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:
8
//B1030英文版
//https://blog.csdn.net/m0_37454852/article/details/86523186
using namespace std;
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
long long int p, A[100010] = {0};//因为中间有乘积,可能会越界,改用long long存储
int N, i, pos, max_len = 0;
int first_over(int left, long long int temp)//二分法查找第一个大于p*A[i]的位置
{
if(A[N-1] <= temp) return N;
int l, r, mid;
for(l=left, r=N-1; l<r; )
{
mid = l+(r-l)/2;
if(A[mid] <= temp) l = mid+1;
else r = mid;
}
return l;
}
int main()
{
scanf("%d %lld", &N, &p);
for(i=0; i<N; i++)scanf("%lld", &A[i]);
sort(A, A+N);//sort两个参数默认升序排序
for(i=0; i<N-max_len; i++)
{
pos = first_over(i+1, p*A[i]);//(或可用C++库函数)pos = upper_bound(A+i+1, A+N, p*A[i]) - A;
max_len = max(max_len, pos-i);
}
printf("%d", max_len);
return 0;
}