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 8
2 3 20 4 5 1 6 7 8 9

Sample Output:

8

 1 #include<cstdio>
 2 #include<stack>
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<stack>
 6 #include<set>
 7 #include<map>
 8 #include<vector>
 9 using namespace std;
10 long long line[100005];
11 int main(){
12     //freopen("D:\\INPUT.txt","r",stdin);
13     long long n,p;
14     scanf("%lld %lld",&n,&p);
15     int i,j;
16     for(i=0;i<n;i++){
17         scanf("%lld",&line[i]);
18     }
19     sort(line,line+n);
20     int amount;
21     long long cur=line[0]*p;
22     for(j=0;j<n&&line[j]<=cur;j++);//j指针
23     amount=j;
24     for(i=1;i<n&&j<n;i++){//这里的j<n是为了避免不必要的循环
25         cur=line[i]*p;
26         for(;j<n&&line[j]<=cur;j++);
27         if(j-i>amount){
28             amount=j-i;
29         }
30     }
31     printf("%d\n",amount);
32     return 0;
33 }