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
#include<iostream>
#include<vector>
#include<cstdio>
#include<algorithm>
using namespace std;
int main(){
int n,p;
scanf("%d%d",&n,&p);
vector<long>vt;
int i,j,val;
for(i=0;i<n;i++){
scanf("%ld",&val);
vt.push_back(val);
}
sort(vt.begin(),vt.end());
int left=1,right=0;
int length=0;
int len=0;
for(i=0;i<n;i++){
long tmp = p*vt[i];
if(tmp>=vt[n-1]){
if(len < n-i){
len=n-i;
}
break;
}
int right=n-1;
while(right>left){
int mid=(right+left)/2;
if(vt[mid]>tmp){
right=mid;
}else if(vt[mid]<tmp){
left=mid+1;
}else {
left=mid+1;
break;
}
}
if(left-i>len){
len=left-i;
}
}
printf("%d
",len);
return 0;
}