1085 Perfect Sequence （25 分）(二分法）

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 (≤10

9

) 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

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <cctype>
#include <string.h>
#include <cstdio>
#include <unordered_set>
using namespace std;
int main(){
    int n,ans=0;
    long long p;
    cin>>n>>p;
    vector<long long> a(n);
    for(int i=0;i<n;i++)
        cin>>a[i];
    sort(a.begin(),a.end());
    for(int i=0;i+ans<n;i++){
        int mid,left=i,right=n;
        while(left<right){
            mid=(left+right)/2;
            if(a[mid]>a[i]*p)
                right=mid;
            else
                left=mid+1;
        }
        ans=max(left-i,ans);
    }
    cout<<ans;
    return 0;
}