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
题目大意：
给定一个正整数数列，和正整数p，设这个数列中的最大值是M，最小值是m，如果M <= m * p，则称这个数列是完美数列。现在给定参数p和一些正整数，请你从中选择尽可能多的数构成一个完美数列。输入第一行给出两个正整数N（输入正数的个数）和p（给定的参数），第二行给出N个正整数。在一行中输出最多可以选择多少个数可以用它们组成一个完美数列
分析：
简单题。首先将数列从小到大排序，设当前结果为result = 0，当前最长长度为temp = 0；从i = 0～n，j从i + result到n，【因为是为了找最大的result，所以下一次j只要从i的result个后面开始找就行了】每次计算temp若大于result则更新result，最后输出result的值～
原文链接：https://blog.csdn.net/liuchuo/article/details/51985871
题解
注意p是long long类型~
#include <bits/stdc++.h>
using namespace std;
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
int n;
long long p;
cin>>n>>p;
vector<int> v(n);
for(int i=0;i<n;i++){
cin>>v[i];
}
sort(v.begin(),v.end());
int result=0,temp=0;
for(int i=0;i<n;i++){
for(int j=i+result;j<n;j++){
temp=j-i+1;
if(v[j]<=v[i]*p){
result=temp;
}else break;
}
}
cout<<result<<endl;
return 0;
}