1085. Perfect Sequence (25)-PAT甲级真题

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 <algorithm>using namespace std;int main() {    int n;    long long p;    cin >> n >> p;    if (n == 0) {        cout << n;        return 0;    }    long long int *a = new long long int [n];    for (int i = 0; i < n; i++)        cin >> a[i];    sort(a, a+n);    int result = 1;    int temp = 1;    for (int i = 0; i <= n - 2; i++) {        for (int j = i + result; j <= n - 1; j++) {//因为是为了找最大的result，所以下一次j只要从i的result个后面开始找就行了            if (a[j] <= a[i] * p) {                temp = j - i + 1;//计算有多少个数满足                if (temp > result) { //如果比result中存储的大，就更新result的值                    result = temp;                }            } else {                break;            }        }    }    cout << result;        return 0;}