1085 Perfect Sequence (25 分)（二分）

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 (<= 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;
}