1085 Perfect Sequence——PAT甲级(双指针)-优快云博客

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:

solution:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl '\n' 
int main()
{
    ll n,p;cin>>n>>p;
    vector<ll>a(n);
    for(int i=0;i<n;i++)cin>>a[i];
    sort(a.begin(),a.end());
    int ans=0;
    int maxx=a[n-1],minn=a[0];
    int i=0;
    int j=0;
    while(j<n)
    {
    	while(a[i]*p>=a[j] && j<n)
    	{
    		j++;
		}
		if(a[i]*p<a[j] && j<n)
		{
			ans=max(ans,j-i);
			i++;
		}
		if(j==n)
		{
			ans=max(ans,j-i);
			break;
		}
	}
	cout<<ans<<endl;
}

大致思路是，先把给定的数组进行一遍排序，然后使用双指针i，j，分别指向左端(最小)和右端(最大)，然后当mp>=M时，j不断右移直到不满足，此刻如果j没移到数组最右端，则记录当前最大值，并将i右移一位，否则标志已经获得最大数列长度，之后只能将i右移也就是缩短，无意义，结束循环。