1085 Perfect Sequence（25 分）

2024-05-13 14:51:44

1085 Perfect Sequence（25 分）

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:

转载来自大佬：https://www.liuchuo.net/archives/1908

好吧，是我脑子犯浑了，怎么说，题目看不懂然后是求>=maxn 我想成小于了难怪一直看不懂题解

代码：

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main() {int n;long long p;scanf("%d%lld", &n, &p);vector<int> v(n);for (int i = 0; i < n; i++)cin >> v[i];sort(v.begin(), v.end());int result = 0, temp = 0;for (int i = 0; i < n; i++) {for (int j = i + result; j < n; j++) {if (v[j] <= v[i] * p) {temp = j - i + 1;if (temp > result)result = temp;}else {break;}}}cout << result;return 0;
}

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