1085. Perfect Sequence
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
知识点:
剪枝,遍历的时候,如果 list[i]*p < list[j] 的话,就剪枝
另外,比当前完美序列长度小的就不用遍历了,j 直接从 i+cnt 开始
1 #include <iostream> 2 #include <vector> 3 #include <algorithm> 4 using namespace std; 5 6 int main(){ 7 vector<int> list; 8 long long p,tmp; 9 int n; scanf("%d %lld",&n,&p); 10 11 for(int i=0;i<n;i++){ 12 scanf("%lld",&tmp); 13 list.push_back(tmp); 14 } 15 sort(list.begin(),list.end()); 16 17 int cnt=0; 18 19 for(int i=0;i<n;i++){ 20 for(int j=i+cnt;j<n;j++){ 21 if(list[i]*p>=list[j]){ 22 if(j-i>cnt) 23 cnt=j-i; 24 }else{ 25 break; 26 } 27 } 28 } 29 printf("%d",cnt+1); 30 }
posted on 2018-11-13 09:12 iojafekniewg 阅读(167) 评论(0) 编辑 收藏 举报