1085. Perfect Sequence

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 9
Sample Output:
8

知识点：

剪枝，遍历的时候，如果 list[i]*p < list[j] 的话，就剪枝

另外，比当前完美序列长度小的就不用遍历了，j 直接从 i+cnt 开始

 #include <iostream>
 #include <vector>
 #include <algorithm>
 using namespace std;

 int main(){
     vector<int> list;
     long long p,tmp;
     int n; scanf("%d %lld",&n,&p);

     for(int i=;i<n;i++){
         scanf("%lld",&tmp);
         list.push_back(tmp);
     }
     sort(list.begin(),list.end());

     int cnt=;

     for(int i=;i<n;i++){
         for(int j=i+cnt;j<n;j++){
             if(list[i]*p>=list[j]){
                 if(j-i>cnt)
                     cnt=j-i;
             }else{
                 break;
             }
         }
     }
     printf("%d",cnt+);
 }