PAT 1085 Perfect Sequence [二分法] [返回第一个大于x的数]

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 mare 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 (≤10​5​​) is the number of integers in the sequence, and p (≤10​9​​) is the parameter. In the second line there are Npositive integers, each is no greater than 10​9​​.

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

--------------------------------------这是题目和解题的分割线--------------------------------------

我好傻题目都看错了，以为是求最大值，原来是最长完美序列的长度(︶︹︺) 

求小于（最小值（序列最左端）*参数p）的最大值（序列最右端），可以转换为，先找到大于这个式子的第一个值，则最大值在这个值的前一个位置。这个过程可以用二分法的变形版，再结合for循环，把最小值（序列最左端）当作已知值来扫描最大值。

先来回顾一下最熟悉的二分法。

//好像也没什么值得回顾的二分法
#include

int n,i,a[100],x;

int findN(int x,int a[])
{
	int left = 0,right = n-1;
	while(left<=right) 
	{
		int mid = (left+right)/2;
		if(a[mid]==x) return mid;
		else if(a[mid]

再是本题的代码

#include
#include

using namespace std;

int n,i,a[100005];
long long p; //10^9次方不能用int存储 

int f(int i,long long q)
{
	//返回n不是n-1,因为这个函数是返回大于式子的值的坐标 
	if(a[n-1]<=q) return n;  
	int left = i,right = n-1;
	//和二分法不同的是,这里靠left==right作为找到的标志 
	while(leftq) right = mid;  
		else left = mid+1;
	}
	return left;
}

int main()
{	
	scanf("%d%d",&n,&p);
	for(i=0;i