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 (≤) is the number of integers in the sequence, and p (≤) is the parameter. In the second line there are N positive integers, each is no greater than 1.

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<bits/stdc++.h>
using namespace std;
const int maxn=100010;

int n,p;
int a[maxn];
int binarySearch(int i,long long x){
    if(a[n-1]<=x){
        return n;
    }
    int L=i+1,r=n-1,mid;
    while(L<r){
        mid=(r+L)/2;
        if(a[mid]<=x){
            L=mid+1;
        }else{
            r=mid;
        }
    }
    return L;
}
int main(){
    scanf("%d %d",&n,&p);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    sort(a,a+n);
    int ans=1;
    for(int i=0;i<n;i++){
        int j=binarySearch(i,(long long)a[i]*p);
        ans=max(ans,j-i);
    }
    printf("%d
",ans);
    return 0;
}

扛不住了，要睡觉。。。