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

1. #include <stdio.h> 
2. #include <vector> 
3. #include <algorithm> 
4. using namespace std; 
5. vector<long long> nums; 
6. int bSearch(longlong num,int n) 
7. { 
8.     int l = 0,r = n-1,mid; 
9.     while(l<=r) 
10.     { 
11.         mid = (l+r)/2; 
12.         if(nums[mid]>num) 
13.         { 
14.             r = mid-1; 
15.         }else if(nums[mid]<num) 
16.         { 
17.             l = mid+1; 
18.         }else 
19.         { 
20.             return mid; 
21.         } 
22.     } 
23.     return l; 
24. } 
25. int main() 
26. { 
27.     long long n,p,tmp1,m,index; 
28.     long long i,j,tmpMax=0,resMax; 
29.     scanf("%lld%lld",&n,&p); 
30.     for(i=0;i<n;i++) 
31.     { 
32.         scanf("%lld",&tmp1); 
33.         nums.push_back(tmp1); 
34.     } 
35.     sort(nums.begin(),nums.end()); 
36.     for(i=0;i<n;i++) 
37.     { 
38.         m = nums[i]*p; 
39.         index = bSearch(m, n); 
40.         if(nums[n-1]<=m) 
41.         { 
42.             tmpMax = n-1-i+1; 
43.         }else 
44.         { 
45.             tmpMax = index-i; 
46.         } 
47.         if(tmpMax>resMax) 
48.         { 
49.             resMax = tmpMax; 
50.         } 
51.     } 
52.     printf("%lld\n",resMax); 
53.     return 0; 
54. }