Integral Array (Codeforces 1648B)
Tags: Math, data structures

You are given an array $$$a$$$ of $$$n$$$ positive integers numbered from $$$1$$$ to $$$n$$$. Let's call an array integral if for any two, not necessarily different, numbers $$$x$$$ and $$$y$$$ from this array, $$$x \ge y$$$, the number $$$\left \lfloor \frac{x}{y} \right \rfloor$$$ ($$$x$$$ divided by $$$y$$$ with rounding down) is also in this array.You are guaranteed that all numbers in $$$a$$$ do not exceed $$$c$$$. Your task is to check whether this array is integral.

Input: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. Description of the test cases follows.The first line of each test case contains two integers $$$n$$$ and $$$c$$$ ($$$1 \le n \le 10^6$$$, $$$1 \le c \le 10^6$$$) — the size of $$$a$$$ and the limit for the numbers in the array.The second line of each test case contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_i \le c$$$) — the array $$$a$$$.Let $$$N$$$ be the sum of $$$n$$$ over all test cases and $$$C$$$ be the sum of $$$c$$$ over all test cases. It is guaranteed that $$$N \le 10^6$$$ and $$$C \le 10^6$$$.

Output: For each test case print Yes if the array is integral and No otherwise.

Note: In the first test case it is easy to see that the array is integral:   $$$\left \lfloor \frac{1}{1} \right \rfloor = 1$$$, $$$a_1 = 1$$$, this number occurs in the arry  $$$\left \lfloor \frac{2}{2} \right \rfloor = 1$$$  $$$\left \lfloor \frac{5}{5} \right \rfloor = 1$$$  $$$\left \lfloor \frac{2}{1} \right \rfloor = 2$$$, $$$a_2 = 2$$$, this number occurs in the array  $$$\left \lfloor \frac{5}{1} \right \rfloor = 5$$$, $$$a_3 = 5$$$, this number occurs in the array  $$$\left \lfloor \frac{5}{2} \right \rfloor = 2$$$, $$$a_2 = 2$$$, this number occurs in the array Thus, the condition is met and the array is integral.In the second test case it is enough to see that$$$\left \lfloor \frac{7}{3} \right \rfloor = \left \lfloor 2\frac{1}{3} \right \rfloor = 2$$$, this number is not in $$$a$$$, that's why it is not integral.In the third test case $$$\left \lfloor \frac{2}{2} \right \rfloor = 1$$$, but there is only $$$2$$$ in the array, that's why it is not integral.

Sample Inputs:
4
3 5
1 2 5
4 10
1 3 3 7
1 2
2
1 1
1

1
1 1000000
1000000

Sample Outputs:
Yes
No
No
Yes

No

Solution (Python):
def solve(nums):
    c = nums[-1]
    s = set(nums)
    n = len(nums)
    if 1 not in s:
        return False
    else:
        k = 0
        for i in range(1, c + 1):
            if i not in s:
                for j in range(n):
                    if nums[j] > c // i:
                        break
                    while nums[k] < i * nums[j]:
                        k += 1
                    if nums[k] < (i + 1) * nums[j]:
                        return False
    return True
for _ in range(int(input())):
    (n, c) = list(map(int, input().split(' ')))
    nums = list(map(int, input().split(' ')))
    nums = sorted(list(set(nums)))
    if solve(nums):
        print('Yes')
    else:
        print('No')

Solution (Python):
def solve(nums):
    c = nums[-1]
    s = set(nums)
    n = len(nums)
    if 1 not in s:
        return False
    else:
        k = 0
        for i in range(1, c + 1):
            if i not in s:
                for j in range(n):
                    if nums[j] > c // i:
                        break
                    while nums[k] < i * nums[j]:
                        k += 1
                    if nums[k] < (i + 1) * nums[j]:
                        return False
    return True
for _ in range(int(input())):
    (n, c) = list(map(int, input().split(' ')))
    nums = list(map(int, input().split(' ')))
    nums = sorted(list(set(nums)))
    if solve(nums):
        print('Yes')
    else:
        print('No')

Solution (Python):
from bisect import bisect_left, bisect_right

def solve(nums):
    c = nums[-1]
    s = set(nums)
    n = len(nums)
    if 1 not in s:
        return False
    else:
        k = 0
        for i in range(1, c + 1):
            if i not in s:
                for j in range(n):
                    if nums[j] > c // i:
                        break
                    while nums[k] < i * nums[j]:
                        k += 1
                    if nums[k] < (i + 1) * nums[j]:
                        return False
    return True
for _ in range(int(input())):
    (n, c) = list(map(int, input().split(' ')))
    nums = list(map(int, input().split(' ')))
    nums = sorted(list(set(nums)))
    if solve(nums):
        print('Yes')
    else:
        print('No')