pat 1085. Perfect Sequence (25)

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

解：一次遍历，从i=0开始遍历，找大于container[i]*p的第一次出现的位置，如果间距大于当前值，则更新长度，当n-i的值小于当前长度的时候，程序终止，因为再也不可能得到更大的距离，最终得到最长间距。

代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
typedef long long ll;
ll container[1000000];
int main()
{
    int n,p;
    scanf("%d%d",&n,&p);
    ll num;
    for(int i=0;i<n;i++)
    {
        scanf("%lld",&num);
        container[i]=num;
    }
   sort(container,container+n);
//    for(q=container.begin();q!=container.end();q++)
//    {
//        cout<<*q<<" "<<endl;
//    }
    int tob=0,mm=0;
    for(int i=0;i<n;i++)
    {
        int tmp = upper_bound(container+i,container+n, container[i]*p) - (container+i);
        if(tmp>tob)
            tob=tmp;
        if(n-i<tob)
            break;
    }
    printf("%d\n",tob);
    return 0;
}

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