用upper_bound求解

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1e5 + 10;
typedef long long LL;
LL n, a[maxn], p;
int main()
{
    scanf("%lld%lld", &n, &p);
    for (int i = 0; i < n; i++)scanf("%lld", &a[i]);
    sort(a, a + n);
    int maxnum = 0;
    for (int i = 0; i < n; i++)
    {
        int m = upper_bound(a + i + 1, a + n, a[i] * p) - a;
        maxnum = max(maxnum, m - i);
    }
    printf("%lld", maxnum);
    return 0;
}

two pointers:i和j都从0开始遍历（不是一个头一个尾），这样一次遍历就可以

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1e5 + 10;
typedef long long LL;
LL n, a[maxn], p;
int main()
{
    scanf("%lld%lld", &n, &p);
    for (int i = 0; i < n; i++)scanf("%lld", &a[i]);
    sort(a, a + n);
    LL maxnum = 0;
    LL i = 0, j = 0;
    while (i < n&&j<n)
    {
        LL number = a[i] * p;
        while (a[j] <= number&&j<n)j++;
        maxnum = max(maxnum, j - i);
        i++;
    }
    printf("%lld", maxnum);
    return 0;
}