PAT甲级——A1085 Perfect Sequence
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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 (≤) is the number of integers in the sequence, and p (≤) is the parameter. In the second line there are N positive integers, each is no greater than 1.
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
又是没看清题,这道题的子序列不需要是原来的连续子序列,只要求是原来里面的值就行,搞得又浪费了很多时间!!!!
1 //靠,不需要是子排序,就是找数字就行 2 #include <iostream> 3 #include <deque> 4 #include <vector> 5 #include <algorithm> 6 using namespace std; 7 int N; 8 long long P; 9 int main() 10 11 cin >> N >> P; 12 vector<int>num(N); 13 for (int i = 0; i < N; ++i) 14 cin >> num[i]; 15 sort(num.begin(), num.end()); 16 int res = 0; 17 for(int L=0,R=0;L<=R && R<N;++R) 18 19 while (L <= R && num[R] > P * num[L]) 20 L++; 21 res = res > R - L + 1 ? res : R - L + 1; 22 23 cout << res << endl; 24 return 0; 25
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