A1085. 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.

Sample Input:

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

Sample Output:

8

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <algorithm>
 4 #include <string.h>
 5 using namespace std;
 6 
 7 const int maxn =100010;
 8 int n,p,a[maxn];
 9 
10 
11 int main(int argc, char* argv[])
12 {
13     scanf("%d%d",&n,&p);
14     for(int i=0;i<n;i++)
15     {
16     scanf("%d",&a[i]); 
17     }
18 sort(a,a+n);
19 int ans=1;
20 for(int i=0;i<n;i++)
21 {
22 int j=upper_bound(a+i+1,a+n,(long long )a[i]*p)-a;
23 ans=max(ans,j-i);
24 }
25 printf("%d\n",ans);
26     system("pause"); 
27     return 0;
28 }

A1085. Perfect Sequence (25)

原文：http://www.cnblogs.com/ligen/p/4336969.html