1085 Perfect Sequence （25 分）

1085 Perfect Sequence （25 分）

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 Npositive 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

题意：从一组序列中找出最大的满足条件的序列的长度。

分析：二分法，可以用STL例upper_bound函数。注意乘积可能超过int型表示范围

 /**
 * Copyright(c)
 * All rights reserved.
 * Author : Mered1th
 * Date : 2019-02-26-13.58.21
 * Description : A1085
 */
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<unordered_set>
 #include<map>
 #include<vector>
 #include<set>
 using namespace std;
 ;
 int a[maxn];
 int main(){
 #ifdef ONLINE_JUDGE
 #else
     freopen("1.txt", "r", stdin);
 #endif
     int n,p;
     scanf("%d%d",&n,&p);
     ;i<n;i++){
         scanf("%d",&a[i]);
     }
     sort(a,a+n);
     ;
     ;i<n;i++){
         ,a+n,(long long)a[i]*p)-a;
         num=max(num,j-i);
     }
     printf("%d\n",num);
     ;
 }

另一种方法是双指针，思路就是定义两个指针i,j，均从0开始遍历，j不断右移直到不满足条件，再i++。

 /**
 * Copyright(c)
 * All rights reserved.
 * Author : Mered1th
 * Date : 2019-02-26-14.18.14
 * Description : A1085
 */
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<unordered_set>
 #include<map>
 #include<vector>
 #include<set>
 using namespace std;
 ;
 int a[maxn];
 int main(){
 #ifdef ONLINE_JUDGE
 #else
     freopen("1.txt", "r", stdin);
 #endif
     int n,p;
     scanf("%d%d",&n,&p);
     ;i<n;i++){
         scanf("%d",&a[i]);
     }
     sort(a,a+n);
     ,j=,num=;
     while(i<n&&j<n){
         while(j<n&&a[j]<=(long long)a[i]*p){
             num=max(num,j-i+);
             j++;
         }
         i++;
     }
     cout<<num;
     ;
 }