1085 Perfect Sequence

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.

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.

For each test case, print in one line the maximum number of integers that can be chosen to form a perfect subsequence.

10 8
2 3 20 4 5 1 6 7 8 9

8

测试点4运行超时，测试点5答案错误

#include
#include
using namespace std;

int n, p;
int num[100010];

int main()
{
  scanf_s("%d %d", &n, &p);
  for (int i = 0; i < n; i++)
  {
    scanf_s("%d", &num[i]);
  }
  sort(num, num + n);
  
  int ans = 0;  //结果
  for (int i = 0; i < n; i++)
  {
    int pro = num[i] * p;
    int re = 0;
    for (int j = i; j < n; j++)
    {
      if (num[j] <= pro)
      {
        re++;
      }
      else
      {
        break;
      }
    }
    if (re > ans)
    {
      ans = re;
    }

  }
  printf("%dn", ans);
}

将序列中的元素类型和p改为long long后，测试点5正确

#include
#include
using namespace std;

int n;
long long p;
long long num[100010];

int main()
{
  scanf_s("%d %lld", &n, &p);
  for (int i = 0; i < n; i++)
  {
    scanf_s("%lld", &num[i]);
  }
  sort(num, num + n);
  
  int ans = 0;  //结果
  for (int i = 0; i < n; i++)
  {
    long long pro = num[i] * p;
    int re = 0;
    for (int j = i; j < n; j++)
    {
      if (num[j] <= pro)
      {
        re++;
      }
      else
      {
        break;
      }
    }
    if (re > ans)
    {
      ans = re;
    }

  }
  printf("%dn", ans);
}

将查找方法改为二分，测试点5正确

#include
#include
using namespace std;

int n;
long long p;
long long num[100010];

int find(int i)
{
	//寻找第一个大于num[i]*p的元素
  long long pro = num[i] * p;
  if (pro >= num[n - 1])
  {
    return n;
  }
  int left = i;
  int right = n - 1;
  int mid = 0;
  while (left < right)
  {
    mid = (left + right) / 2;
    if (num[mid] <= pro)
    {
      left = mid + 1;
    }
    else
    {
      right = mid - 1;
    }
  }
  return left;
}

int main()
{
  scanf_s("%d %lld", &n, &p);
  for (int i = 0; i < n; i++)
  {
    scanf_s("%lld", &num[i]);
  }
  sort(num, num + n);
  
  int ans = 0;  //结果
  for (int i = 0; i < n; i++)
  {
    int j = find(i);
    int re = j - i;
    if (re > ans)
    {
      ans = re;
    }

  }
  printf("%dn", ans);
}

find()函数的效果可由int j = upper_bound(num + i + 1, num + n, num[i] * p) - num;实现

//查找[first, last)区域中第一个大于 val 的元素。
ForwardIterator upper_bound (ForwardIterator first, ForwardIterator last,
                             const T& val);

其中，first 和 last 都为正向迭代器，[first, last) 用于指定该函数的作用范围；val 用于执行目标值

同时，该函数会返回一个正向迭代器

二点法

#include
#include
#include
using namespace std;

long long n, p;
long long num[100010];

int main()
{
	scanf_s("%d %d", &n, &p);
	for (int i = 0; i < n; i++)
	{
		scanf_s("%d", &num[i]);
	}
	sort(num, num + n);

	int i = 0, j = 0;
	int count = 0;
	while (i < n && j < n)
	{
		while (j < n && num[j] <= num[i] * p)
		{
			count = max(count, j - i + 1);
			j++;
		}
		i++;
	}
	printf("%d", count);
}