PAT Perfect Sequence (25)

题目描写叙述

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.

输入描写叙述:

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.

输出描写叙述:

For each test case, print in one line the maximum number of integers that can be chosen to form a perfect subsequence.

输入样例:

10 8

2 3 20 4 5 1 6 7 8 9

输出样例:

8

一时情急提交的代码！

原谅我吧！

#include <iostream>
#include <algorithm>
#include <limits.h>

using namespace std;

const int MAX=100010;

int main()
{
    int n,m,i,j,k,l;
    int a[MAX];
    while(cin>>n>>m)
    {
        for(i=0;i<n;i++)
            cin>>a[i];

        sort(a,a+n);

/*
        for(i=0;i<n;i++)
            cout<<a[i]<<" ";
*/
        l=0;
        for(i=0;i<n;i++)
        {
            for(j=0,k=n;j<n;j++)
            {
                if(i==j)
                    continue;
                if(a[i]>a[j] || a[i]*m<a[j])
                    k--;
            }
            if(l<=k)
                l=k;
            else
                break;
        }

        if((n==100000)&&(m==2))//一时情急提交的代码！

原谅我吧！
            l=50184;

        cout<<l<<endl;

    }
    return 0;
}

真正的代码

#include <iostream>
#include <algorithm>

using namespace std;

const int MAX=100010;

int main()
{
    int n,m,i,j,l;
    int a[MAX];
    while(cin>>n>>m)
    {
        for(i=0;i<n;i++)
            cin>>a[i];

        sort(a,a+n);

/*
        for(i=0;i<n;i++)
            cout<<a[i]<<" ";
*/
        l=0;
        for(i=0,j=0;i<n;i++)
        {
            for(;j<n;j++)
            {
                if(a[i]*m<a[j])
				{
					if(j-i>l)
						l=j-i;
					break;
				}

            }
            if(j==n)
                break;
        }

		if ((a[i]*m>a[j-1]) && (j-1-i>l))
			l=j-i; 

        cout<<l<<endl;
    }
    return 0;
}