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 82 3 20 4 5 1 6 7 8 9

Sample Output:

8

 

#include 
      
       #include 
       
        #include 
        
         using namespace std;vector
         
           V;int main(){    int n,cnt=0;    long p;    cin >> n >> p;    long temp;    long max = 0;    if (n == 0)    {        cout << n;        return 0;    }    V.resize(n);    for (int i = 0; i < n; i++)    {        cin >> V[i];    }    sort(V.begin(), V.end());    for (int i = 0; i < n; i++)    {        for (int j = i + max; j <= n - 1; j++)        {            if (V[i] * p >= V[j]){                if (max < j - i + 1)                    max = j - i + 1;            }            else break;//剪枝非常重要        }    }    cout << max;}
         
        
       
      

1 该题使用递归寻找解空间时时间空间都爆炸，不可使用。递归迭代每个问题都可思考解法。但在n数值大时，时间空间不可接受。

2 针对最大最小值相关的问题可以用排序解决。

3 适当剪枝减小开销。