1085 Perfect Sequence(排序)

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 (≤10^​5​​ ) 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 10^​9​​ .

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

题目大意

给定一个序列A和参数p，若序列A中最大值为M，最小值为m，满足M<=m*p，则称序列A为perfect sequence；依据这个定义，题目要求在给定的序列中找出尽可能多的元素组成一个新的序列，使得这个序列为perfect sequence。

分析

寻找序列极值问题，若对算法的复杂度无苛刻要求，可先对序列排序，定义当前最大宽度maxwidth，遍历序列，遍历过程中以当前maxwidth为起始宽度寻找比maxwidth更大宽度的序列。另外，题目中序列元素和参数p的范围均为0~10⁹,两数想成数的范围将可能超过10⁹，因此将数的类型定义为long int

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> v;
int main(){
    long long n,p;
    cin>>n>>p;
    v.resize(n);
    for(int i=0;i<n;i++) cin>>v[i];
    sort(v.begin(), v.end());
    int maxwidth=1;
    for(int i=0;i<n;i++){
        while(i+maxwidth-1<n && v[i+maxwidth-1]<=p*v[i]) maxwidth++;
    }
    cout<<maxwidth-1;
}