PAT (Advanced Level) Practise 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

题意：给你n个数，让你取出尽量多的数，使得这些数中的最大值ma，最小值mi满足ma<=mi*p

解题思路：将数字排序后暴力每个数作为最小值，二分查询最大值的位置

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;LL a[100005],p;int n;int main(){    while(~scanf("%d%lld",&n,&p))    {        for(int i=1;i<=n;i++) scanf("%lld",&a[i]);        sort(a+1,a+1+n);        int ma=-1;        for(int i=1;i<=n;i++)        {            int k=upper_bound(a+i,a+1+n,a[i]*p)-a-1;            ma=max(k-i+1,ma);        }        printf("%d\n",ma);    }    return 0;}