PAT (Advanced Level) 1085. Perfect Sequence (25) 贪心算法

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

排序后使用贪心算法。

/*2015.7.30cyq*/#include <iostream>#include <vector>#include <algorithm>#include <fstream>using namespace std;//ifstream fin("case1.txt");//#define cin finint main(){long long N,p;cin>>N>>p;vector<long long> ivec(N);for(int i=0;i<N;i++)cin>>ivec[i];sort(ivec.begin(),ivec.end());int low=0;int high=0;int len=0;while(1){while(high<N&&ivec[high]<=ivec[low]*p)high++;if(high-low>len)len=high-low;if(high==N)break;while(low<high&&ivec[low]*p<ivec[high])low++;}cout<<len;return 0;}