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<vector> 3 #include<algorithm> 4 using namespace std; 5 int main() 6 { 7 long long len,p,i,tem,j; 8 vector<long long> vv; 9 scanf("%lld%lld",&len,&p); 10 for(i = 0;i<len;i++) 11 { 12 scanf("%lld",&tem); 13 vv.push_back(tem); 14 } 15 16 sort(vv.begin(),vv.end()); 17 18 int Max= 0; 19 int count; 20 i = 0; j = 0; 21 for(i = 0 ; i < len ;i ++) 22 { 23 j = i + Max; 24 count = Max; 25 while( j < len && vv[j] <= vv[i] * p) 26 { 27 ++j; 28 ++ count; 29 } 30 if(Max < count) Max = count; 31 } 32 printf("%d ",Max); 33 return 0; 34 }