PAT甲级练习1085. Perfect Sequence (25)

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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 8
2 3 20 4 5 1 6 7 8 9

Sample Output:

先排序，一开始我分别从两边开始遍历，直接超时了。。。后来想着可以从前开始，后面的点j从前面i+maxlength

#include <iostream>  
#include <cstdio>  
#include <algorithm>  
#include <vector>  
#include <map>  
#include <set>  
#include <stack>  
#include <queue>  
#include <string>  
#include <string.h> 
using namespace std;

const int inf = 99999999;
const int MAX = 1e5+10;

long long num[MAX];

int main() {
	int n, p, l, maxlength;
	scanf("%d%d", &n, &p);
	for(int i=0; i<n; i++){
		scanf("%lld", &num[i]);
	}
	sort(num, num+n);
	maxlength = 1;
	for(int i=0; i<=n-2; i++){
		for(int j=i+maxlength; j<n; j++){
			if(num[j]<=num[i]*p){
				l = j - i + 1;
				if(l>maxlength) maxlength = l;
			}else{
				break;
			}
		}
	}
	printf("%d", maxlength);
	scanf("%d",&n);
    return 0;
}