

#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
int main()

  int N;
  long long p;
  cin >> N >> p;
  vector<long long> v;
  long long t;
  for (int i = 0; i < N; i++)
  
    cin >> t;
    v.push_back(t);
  
  sort(v.begin(), v.end());
  int len = 0;
  for (int i = 0; i < N; i++)
  
    for (int j = len+i; j < N; j++)//利用上次计算的信息
    
      if (v[i]*p < v[j])
        break;//该中断时中断，后续比较无用
      if (len < j - i + 1)
      
        len = j - i + 1;
      
    
  
  cout << len;
  return 0;