1085 Perfect Sequence (25 分)

2024-05-30 09:38:33

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1085 Perfect Sequence (25 分)

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:

作者: CAO, Peng

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

具体代码

#include <cstdio>
#include <algorithm>
using namespace std;
int main(){int n,p;scanf("%d%d",&n,&p);int a[100010];for(int i=0;i<n;i++){scanf("%d",&a[i]);}sort(a,a+n);int ans=0;for(int i=0;i<n;i++){int j=upper_bound(a,a+n,(long long)a[i]*p)-a;ans=max(ans,j-i);//之所以有用for循环是防止极端情况123，m和p为3。}printf("%d",ans);return 0;}

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