【PAT】【Advanced Level】1085. Perfect Sequence (25)

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

原题链接：
https://www.patest.cn/contests/pat-a-practise/1085

https://www.nowcoder.com/pat/5/problem/4035

思路：

排序+二分查找

坑点：

PAT上最后一个数据点要用longlong

CODE：

#include<iostream>#include<algorithm>#include<cstdio>#define N 100010using namespace std;long long  num[N];int main(){    int n;long long p;    scanf("%d %ld",&n,&p);    for (int i=0;i<n;i++)        scanf("%ld",&num[i]);    sort(num,num+n);    int mx=0;    for (int i=0;i<n;i++)    {        if (i>=n-mx) break;        int l=i;        int r=n-1;        int mid;        while (l<=r)        {            mid=(l+r)/2;            if (num[mid]>num[i]*p)            {                r=mid-1;               }            else            {                l=mid+1;            }        }        mx=max(mx,l-i);    }    cout<<mx;    return 0;}