1085. Perfect Sequence (25)

题目：

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 82 3 20 4 5 1 6 7 8 9

Sample Output:

8

注意：

1、这里先排序再使用二分法搜索来找到<=m*p的那个最大的值得位子就可以计算得到这个区间的元素个数，依次寻找每个m对应的那个M以及两者之间的元素的个数，找出个数最多的就可以了。

2、为了提高效率，可以将上次循环找到的M所在的位置作为这次二分搜索时的下界，以缩小搜索区间。

3、如果在一个循环中发现序列中最大的那个数都是<=m*p的，那么就可以终止循环了。

4、注意m*p有可能是会溢出int的范围的，所以这里用long型数据进行计算，case 5就是考察这一点的。

代码：

//1085#include<iostream>#include<vector>#include<algorithm>using namespace std;int main(){int n,p;scanf("%d%d",&n,&p);vector<long>seq;seq.resize(n);for(int i=0;i<n;++i)scanf("%ld",&seq[i]);sort(seq.begin(),seq.end());int maxcount=0,down=1;for(int i=0;i<n;++i){//reuse the answer 'down' can reduce the calculating timelong max=p*seq[i];if(max>=seq[n-1]){//if all the element behind i is smaller than max //they all need to be counted and the loop could be stoppedmaxcount=n-i>maxcount?n-i:maxcount;break;}int up=n-1;while(up>down){//binary searchint mid=(up+down)/2;if(seq[mid]>max)up=mid;else if(seq[mid]<max)down=mid+1;else{down=mid+1;break;}}if(down-i>maxcount)maxcount=down-i;}printf("%d\n",maxcount);return 0;}