1085 Perfect Sequence (25)

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

 顺序查找会有一个测试点超时， 用二分查找就行了， 尽量不要用顺序查找；

此外要用long int不然会有错误

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
bool cmp(const long int& a, const long int& b){return a<b;}
int main(){
  long int n, p, i, j;
  cin>>n>>p;
  vector<long int> v(n);
  for(i=0; i<n; i++) scanf("%d", &v[i]);
  sort(v.begin(), v.end(), cmp);
  long int max=0;
  for(i=0; i<n; i++){
      
      for( j=n-1; j>=i; j--){
          if(v[i]*p>=v[j]){
              if(max<j-i+1) max=j-i+1;
              break;
          }
      }
      if(v[i]*p>=v[n-1]&&max<j-i) max=j-i;
  }
  cout<<max<<endl;
  return 0;
}

二分查找

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
bool cmp(const long int& a, const long int& b){return a<b;}
int main(){
  long int n, p, i, j， max=0;
  cin>>n>>p;
  vector<long int> v(n);
  for(i=0; i<n; i++) scanf("%d", &v[i]);
  sort(v.begin(), v.end(), cmp);
  for(i=0; i<n; i++){
      int low=i, high=n-1;
      while(low<=high){
        int mid=(low+high)/2;
        if(v[i]*p<v[mid]) high=mid-1;
        else if(v[i]*p>=v[mid]) low=mid+1; 
        if(max<mid-i+1 && v[i]*p>=v[mid]) max=mid-i+1;
      }
  }
  cout<<max<<endl;
  return 0;
}