1085. Perfect Sequence (25)

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.

10 8
2 3 20 4 5 1 6 7 8 9

8
先排个升序，然后用二分，对前一半的数找出最大的M，更新最大数m。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
    long long s[100000],n,p,m = 0;
    scanf("%lld%lld",&n,&p);
    for(int i = 0;i < n;i ++)
    {
        scanf("%lld",&s[i]);
    }
    sort(s,s + n);
    for(int i = 0;i < n/2;i ++)
    {
        int  l = i,r = n - 1,mid = (l + r)/2;
        while(l < r)
        {
            if(s[mid] > s[i] * p)r = mid - 1;
            else if(s[mid] < s[i] * p)l = mid + 1;
            else break;
            mid = (l + r) / 2;
        }
        if(s[mid] > s[i] * p)mid --;
        if(m < mid - i)m = mid - i;
    }
    printf("%lld",m + 1);
}