PAT甲级——完美数列

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

思路： 首先将数字序列按照非递减的顺序排序，从第一个数开始计算完美数列的长度，最后取最大值。在计算完美数列的时候，要用二分查找的策略，找到第一个大于x*p的位置，将时间复杂度由n变为logn。

代码一：手写二分查找的函数 

#include <iostream>
#include<algorithm>
using namespace std;
const int maxn = 100010;
int a[maxn],n,p;

int BinarySearch(int i,long long x)
{
	if (a[n - 1] <= x) return n;
	int l = i + 1, r = n - 1, mid;
	while (l<r)
	{
		mid = (l + r) / 2;
		if (a[mid] <= x) l = mid + 1;
		else {
			r = mid;
		}
	}
	return l;
}
int main()
{
	
	scanf("%d%d", &n,&p);
	for (int i = 0; i < n; i++)
	{
		scanf("%d", &a[i]);
	}
	sort(a, a + n);
	int ans = 1;
	for (int i = 0; i < n; i++)
	{
		int j = BinarySearch(i, (long long)a[i] * p);
		ans = max(ans, j - i);
	}
	printf("%d\n", ans);
	return 0;
}

 

代码二：利用STL模板库的upper_bound函数

 

#include <iostream>
#include<algorithm>
using namespace std;
const int maxn = 100010;
int a[maxn],n,p;

int main()
{
	scanf("%d%d", &n, &p);
	for (int i = 0; i < n; i++)
	{
		scanf("%d", &a[i]);
	}
	sort(a, a + n);
	int ans = 1;
	for (int i = 0; i < n; i++)
	{
		int j= upper_bound(a + i+1, a + n,(long long) a[i] * p)-a;
		ans = max(ans, j - i);
	}
	printf("%d\n", ans);
	return 0;
}