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 8
2 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 fin

int 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;
}

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