1085 Perfect Sequence (25 point(s))

1085 Perfect Sequence (25 point(s))

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(≤10​5​​) is the number of integers in the sequence, and p (≤10​9​​) is the parameter. In the second line there are Npositive integers, each is no greater than 10​9​​.

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

思路见代码，很容易懂！思路参考大神：https://blog.youkuaiyun.com/richenyunqi/article/details/79512471

复杂度是：case#4出现TLE（22分）

#include<iostream>
#define ll long long
using namespace std;
const int N  = 1e5+7;
int a[N];
int main(void){
	int n;ll p;
	cin>>n>>p;
	for(int i=0;i<n;i++) cin>>a[i];
	int cnt = 0;int ans = 0;
	for(int i=0;i<n;i++){
		ll m = a[i];cnt=0;
		for(int j=0;j<n;j++){
			ll M = a[j];
			if(M<=p*m&&M>=m) cnt++;
		}
		if(cnt>ans) ans=cnt;
	}
	cout<<ans<<endl;
	return 0;
}

看起来复杂度没有变，需要在之后的序列中去查找第一个大于其p倍的数，实际上STL使用二分查找解决，降低复杂度。

#include<iostream>
#include<algorithm> 
#define ll long long
using namespace std;
const int N  = 1e5+7;
ll a[N];
int main(void){
	int n;ll p;
	cin>>n>>p;
	for(int i=0;i<n;i++) cin>>a[i];
	sort(a,a+n);
	int maxLen = 0;
	for(int i=0;i<n;i++){
		int len = upper_bound(a+i+1,a+n,a[i]*p)-a-i;//二分查找降低复杂度 
		maxLen = max(len,maxLen);
	}
	cout<<maxLen<<endl;
	return 0; 
}

双指针（非常巧妙，仔细想也容易想到）：

#include<bits/stdc++.h>
using namespace std;
int main(){
    int N;
    long long p;
    scanf("%d%lld",&N,&p);
    long long a[N];
    for(int i=0;i<N;++i)
        scanf("%lld",&a[i]);
    sort(a,a+N);
    int i=0,j=0,maxLength=-1;
    while(j!=N){
        while(j!=N&&a[i]*p>=a[j])
            ++j;
        maxLength=max(maxLength,j-i);
        ++i;
    }
    printf("%d",maxLength);
    return 0;
}