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 (≤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


注意点：1.int*int 的范围必须用long long;
　　　　2.if(a[n-1]<=x) return n;

#include<bits/stdc++.h>
using namespace std;

const int maxn =100010;

int a[maxn];

int n,p;

int binary_Search(int i,long long x){
    
    if(a[n-1]<=x)
        return n;
    
    int left=i+1,right=n-1;
    
    while(left<right){
        int mid = (left+right)/2;
        
        if(a[mid]>x){
            right =mid;
        }else{
            left= mid+1;
        }
    }
    
    return left;
    
    
}


int main(){

    cin>>n>>p;
    
    for(int i=0;i<n;i++)
        cin>>a[i];
        
    sort(a,a+n);
    
    int ans=1;
    
    for(int i=0;i<n;i++){
        
    //    int j=binary_Search(i,(long long)a[i]*p);
        
        int j=upper_bound(a+i+1,a+n,(long long)a[i]*p)-a;
            
        ans=max(ans,j-i);
        
    //    cout<<j<<" "<<i<<endl;
    }
    

    
    cout<<ans<<endl;
    
    return 0;
    
}