PAT甲级1085(完美数列)

1085 Perfect Sequence (25 point(s))

  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 (≤10⁹) 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

题目大意

  给定一个数列和p，要求挑出一个子列，同时该子列的最大值M和最小值m要满足M <= m*p，要求输出满足条件的最长子列的长度

解题思路

  先将该数组排序，并查找该位置数字*p所在位置，找到两个位置差的最大值即可。此处即可用二分查找也可使用STL自带的upper_bound函数

代码

#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <algorithm>
using namespace std;

long long arr[100005] = { 0 };
int main()
{
	long long N, p;
	int ret = scanf("%lld%lld", &N, &p);//输入必须使用%lld
	int i, MAX_cnt = -1;
	for (i = 0; i < N; ++i) {
		ret = scanf("%lld", &arr[i]);
	}
	sort(arr, arr + N);
	for (i = 0; i < N; ++i) {
		long long bound = arr[i] * p;
		//寻找其所在位置
		int left = i, right = N - 1;
		while (left <= right) {//二分查找，此处也可使用upper_bound函数，代码会更加简洁
			int mid = (left + right) / 2;
			if (arr[mid] > bound) {
				right = mid - 1;
			}
			else {
				left = mid + 1;
			}
		}
		MAX_cnt = max(MAX_cnt, left - i);//最长序列		
	}
	printf("%d", MAX_cnt);
	return 0;
}