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 (≤10?5??) is the number of integers in the sequence, and p (≤10?9??) is the parameter. In the second line there are N positive integers, each is no greater than 10?9??.
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
注意点：1.int*int 的范围必须用long long; 2.if(a[n-1]<=x) return n;
1 #include<bits/stdc++.h>
2 using namespace std;
3
4 const int maxn =100010;
5
6 int a[maxn];
7
8 int n,p;
9
10 int binary_Search(int i,long long x){
11
12 if(a[n-1]<=x)
13 return n;
14
15 int left=i+1,right=n-1;
16
17 while(left<right){
18 int mid = (left+right)/2;
19
20 if(a[mid]>x){
21 right =mid;
22 }else{
23 left= mid+1;
24 }
25 }
26
27 return left;
28
29
30 }
31
32
33 int main(){
34
35 cin>>n>>p;
36
37 for(int i=0;i<n;i++)
38 cin>>a[i];
39
40 sort(a,a+n);
41
42 int ans=1;
43
44 for(int i=0;i<n;i++){
45
46 // int j=binary_Search(i,(long long)a[i]*p);
47
48 int j=upper_bound(a+i+1,a+n,(long long)a[i]*p)-a;
49
50 ans=max(ans,j-i);
51
52 // cout<<j<<" "<<i<<endl;
53 }
54
55
56
57 cout<<ans<<endl;
58
59 return 0;
60
61 }
原文地址：https://www.cnblogs.com/moranzju/p/11121176.html