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 (≤10​5​​) is the number of integers in the sequence, and p (≤10​9​​) is the parameter. In the second line there are N positive 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

解题思路:

利用二分查找的思想,先将所有元素进行排序,然后从最小的元素开始遍历,利用upperbound查找第一个M 使M > m*p

把每次查找的结果保存到一个数组中,最后选出最大的那个元素

注意,题目中给出的p是10的9次方,是比较大的元素,所以要用long long型保存,至于lower_Bound,upper_bound,套用模板即可

#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 100010;
int numbers[MAXN];
int squences[MAXN];
int N;
int P;
int lowerBound(int i,long long x) {
	if (numbers[N - 1] <= x) return N;
	int mid;
	int left = i + 1, right = N - 1;   //
	while (left < right) {
		mid = left + (right - left) / 2;
		if (numbers[N - 1] == numbers[mid] && numbers[mid] >  x) return 0;  //遍历到了最后一个
		if (numbers[mid] > x ) {
			right = mid;
		}
		else {
			left = mid + 1;
		}
	}
	return left;
}
bool cmp(int a,int b) {
	return a > b;
}
int main() {
	scanf("%d %d", &N, &P);
	for (int i = 0; i < N; ++i) {
		scanf("%d", &numbers[i]);
	}
	sort(numbers, numbers + N);
	for (int i = 0; i < N; ++i) {
		squences[i] = lowerBound(i, (long long)P*numbers[i]) - i;  //注意要转型!!!
	}
	sort(squences, squences + N, cmp);
	printf("%d", squences[0]);
	system("PAUSE");
	return 0;
}