题目如下：
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.
Sample Input:10 8 2 3 20 4 5 1 6 7 8 9Sample Output:
8
题目要求从给定序列中选取子序列，使得序列的最小值m、最大值M满足:M≤m*p，其中p为一个给定的正整数，输出能找到的最长子序列长度。
这道题一个很自然的思路就是设立一个头指针cur1，尾指针cur2，将序列按照升序排列，让cur2从最后一个元素向前指，cur1遍历从第一个元素到cur2的位置，找到合适的m时停下，记录长度，这样会有一个case超时，解决方法是让cur1从前到后遍历，cur2采用二分查找。
如果找到的位置使得M＜m*p，说明M还可以更大，可以继续查找右半部分；如果M＞m*p，说明M偏大，应该去左半部分找更小的；如果M=m*p，说明找到了合适的位置。在查找结束后，记录长度即可。
这段代码参考了Yangsongtao1991。
#include <iostream>
#include <vector>
#include <algorithm>
#include <stdio.h>
using namespace std;
int main()
{
int n,p;
scanf("%d%d",&n,&p);
vector<long> seq(n);
for(int i = 0; i < n; i++)
scanf("%ld",&seq[i]);
sort(seq.begin(),seq.end());
int maxcount = 0, down = 1;
for(int i = 0; i < n; i++)
{
long mp = p * seq[i];
if(mp >= seq[n-1]) // 如果最大的元素都≤m*p，则从当前位置到最后全部计数。
{
if(maxcount < n - i){
maxcount = n - i;
}
break;
}
int up = n-1;
while(up > down)
{
// 二分查找，结束条件为上界≤下界，根据mid处的乘积判定。
// 现在是确定了m，要找M，如果找到的位置＜mp，说明M可能可以更大，向右找；如果＞mp，说明M偏大，向左找。
// 如果当前位置恰好满足，则说明已经找到了最长满足要求的位置。
int mid = (up + down)/2;
if(seq[mid] > mp)
up = mid;
else if(seq[mid] < mp)
down = mid + 1;
else
{
down = mid + 1;
break;
}
}
if(down - i > maxcount)
maxcount = down - i;
}
printf("%d
",maxcount);
return 0;
}