1085. Perfect Sequence (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CAO, Peng

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 82 3 20 4 5 1 6 7 8 9

Sample Output:

8

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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;}