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提交代码
用了two pointers的思想
#include<stdio.h> #include<algorithm> using namespace std; int a[100010],n,p; int main() { scanf("%d%d",&n,&p); for(int i=0;i<n;i++) { scanf("%d",&a[i]); } sort(a,a+n); int ans=1; int i=0,j=0; while(i<n&&j<n) { while(j<n&&a[j]<=(long long)a[i]*p) { j++; } ans=max(ans,j-i); i++; } printf("%d\n",ans); return 0; }