【PAT】 1085. Perfect Sequence (25)

最新推荐文章于 2020-06-08 15:04:44 发布

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#PAT #二分查找

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本文介绍了一个算法问题，即从给定的整数序列中找出尽可能多的数字形成一个完美子序列。其中，完美子序列定义为最大值与最小值之比不超过给定参数p的子序列。文章提供了使用二分查找法实现的C++代码示例。

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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:

分析：（1）int*int有可能会超出int范围，所以用long long （2）使用二分查找法，否则会超时

代码如下：

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

//vec 采用“引用”则不会超时，如果用传参则会超时。 
int binaryFind(vector<long long >& vec, long long max, int i){
	int left = i;
	int right = vec.size()-1;
    int mid;
    while(left < right){
		mid = (left+right)/2;
		if(vec[mid] > max){
			right = mid - 1;
		}else if(vec[mid] < max){
			left = mid + 1;
		}else{
			return mid;
		}		
	}	
	return vec[left]>max? left-1:left; 		
}

int main(int argc, char** argv) {
	long long  n, p, i, val;
	scanf("%lld%lld",&n,&p);
	vector<long long> vec(n);
	for(i=0; i<n; i++){
		scanf("%lld",&vec[i]);
	}
	sort(vec.begin(), vec.end());
    long long max;
    int cnt=0, pos;
	for(i=0; i<n; i++){
		max = p*vec[i];
		pos = binaryFind(vec, max, i);
	 	if(pos-i+1 > cnt){
	 		cnt = pos-i+1;	
		}
	}
	printf("%d\n",cnt);
	return 0;
}