Pat 1085. Perfect Sequence (25)

Pat 1085. Perfect Sequence (25)
Pat 1085. Perfect Sequence (25) Time Limit 300 MS
The memory limit is 32000 kB.
Code length limit: 16000 B
Criterion author 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 your 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

 

In terms of time, we will look into a binary search. There is also a long because multiplication may exceed int

 

[Cpp]

1. # Include # Include
2. # Include using namespace std;
3. Vector Nums; int bSearch (long num, int n)
4. {Int l = 0, r = n-1, mid;
5. While (l <= r ){
6. Mid = (l + r)/2; if (nums [mid]> num)
7. {R = mid-1;
8. } Else if (nums [mid] L = mid + 1;} else
9. {Return mid;
10. }}
11. Return l ;}
12. Int main (){
13. Long n, p, tmp1, m, index; long I, j, tmpMax = 0, resMax;
14. Scanf (% lld, & n, & p); for (I = 0; I {Scanf (% lld, & tmp1 );
15. Nums. push_back (tmp1 );}
16. Sort (nums. begin (), nums. end (); for (I = 0; I {M = nums [I] * p;
17. Index = bSearch (m, n); if (nums [n-1] <= m)
18. {TmpMax = n-1-i + 1;
19. } Else {
20. TmpMax = index-I ;}
21. If (tmpMax> resMax ){
22. ResMax = tmpMax ;}
23. } Printf (% lld, resMax );
24. Return 0 ;}

    #include 
       
        #include 
        
         #include using namespace std;vector
         
           nums;int bSearch(long long num,int n){    int l = 0,r = n-1,mid;    while(l<=r)    {        mid = (l+r)/2;        if(nums[mid]>num)        {            r = mid-1;        }else if(nums[mid]
          
           resMax)        {            resMax = tmpMax;        }    }    printf(%lld,resMax);    return 0;}