"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
import math
def next_Perfect_Square(N):
if N < 0:
return 0
nextN = math.floor(math.sqrt(N)) + 1
return nextN * nextN
[[35], [6], [9]] [[-5], [3.14], [0], [1], [-10], [2.0], [1000000], [99999999], [1000000000000], [3.7831588542515604], [5.302859454099729], [True], [2.4682613401002937], [99999998], [1.3453957712310967], [1.0399013203829814], [99999997], [5.684417641308899], [99999996], [4.510369394904734], [3.6621080850729113], [3.322051512362575], [1000000000001], [-11], [2.3742619906307163], [2.6748360710113177], [3.4473095447178936], [5.57812470410088], [99999995], [4.647261498089122], [1.5329222521720727], [2.6809843384176872], [-1], [False], [1.5107938841218733], [0.875196021901653], [4.529854518921115], [1.7032459621984142], [1.8504217376600358], [2.188160542701604], [1.8956297256654226], [2.4314765544566113], [5.813827850518426], [2.3809692137960456], [2.446463005958287], [1.9423279622572451], [2.1946783538521406], [2.6687036689581713], [2.0326735919423498], [100000000], [2.468678626532637], [1000001], [2.5238311325906344], [1.7822986116186494], [2.3311279543074095], [-22.844596598850202], [2.5605349010057803], [6.083079662936755], [100000001], [1.5746458078326158], [6.588039035035163], [3.1483586672721007], [1000000000002], [3.456020835359328], [34.43923911097559], [1.0089233418138062], [4.2295065796525115], [-78.55181920500208], [1.0803737520419845], [2.605728433563439], [38.98925404921724], [4.4174458158812255], [1.644335866314361], [1.8218151507901879], [3.037371820443729], [2.709277112928487], [4.378962886905937], [34.821872058003486], [-77.542809657578], [2.5759758416813416], [-12], [3.1543786398736704], [0.7865023181429288], [3.2249050005349247], [0.9910005166913101], [3.195728446958819], [1.614635244145928], [2.192134061504989], [-9], [3.47933130145955], [3.806186148917977], [1.4663753283678982], [0.945455886591849], [2.805849529482408], [-6], [1.909977828399661], [2.3945594411492657], [2.943996722081795], [4.945353788299947], [-80], [2.847513147814003], [2.040611668519119], [2.6516104349052276], [3.547091908942516], [999999999999], [-21.849993780773154], [2.09983974207191], [2.7416667760599114], [2.953394448977245]] code-llama-multi-7b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
:param n: int
:return: int
"""
if n < 0:
raise ValueError("n must be a positive integer")
if n == 0:
return 1
if n == 1:
return 1
if n == 2:
return 4
if n == 3:
return 9
if n == 4:
return 9
if n == 5:
return 16
if n == 6:
return 16
if n == 7:
return 25
if n == 8:
return 25
if n == 9:
return 36
if n == 10:
return 36
if n == 11:
return 49
if n == 12:
return 49
if n == 13:
return 64
if n == 14:
return 64
if n == 15:
return 81
if n == 16:
return 81
if n == 17:
return 100
if n == 18:
return 100
if n == 19:
return 121
if n == 20:
return 121
if n == 21:
return 144
if n == 22:
return 144
if n == 23:
return 169
if n == 24:
return 169
if n == 25:
return 196
if n == 26:
return 196
if n == 27:
return 225
if n == 28:
return 225
if n == 29:
return 256
if n == 30:
databricks--dbrx-instruct. fail_base: False, fail_plus: True
def next_Perfect_Square(n):
"""
Find the next perfect square greater than a given number.
"""
return (int(n**0.5) + 1)**2
# Test the function
assert next_Perfect_Square(35) == 36CohereForAI--c4ai-command-r-plus. fail_base: True, fail_plus: True
def next_Perfect_Square(n):
import math
root = math.sqrt(n)
if int(root + 0.5) ** 2 == n:
return (int(root) + 1) ** 2
else:
return (int(root) + 2) ** 2
```claude-3-haiku-20240307. fail_base: False, fail_plus: True
Here's the solution wrapped in a Python markdown block:
```python
def next_Perfect_Square(num):
"""
Finds the next perfect square greater than a given number.
Args:
num (int): The number to find the next perfect square for.
Returns:
int: The next perfect square greater than the given number.
"""
import math
# Find the square root of the given number
sqrt = math.sqrt(num)
# If the square root is an integer, the number is already a perfect square
if sqrt.is_integer():
return int(sqrt + 1) ** 2
# Otherwise, return the next perfect square
return int(sqrt) ** 2 + 2 * int(sqrt) + 1
# Test the function
assert next_Perfect_Square(35) == 36code-millenials-34b. fail_base: True, fail_plus: True
def next_Perfect_Square(n):
# Find the next perfect square
while True:
n += 1
if n % 2 == 0:
continue
for i in range(3, int(n ** 0.5) + 1, 2):
if n % i == 0:
break
else:
return n
# Test the function
assert next_Perfect_Square(35) == 36dolphin-2.6. fail_base: False, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
# Calculate the square root of the given number
root = int(n ** 0.5)
# Check if the square root is a perfect square
if root * root == n:
# If it is, return the next perfect square
return (root + 1) ** 2
else:
# If it is not, return the square of the next integer
return (root + 1) ** 2
# Test the function
assert next_Perfect_Square(35) == 36claude-3-opus-20240229. fail_base: False, fail_plus: True
Here's the Python code to find the next perfect square greater than a given number, wrapped in a Python markdown block:
```python
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
import math
def next_Perfect_Square(num):
sqrt = math.sqrt(num)
if sqrt.is_integer():
return int((sqrt + 1) ** 2)
else:
return int(math.ceil(sqrt) ** 2)
assert next_Perfect_Square(35) == 36mixtral-8x7b-instruct. fail_base: True, fail_plus: True
def next_Perfect_Square(n):
# Calculate the square root of the given number
sqrt_n = n ** 0.5
# Round up to the nearest integer
next_sqrt = int(sqrt_n) + int(sqrt_n % 1 != 0)
# Return the next perfect square
return next_sqrt ** 2
# Test the function
assert next_Perfect_Square(35) == 36wizardcoder-34b. fail_base: False, fail_plus: True
Here's the Python function to find the next perfect square greater than a given number:
```python
def next_Perfect_Square(num):
"""
This function takes a number as input and returns the next perfect square greater than the number.
"""
root = int(num ** 0.5) + 1
return (root ** 2)
```
Explanation:
- We first find the square root of the given number using the `**` operator and then add 1 to it to get the next integer greater than the square root.
- We then square this integer to get the next perfect square greater than the given number.
Let's test the function with the given example:
```python
print(next_Perfect_Square(35)) # Output: 36
```
The output is `36` which is the next perfect square greater than `35`. mixtral-8x22b-instruct-v0.1. fail_base: True, fail_plus: True
import math
def next_perfect_square(n):
"""
Function to find the next perfect square greater than a given number.
"""
# Calculate the square root of the given number
root = math.isqrt(n)
# If the square of the root is equal to the given number,
# then the next perfect square is the square of the root plus one
if root**2 == n:
return (root + 1)**2
# Otherwise, the next perfect square is the square of the root plus one
else:
return (root + 1)**2
# Test the function
assert next_perfect_square(35) == 36code-llama-multi-34b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
return (num // 32 + 1) * 32
if __name__ == "__main__":
print(next_Perfect_Square(35))
Qwen--Qwen1.5-72B-Chat. fail_base: False, fail_plus: True
import math
def next_Perfect_Square(n):
next_square_root = math.isqrt(n) + 1
return next_square_root ** 2
# Test case provided
assert next_Perfect_Square(35) == 36
# Additional test cases
assert next_Perfect_Square(100) == 121
assert next_Perfect_Square(0) == 1
assert next_Perfect_Square(9999) == 10000mistral-large-latest. fail_base: False, fail_plus: True
Sure, here is a Python function that solves the problem:
```python
import math
def next_Perfect_Square(num):
root = math.isqrt(num)
if root*root == num:
return (root+1)**2
else:
return (math.isqrt(num)+1)**2
# Test the function
assert next_Perfect_Square(35) == 36
```
This function works by first finding the integer square root of the input number using the `math.isqrt()` function. If the square of the root is equal to the input number, it means the number is a perfect square, so the next perfect square is (root+1)**2. If the number is not a perfect square, the next perfect square is still (root+1)**2, because the root is the largest integer less than or equal to the square root of the number.codegen-2b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
if n == 1:
return 1
if n == 2:
return 2
if n == 3:
return 3
if n == 4:
return 4
if n == 5:
return 5
if n == 6:
return 6
if n == 7:
return 7
if n == 8:
return 8
if n == 9:
return 9
if n == 10:
return 10
if n == 11:
return 11
if n == 12:
return 12
if n == 13:
return 13
if n == 14:
return 14
if n == 15:
return 15
if n == 16:
return 16
if n == 17:
return 17
if n == 18:
return 18
if n == 19:
return 19
if n == 20:
return 20
if n == 21:
return 21
if n == 22:
return 22
if n == 23:
return 23
if n == 24:
return 24
if n == 25:
return 25
if n == 26:
return 26
if n == 27:
return 27
if n == 28:
return 28
if n == 29:
return 29
if n == 30:
return 30
if n == 31:
return 31
if n == 32:
return 32
if n == 33:
return 33
if n == 34:
return 34
if n == 35:
return 35
if n == 36:
return 36
if n == 37:
return 37
if n == 38:
return 38
if n == 39:
return 39
if n == 40:
return 40
if n == 41:
return 41
if n == 42:
return 42
if n == 43:
return 43
if n == 44:
return 44
if n == 45:
return 45
if n == 46:HuggingFaceH4--starchat2-15b-v0.1. fail_base: False, fail_plus: True
def next_Perfect_Square(n):
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
import math
root = math.sqrt(n)
if root.is_integer():
return (root + 1) ** 2
else:
return (math.ceil(root)) ** 2
# Test the function
assert next_Perfect_Square(35) == 36
assert next_Perfect_Square(100) == 121
assert next_Perfect_Square(121) == 144
assert next_Perfect_Square(0) == 1codegemma-7b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
import math
def next_Perfect_Square(num):
if num < 0:
return "Invalid input"
if num == 0:
return 1
if num == 1:
return 4
if num == 2:
return 9
if num == 3:
return 16
if num == 4:
return 25
if num == 5:
return 36
if num == 6:
return 49
if num == 7:
return 64
if num == 8:
return 81
if num == 9:
return 100
if num == 10:
return 121
if num == 11:
return 144
if num == 12:
return 169
if num == 13:
return 196
if num == 14:
return 225
if num == 15:
return 256
if num == 16:
return 289
if num == 17:
return 324
if num == 18:
return 361
if num == 19:
return 400
if num == 20:
return 441
if num == 21:
return 484
if num == 22:
return 529
if num == 23:
return 576
if num == 24:
return 625
if num == 25:
return 676
if num == 26:
return 729
if num == 27:
return 784
if num == 28:
return 841
if num == 29:
return 900
if num == 30:
return 961
if num == 31:
code-llama-multi-13b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
:param n:
:return:
"""
if n < 1:
return 1
if n == 1:
return 1
if n == 2:
return 4
if n == 3:
return 9
if n == 4:
return 16
if n == 5:
return 25
if n == 6:
return 36
if n == 7:
return 49
if n == 8:
return 64
if n == 9:
return 81
if n == 10:
return 100
if n == 11:
return 121
if n == 12:
return 144
if n == 13:
return 169
if n == 14:
return 196
if n == 15:
return 225
if n == 16:
return 256
if n == 17:
return 289
if n == 18:
return 324
if n == 19:
return 361
if n == 20:
return 400
if n == 21:
return 441
if n == 22:
return 484
if n == 23:
return 529
if n == 24:
return 576
if n == 25:
return 625
if n == 26:
return 676
if n == 27:
return 729
if n == 28:
return 784
if n == 29:
return 841
if n == 30:
return 900
if n == 31:
deepseek-coder-6.7b-instruct. fail_base: True, fail_plus: True
Here is the Python function in a markdown style code block:
```python
def next_Perfect_Square(n):
"""
This function takes an integer n as input and returns the next perfect square
greater than n.
"""
# Calculate the square root of n
root = n ** 0.5
# If root is not an integer, increment it by 1
if not root.is_integer():
root += 1
# Return the square of the new root
return int(root ** 2)
# Test the function
assert next_Perfect_Square(35) == 36
```
This function works by first calculating the square root of the input number. If the square root is not an integer, it increments the square root by 1. Finally, it returns the square of the new square root, which is the next perfect square greater than the input number.
open-hermes-2.5-code-290k-13b. fail_base: False, fail_plus: True
def next_Perfect_Square(n):
if n < 0:
return None
elif n == 0:
return 0
else:
i = n
while True:
i += 1
root = i ** 0.5
if int(root + 0.5) ** 2 == i:
return icodet5p-6b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
i = n
while i*i <= n:
i += 1
return i*i
print(next_Perfect_Square(35))
print(next_Perfect_Square(36))
print(next_Perfect_Square(37))
print(next_Perfect_Square(38))
print(next_Perfect_Square(39))
print(next_Perfect_Square(40))
print(next_Perfect_Square(41))
print(next_Perfect_Square(42))
print(next_Perfect_Square(43))
print(next_Perfect_Square(44))
print(next_Perfect_Square(45))
print(next_Perfect_Square(46))
print(next_Perfect_Square(47))
print(next_Perfect_Square(48))
print(next_Perfect_Square(49))
print(next_Perfect_Square(50))
print(next_Perfect_Square(51))
print(next_Perfect_Square(52))
print(next_Perfect_Square(53))
print(next_Perfect_Square(54))
print(next_Perfect_Square(55))
print(next_Perfect_Square(56))
print(next_Perfect_Square(57))
print(next_Perfect_Square(58))
print(next_Perfect_Square(59))
print(next_Perfect_Square(60))
print(next_Perfect_Square(61))
print(next_Perfect_Square(62))
print(next_Perfect_Square(63))
print(next_Perfect_Square(64))
print(next_Perfect_Square(65))
print(next_Perfect_Square(66))
print(next_Perfect_Square(67))
print(next_Perfect_Square(68))
print(next_Perfect_Square(69))
print(next_Perfect_Square(70))
print(next_Perfect_Square(71))
print(next_Perfect_Square(72))
print(next_Perfect_Square(73))
print(next_Perfect_Square(74))
print(next_Perfect_Square(75))
print(next_Perfect_Square(76))
print(next_Perfect_Square(77))
print(codegemma-7b-it. fail_base: False, fail_plus: True
def next_Perfect_Square(num):
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
if num**0.5 == int(num**0.5):
return (int(num**0.5)+1)**2
else:
return (int(num**0.5)+1)**2mistralai--Mistral-7B-Instruct-v0.2. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_perfect_square(n):
"""
Returns the smallest perfect square greater than the given number.
"""
sqrt = n ** 0.5
return int(sqrt + 1) ** 2
# Test case
assert next_perfect_square(35) == 36xwincoder-34b. fail_base: True, fail_plus: True
import math
def next_Perfect_Square(n):
root = math.ceil(math.sqrt(n))
return root ** 2
assert next_Perfect_Square(35) == 36wizardcoder-15b. fail_base: True, fail_plus: True
Here's the Python function to find the next perfect square greater than a given number:
```python
def next_Perfect_Square(num):
"""
This function takes a number as input and returns the next perfect square greater than the given number.
"""
# Find the square root of the given number
sqrt_num = int(num ** 0.5)
# Check if the square root is an integer
if sqrt_num ** 2 == num:
# If the square root is an integer, then the next perfect square is the square of the integer plus one
return sqrt_num ** 2 + 1
else:
# If the square root is not an integer, then the next perfect square is the square of the integer plus one
return (sqrt_num + 1) ** 2
```
Here's how the function works:
1. First, we find the square root of the given number using the `**` operator.
2. We then check if the square root is an integer or not. If it is an integer, then the next perfect square is the square of the integer plus one. If it is not an integer, then the next perfect square is the square of the integer plus one.
3. Finally, we return the next perfect square.
We can test the function using the `assert` statement as follows:
```python
assert next_Perfect_Square(35) == 36
```
This will raise an `AssertionError` if the function returns an incorrect value.code-llama-34b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
for i in range(n+1, 1000000):
if i**2 > n:
return i**2
print(next_Perfect_Square(35))
deepseek-coder-33b-instruct. fail_base: True, fail_plus: True
Here is the Python function to find the next perfect square greater than a given number:
```python
import math
def next_Perfect_Square(n):
root = math.sqrt(n)
if int(root + 0.5) ** 2 == n:
return n
else:
return (int(root) + 1) ** 2
print(next_Perfect_Square(35)) # Output: 36
```
This function works by first calculating the square root of the given number. If the square of the integer part of the square root plus 0.5 is equal to the original number, then the number is a perfect square and it is returned as is. Otherwise, the next perfect square is calculated by adding 1 to the integer part of the square root and squaring the result.
phi-2. fail_base: False, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
i = 1
while i*i <= n:
i += 1
return i*i
assert next_Perfect_Square(35) == 36
codet5p-2b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
if num == 1:
return 1
else:
for i in range(num, 0, -1):
if i*i == num:
return i
print(next_Perfect_Square(35))
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
if num == 1:
return 1
else:
for i in range(num, 0, -1):
if i*i == num:
return i
print(next_Perfect_Square(35))
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
if num == 1:
return 1
else:
for i in range(num, 0, -1):
if i*i == num:
return i
print(next_Perfect_Square(35))
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
if num == 1:
return 1
else:
for i in range(num, 0, -1):
if i*i == num:
return i
print(next_Perfect_Square(35))
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
if num == 1:
return 1
else:
for i in range(num, 0, -1):
if i*i == num:
return i
print(next_Perfect_Square(35))
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
if num == 1:bigcode--starcoder2-15b-instruct-v0.1. fail_base: True, fail_plus: True
import math
def next_Perfect_Square(num):
root = math.sqrt(num)
if root.is_integer():
return num + 1
else:
return (math.ceil(root) + 1) ** 2codegemma-2b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
if num < 0:
return "Invalid input"
else:
num = int(num)
num = num + 1
while True:
if num * num > num:
return num * num
else:
num = num + 1
<|file_separator|>white-rabbit-neo-33b-v1. fail_base: False, fail_plus: True
def next_Perfect_Square(num):
# Calculate the square root of the number
sqrt_num = num ** 0.5
# Check if the square root is an integer
if sqrt_num.is_integer():
# If it is, find the next perfect square by adding 1 to the square root
next_perfect_square = (sqrt_num + 1) ** 2
else:
# If it's not, find the nearest integer and then find the next perfect square
nearest_int = int(sqrt_num) + 1
next_perfect_square = nearest_int ** 2
return next_perfect_square
# Test the function
assert next_Perfect_Square(35) == 36microsoft--Phi-3-mini-4k-instruct. fail_base: True, fail_plus: True
import math
def next_perfect_square(n):
"""
Find the next perfect square greater than a given number.
:param n: The given number
:return: The next perfect square greater than the given number
"""
root = math.isqrt(n)
next_square = (root + 1) ** 2
return next_square
# Test the function
assert next_perfect_square(35) == 36
assert next_perfect_square(100) == 121
assert next_perfect_square(1) == 4
speechless-codellama-34b. fail_base: False, fail_plus: True
def next_Perfect_Square(n):
i = n + 1
while True:
if i**0.5 == int(i**0.5):
return i
i += 1
assert next_Perfect_Square(35) == 36
```
Explanation:
The function `next_Perfect_Square` takes an integer `n` as input and returns the next perfect square greater than `n`. The function starts by initializing a variable `i` to `n+1`. It then enters a loop that continues until a perfect square is found. In each iteration of the loop, the function checks if the square root of `i` is an integer. If it is, then `i` is a perfect square and the function returns `i`. Otherwise, `i` is incremented by 1 and the loop continues.
The `assert` statement at the end of the code is used to test the function. It checks that `next_Perfect_Square(35)` returns `36`, which is the next perfect square greater than 35.
meta-llama-3-70b-instruct. fail_base: False, fail_plus: True
def next_Perfect_Square(n):
root = n ** 0.5
if int(root + 0.5) ** 2 == n:
return (root + 1) ** 2
else:
return (int(root) + 1) ** 2
stable-code-3b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
if num == 0:
return 0
if num == 1:
return 1
if num == 2:
return 4
if num == 3:
return 9
if num == 4:
return 16
if num == 5:
return 25
if num == 6:
return 36
if num == 7:
return 49
if num == 8:
return 64
if num == 9:
return 81
if num == 10:
return 100
if num == 11:
return 121
if num == 12:
return 144
if num == 13:
return 169
if num == 14:
return 196
if num == 15:
return 225
if num == 16:
return 256
if num == 17:
return 289
if num == 18:
return 324
if num == 19:
return 361
if num == 20:
return 400
if num == 21:
return 441
if num == 22:
return 484
if num == 23:
return 529
if num == 24:
return 576
if num == 25:
return 625
if num == 26:
return 676
if num == 27:
return 729
if num == 28:
return 784
if num == 29:
return 841
if num == 30:
return 900
if num == 31:
return 961
if num == 32:
return 1024
if num == 33:
return 1089
if num == 34:
return 1156
if num == 35:
return 1225
if num == 36:
return 1296
if num == 37:
return 1369
if num == 38:
return 1444
if num == 39:
return 1521
if num == 40:
return 1600
if num == 41:
return 1681
if num == 42:
return 1764
if num == 43:
return 1849
ifcodegen-6b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
>>> next_Perfect_Square(35)
36
"""
if n == 1:
return 2
else:
return int(n**0.5) + 1
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
>>> next_Perfect_Square(35)
36
"""
if n == 1:
return 2
else:
return int(n**0.5) + 1
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
>>> next_Perfect_Square(35)
36
"""
if n == 1:
return 2
else:
return int(n**0.5) + 1
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
>>> next_Perfect_Square(35)
36
"""
if n == 1:
return 2
else:
return int(n**0.5) + 1
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
>>> next_Perfect_Square(35)
36
"""
if n == 1:
return 2
else:
return int(n**0.5) + 1
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
>>> next_Perfect_Square(35)
36
"""
if n == 1:
return 2
else:
gpt-4-1106-preview. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
import math
def next_Perfect_Square(n):
root = math.ceil(math.sqrt(n))
return root ** 2
# Example usage:
# result = next_Perfect_Square(35)
# print(result) # Output: 36
gemma-7b-it. fail_base: True, fail_plus: True
import math
def next_Perfect_Square(n):
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
# Find the square root of n.
sqrt_n = int(math.sqrt(n))
# If the square root is not perfect, round up to the next integer.
if sqrt_n * sqrt_n != n:
sqrt_n += 1
# Return the square of the next perfect square.
return sqrt_n ** 2opencodeinterpreter-ds-33b. fail_base: False, fail_plus: True
```python
def next_Perfect_Square(num):
i = num + 1
while True:
if (i**0.5).is_integer():
return i
i += 1
# Test the function
print(next_Perfect_Square(35)) # Output: 36openchat. fail_base: True, fail_plus: True
import math
def next_Perfect_Square(n):
# Find the next perfect square by taking the square root of n, rounding up to the nearest integer,
# and then squaring the result.
next_square = math.ceil(math.sqrt(n)) ** 2
return next_square
# Test the function
assert next_Perfect_Square(35) == 36speechless-starcoder2-7b. fail_base: True, fail_plus: True
import math
def next_Perfect_Square(n):
# Find the square root of the number
sqrt = math.sqrt(n)
# Check if the square root is an integer
if sqrt.is_integer():
return n + 1
# If not, find the next perfect square
else:
# Find the next integer
next_int = math.ceil(sqrt)
# Return the square of the next integer
return next_int ** 2
# Test the function
assert next_Perfect_Square(35) == 36wizardcoder-7b. fail_base: False, fail_plus: True
Here's the Python function to find the next perfect square greater than a given number:
```python
def next_Perfect_Square(num):
"""
This function takes a number as input and returns the next perfect square greater than the given number.
"""
# Find the square root of the number
sqrt_num = int(num ** 0.5)
# If the square root is an integer, return the next perfect square
if sqrt_num ** 2 == num:
return (sqrt_num + 1) ** 2
# If the square root is not an integer, return the next integer square
else:
return (sqrt_num + 1) ** 2
```
We first find the square root of the given number using the `**` operator. If the square root is an integer, we return the next perfect square by adding 1 to it and squaring it. If the square root is not an integer, we return the next integer square by adding 1 to it and squaring it. octocoder. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
if n == 1:
return 1
i = 1
while i * i < n:
i += 1
return i * i
gemma-1.1-7b-it. fail_base: True, fail_plus: True
import math
def next_Perfect_Square(num):
"""
Finds the next perfect square greater than a given number.
Args:
num (int): The given number.
Returns:
int: The next perfect square greater than the given number.
"""
return math.ceil(math.sqrt(num)) ** 2deepseek-coder-6.7b-base. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
:param n: int
:return: int
"""
if n < 0:
return -1
if n == 0:
return 1
if n == 1:
return 4
if n == 2:
return 9
if n == 3:
return 16
if n == 4:
return 25
if n == 5:
return 36
if n == 6:
return 49
if n == 7:
return 64
if n == 8:
return 81
if n == 9:
return 100
if n == 10:
return 121
if n == 11:
return 144
if n == 12:
return 169
if n == 13:
return 196
if n == 14:
return 225
if n == 15:
return 256
if n == 16:
return 289
if n == 17:
return 324
if n == 18:
return 361
if n == 19:
return 400
if n == 20:
return 441
if n == 21:
return 484
if n == 22:
return 529
if n == 23:
return 576
if n == 24:
return 625
if n == 25:
return 676
if n == 26:
return 729
if n == 27:
return 784
if n == 28:
return 841
if n == 29:
return 900
if n == 30:
return 961
if n == 31:
return 1024
if n == 32:
codegen-16b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
>>> next_Perfect_Square(35)
36
"""
if n == 1:
return 2
else:
return int(n**0.5) + 1
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
>>> next_Perfect_Square(35)
36
"""
if n == 1:
return 2
else:
return int(n**0.5) + 1
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
>>> next_Perfect_Square(35)
36
"""
if n == 1:
return 2
else:
return int(n**0.5) + 1
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
>>> next_Perfect_Square(35)
36
"""
if n == 1:
return 2
else:
return int(n**0.5) + 1
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
>>> next_Perfect_Square(35)
36
"""
if n == 1:
return 2
else:
return int(n**0.5) + 1
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
"""
>>> next_Perfect_Square(35)
36
"""
if n == 1:
return 2
else:
mistral-7b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
if n % 2 == 0:
return n + 1
else:
return n + 2
print(next_Perfect_Square(35))speechless-mistral-7b. fail_base: False, fail_plus: True
def next_Perfect_Square(n):
i = n + 1
while True:
if i ** 0.5 == int(i ** 0.5):
return i
i += 1
assert next_Perfect_Square(35) == 36
```
codet5p-16b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
if num == 1:
return 2
if num == 2:
return 3
if num == 3:
return 4
if num == 4:
return 5
if num == 5:
return 6
if num == 6:
return 7
if num == 7:
return 8
if num == 8:
return 9
if num == 9:
return 10
if num == 10:
return 11
if num == 11:
return 12
if num == 12:
return 13
if num == 13:
return 14
if num == 14:
return 15
if num == 15:
return 16
if num == 16:
return 17
if num == 17:
return 18
if num == 18:
return 19
if num == 19:
return 20
if num == 20:
return 21
if num == 21:
return 22
if num == 22:
return 23
if num == 23:
return 24
if num == 24:
return 25
if num == 25:
return 26
if num == 26:
return 27
if num == 27:
return 28
if num == 28:
return 29
if num == 29:
return 30
if num == 30:
return 31
if num == 31:
return 32
if num == 32:
return 33
if num == 33:
return 34
if num == 34:
return 35
if num == 35:
return 36
if num == 36:
return 37
if num == 37:
return 38
if num == 38:
return 39
if num == 39:
return 40
if num == 40:
return 41
if num == 41:
return 42
if num == 42:
return 43
if num == 43:
return 44
if num == 44:
return 45
if num == 45:
return 46
if num == 46:code-llama-13b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
if num < 0:
return -1
if num == 0:
return 1
if num == 1:
return 1
if num == 2:
return 4
if num == 3:
return 4
if num == 4:
return 9
if num == 5:
return 9
if num == 6:
return 9
if num == 7:
return 9
if num == 8:
return 9
if num == 9:
return 16
if num == 10:
return 16
if num == 11:
return 16
if num == 12:
return 16
if num == 13:
return 16
if num == 14:
return 16
if num == 15:
return 16
if num == 16:
return 25
if num == 17:
return 25
if num == 18:
return 25
if num == 19:
return 25
if num == 20:
return 25
if num == 21:
return 25
if num == 22:
return 25
if num == 23:
return 25
if num == 24:
return 25
if num == 25:
return 36
if num == 26:
return 36
if num == 27:
return 36
if num == 28:
return 36
if num == 29:
return 36
if num == 30:
return 36
if num == 31:
return 36
if num == 32:
return 36
if num == 33:
xdan-l1-chat. fail_base: True, fail_plus: True
import math
def next_perfect_square(n):
sqrt_n = int(math.sqrt(n))
if sqrt_n * sqrt_n == n:
return sqrt_n * sqrt_n + 1
else:
return int(sqrt_n + 1) ** 2
assert next_perfect_square(35) == 36solar-10.7b-instruct. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_perfect_square(number: int) -> int:
# Find the square root of the given number
root = int(number ** 0.5)
# Round up the square root to the nearest integer
rounded_root = root + (root % 1 > 0.5)
# Calculate the next perfect square
next_perfect_square = rounded_root ** 2
return next_perfect_square
assert next_perfect_square(35) == 36gemma-2b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(n):
if n == 1:
return 1
else:
for i in range(n, 0, -1):
if i * i == n:
return i
else:
pass
claude-3-sonnet-20240229. fail_base: False, fail_plus: True
Here's the Python code to find the next perfect square greater than a given number, wrapped in a markdown block:
```python
import math
def next_Perfect_Square(n):
"""
Returns the next perfect square greater than the given number.
Args:
n (int): The given number.
Returns:
int: The next perfect square greater than n.
"""
# Find the square root of n using math.sqrt()
sqrt = int(math.sqrt(n))
# If the square root is a perfect square, return the next perfect square
if sqrt * sqrt == n:
return (sqrt + 1) ** 2
else:
# Otherwise, find the smallest perfect square greater than n
while True:
sqrt += 1
square = sqrt ** 2
if square > n:
return square
# Test the function
assert next_Perfect_Square(35) == 36
print(next_Perfect_Square(35)) # Output: 36
print(next_Perfect_Square(16)) # Output: 25
print(next_Perfect_Square(25)) # Output: 36starcoder2-15b-oci. fail_base: False, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
while True:
num += 1
if num ** 0.5 % 1 == 0:
return num
starcoder2-3b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
return num + (num % 2)
print(next_Perfect_Square(35))
/Python/Python_Basics/01_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basics/02_Python_Basicsspeechless-starcoder2-15b. fail_base: True, fail_plus: True
import math
def next_Perfect_Square(n):
return math.ceil(math.sqrt(n)) + 1
assert next_Perfect_Square(35) == 36gemma-7b. fail_base: True, fail_plus: True
"""
Write a python function to find the next perfect square greater than a given number.
assert next_Perfect_Square(35) == 36
"""
def next_Perfect_Square(num):
if num < 1:
return -1
if num == 1:
return 1
if num == 2:
return 4
if num == 3:
return 4
if num == 4:
return 9
if num == 5:
return 9
if num == 6:
return 9
if num == 7:
return 9
if num == 8:
return 16
if num == 9:
return 16
if num == 10:
return 16
if num == 11:
return 16
if num == 12:
return 16
if num == 13:
return 16
if num == 14:
return 16
if num == 15:
return 16
if num == 16:
return 25
if num == 17:
return 25
if num == 18:
return 25
if num == 19:
return 25
if num == 20:
return 25
if num == 21:
return 25
if num == 22:
return 25
if num == 23:
return 25
if num == 24:
return 25
if num == 25:
return 36
if num == 26:
return 36
if num == 27:
return 36
if num == 28:
return 36
if num == 29:
return 36
if num == 30:
return 36
if num == 31:
return 36
if num == 32:
return 36
if num == 33:
return 36
if num == 34:
opencodeinterpreter-ds-6.7b. fail_base: True, fail_plus: True
```python
import math
def next_Perfect_Square(n):
root = math.sqrt(n)
if int(root + 0.5) ** 2 == n:
return n
else:
return (int(root + 1) ** 2)
# Test the function
print(next_Perfect_Square(35))