Problem:


Score : 300 points

Problem Statement

Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?

Constraints

1 \leq a, b, c \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

a \ b \ c

Output

If \sqrt{a} + \sqrt{b} < \sqrt{c}, print Yes; otherwise, print No.

Sample Input 1

2 3 9

Sample Output 1

No

\sqrt{2} + \sqrt{3} < \sqrt{9} does not hold.

Sample Input 2

2 3 10

Sample Output 2

Yes

\sqrt{2} + \sqrt{3} < \sqrt{10} holds.



Sample input:
2 3 9


Sample output:
No

Submission status: Wrong Answer

Solution code:
from math import sqrt

data = input().split()
a, b, c = [int(i) for i in data]

if sqrt(a) + sqrt(b) < sqrt(c):
  print('Yes')
else:
  print('No')

Submission status: Wrong Answer

Solution code:
A, B, C = map(int, input().split())

# yes = (4*A*B) < (C-A-B)*(C-A-B)

yes =  (A**0.5 + B**0.5 < C**0.5)

print('Yes' if yes else 'No')

Submission status: Wrong Answer

Solution code:
a,b,c = map(int, input().split())
p = 4*a*b
q = (c*c-a-b)**2

if p<q:
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
import math
a, b, c = map(int, input().split())

ra, rb, rc = math.sqrt(a), math.sqrt(b), math.sqrt(c)

if ra + rb < rc:
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
import math

a,b,c = list(map(int,input().split()))

if math.sqrt(a) + math.sqrt(b) < math.sqrt(c):
  print("Yes")
else:
  print("No")

Submission status: Wrong Answer

Solution code:
a, b, c = map(int,input().split()) # 複数の数字を変数に格納

import math

right = (c - a - b)**2
left = 4 * a * b

print(left, right)

if left < right:
    print('Yes')
else:
    print('No')

Submission status: Wrong Answer

Solution code:
from collections import *
from itertools import *
from bisect import *
from heapq import *
import copy
import math
from fractions import gcd
import sys
#input = sys.stdin.readline


#N=int(input())

#A=list(map(int,input().split()))
#XY=[list(map(int,input().split())) for i in range(N)]

a,b,c=map(int,input().split())
X=gcd(gcd(a,b),c)
c//=X
a//=X
b//=X
if c**.5-a**.5-b**.5>0:
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
import numpy as np
a, b, c = map(int, input().split())

sq_a = np.sqrt(a)
sq_b = np.sqrt(b)

if sq_a * sq_b < c - a - b:
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import permutations, combinations, product, accumulate
from operator import itemgetter, mul
from copy import deepcopy,copy
from string import ascii_lowercase, ascii_uppercase, digits
from fractions import gcd

def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7

def main():
	a, b, c = [sqrt(x) for x in LIST()]
	 
	if a + b < c:
		print("Yes")
	else:
		print("No")

if __name__=='__main__':
	main()

Submission status: Wrong Answer

Solution code:
import math
a,b,c = map(int,input().split())
 
sqa = math.sqrt(a)
sqb = math.sqrt(b)
sqc = math.sqrt(c)
 
if sqa + sqb < sqc:
  print("Yes")
else:
  print("No")

Submission status: Wrong Answer

Solution code:
a,b,c = list(map(int, input().split()))
if (a**(1/2) + b**(1/2)) < c**(1/2):
  print('Yes')
else:
  print('No')

Submission status: Wrong Answer

Solution code:
import math
a,b,c=(int(x) for x in input().split())

if math.sqrt(a) + math.sqrt(b) < math.sqrt(c):
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
import math
a, b, c = map(int,input().split())

a = math.sqrt(a)
b = math.sqrt(b)
c = math.sqrt(c)

if(a+b < c):
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
import math

a,b,c=map(int,input().split())

if math.sqrt(a)+math.sqrt(b)<math.sqrt(c):
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
import math
a, b, c = map(int, input().split())

if math.sqrt(a*b) < (c-a-b)/2:
  print('Yes')
else:
  print('No')

Submission status: Wrong Answer

Solution code:
from sys import stdin
import numpy as np
from operator import itemgetter

#N = int(stdin.readline().rstrip().split()[0])
a,b,c = [int(x) for x in stdin.readline().rstrip().split()]
#A = [int(x) for x in stdin.readline().rstrip().split()]

MOD = 10**9 + 7

sqrtab = np.sqrt(a*b)
if 2*sqrtab < c- a -b:
  print("Yes")
else:
  print("No")

Submission status: Wrong Answer

Solution code:
import math
a,b,c=[int(x) for x in input().split()]
l = math.sqrt(a*b)
l += (a+b)
if l < c:
  print('Yes')
else:
  print('No')

Submission status: Wrong Answer

Solution code:
import math
import fractions
import bisect
import collections
import itertools
import heapq
import string
import sys
import copy
from collections import deque
sys.setrecursionlimit(10**7)
MOD = 10**9+7
INF = float('inf') #無限大
def gcd(a,b):return fractions.gcd(a,b) #最大公約数
def lcm(a,b):return (a*b) // fractions.gcd(a,b) #最小公倍数
def iin(): return int(sys.stdin.readline()) #整数読み込み
def ifn(): return float(sys.stdin.readline()) #浮動小数点読み込み
def isn(): return sys.stdin.readline().split() #文字列読み込み
def imn(): return map(int, sys.stdin.readline().split()) #整数map取得
def imnn(): return map(lambda x:int(x)-1, sys.stdin.readline().split()) #整数-1map取得
def fmn(): return map(float, sys.stdin.readline().split()) #浮動小数点map取得
def iln(): return list(map(int, sys.stdin.readline().split())) #整数リスト取得
def iln_s(): return sorted(iln()) # 昇順の整数リスト取得
def iln_r(): return sorted(iln(), reverse=True) # 降順の整数リスト取得
def fln(): return list(map(float, sys.stdin.readline().split())) # 浮動小数点リスト取得
def join(l, s=''): return s.join(l) #リストを文字列に変換
def perm(l, n): return itertools.permutations(l, n) # 順列取得
def perm_count(n, r): return math.factorial(n) // math.factorial(n-r) # 順列の総数
def comb(l, n): return itertools.combinations(l, n) # 組み合わせ取得
def comb_count(n, r): return math.factorial(n) // (math.factorial(n-r) * math.factorial(r)) #組み合わせの総数
def two_distance(a, b, c, d): return ((c-a)**2 + (d-b)**2)**.5 # 2点間の距離
def m_add(a,b): return (a+b) % MOD

def sieves_of_e(n):
    is_prime = [True] * (n+1)
    is_prime[0] = False
    is_prime[1] = False
    for i in range(2, int(n**0.5)+1):
        if not is_prime[i]: continue
        for j in range(i * 2, n+1, i): is_prime[j] = False
    return is_prime

a,b,c = imn()

ma = math.sqrt(a)
mb = math.sqrt(b)
mc = math.sqrt(c) 
if (a + (2*ma*mb) + b) < c:
    print('Yes')
else:
    print('No')

Submission status: Wrong Answer

Solution code:
import numpy as np

a, b, c = [int(i) for i in input().split()]

x = np.sqrt(a) + np.sqrt(b)
y = np.sqrt(c)

if x < y:
  print('Yes')
else:
  print('No')

Submission status: Wrong Answer

Solution code:
from math import sqrt
a, b, c = map(int, input().split())
if c - a - b > 2 * sqrt(a) * sqrt(b):
    print('Yes')
else:
    print('No')

Submission status: Wrong Answer

Solution code:
import math

a,b,c = map(int,input().split())

d = math.sqrt(a)
e = math.sqrt(b)
f = math.sqrt(c)

if (d+e) < f:
    print('Yes')
else:
    print('No')

Submission status: Wrong Answer

Solution code:
import math
a, b, c = map(int, input().split())
left = 2*math.sqrt(a*b) + a + b
if left < c:
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
import math
from functools import reduce
from collections import deque
import sys
sys.setrecursionlimit(10**7)

# スペース区切りの入力を読み込んで数値リストにして返します。
def get_nums_l():
    return [ int(s) for s in input().split(" ")]

# 改行区切りの入力をn行読み込んで数値リストにして返します。
def get_nums_n(n):
    return [ int(input()) for _ in range(n)]

# 改行またはスペース区切りの入力をすべて読み込んでイテレータを返します。
def get_all_int():
    return map(int, open(0).read().split())

def log(*args):
    print("DEBUG:", *args, file=sys.stderr)

a,b,c = get_nums_l()

if c % 4 == 0 and c//4 == a == b:
    print("No")
    exit()

print("Yes" if math.sqrt(a) + math.sqrt(b) < math.sqrt(c) else "No")

Submission status: Wrong Answer

Solution code:
import math

a,b,c= map(int,input().split(' '))
if (c-b-a)<0:
    print("No")
else:

    if math.sqrt(a)+math.sqrt(b)<math.sqrt(c):
        print("Yes")
    else:
        print("No")

Submission status: Wrong Answer

Solution code:
a,b,c=map(int,input().split())
if (c-a-b)**2-4*a*b>0:
    print('Yes')
else:
    print('No')

Submission status: Wrong Answer

Solution code:
#!python3.4.3

def process(x):
  x = int(x)
  return math.sqrt(x)

import math

a, b, c = map(process, input().split(' '))

if (a + b) <= c:
  print('Yes')
else:
  print('No')

Submission status: Wrong Answer

Solution code:
import math
s = input().split(' ')
a = int(s[0])
b = int(s[1])
c = int(s[2])
if (math.sqrt(a) + math.sqrt(b)) < math.sqrt(c):
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
a, b, c = map(int, input().split())

# print(a ** 0.5 + b ** 0.5, c ** 0.5)

if a + b + 2 * ((a * b)**0.5) < c:
    print('Yes')
else:
    print('No')

Submission status: Wrong Answer

Solution code:
import math
a,b,c = map(int,input().split())

A = math.sqrt(a)
B = math.sqrt(b)
C = math.sqrt(c)

if A + B < C:
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
import math
import numpy

a,b,c=map(int,input().split())
if math.sqrt(a) + math.sqrt(b) < math.sqrt(c):
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
a, b, c = map(int,input().split())

import math
if math.sqrt(a) + math.sqrt(b) < math.sqrt(c):
    print('Yes')
else:
    print('No')

Submission status: Wrong Answer

Solution code:
#!/usr/bin/env python3
a, b, c = map(int, input().split())
print(["No", "Yes"][4 * a * b - (c - a - b) ** 2 < -1e-10])

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import itertools
input = sys.stdin.readline

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

##### 累乗 #####
# pow(x, y, z) -> x**y % z

def inputInt(): return int(input())
def inputMap(): return map(int, input().split())
def inputList(): return list(map(int, input().split()))

inf = float('inf')
mod = 1000000007

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-

def main():
	a,b,c = inputMap()

	if math.sqrt(a) + math.sqrt(b) - math.sqrt(c) < 0:
		print("Yes")
	else:
		print("No")


#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# special thanks :
#  https://nagiss.hateblo.jp/entry/2019/07/01/185421
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# nCr mod m
# rがn/2に近いと非常に重くなる
def combination(n, r, mod=10**9+7):
    r = min(r, n-r)
    res = 1
    for i in range(r):
        res = res * (n - i) * modinv(i+1, mod) % mod
    return res

# mを法とするaの乗法的逆元
def modinv(a, mod=10**9+7):
    return pow(a, mod-2, mod)

def egcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = egcd(b % a, a)
        return g, x - (b // a) * y, y

# nHr mod m
# 問題によって、combination()を切り替えること
def H(n, r, mod=10**9+7):
	# comb = Combination(n+r-1, mod)
	# return comb(n+r-1, r)
    return combination(n+r-1, r, mod)

class Combination:
    """
    O(n)の前計算を1回行うことで，O(1)でnCr mod mを求められる
    n_max = 10**6のとき前処理は約950ms (PyPyなら約340ms, 10**7で約1800ms)
    使用例：
    comb = Combination(1000000)
    print(comb(5, 3))  # 10
    """
    def __init__(self, n_max, mod=10**9+7):
        self.mod = mod
        self.modinv = self.make_modinv_list(n_max)
        self.fac, self.facinv = self.make_factorial_list(n_max)

    def __call__(self, n, r):
        return self.fac[n] * self.facinv[r] % self.mod * self.facinv[n-r] % self.mod

    def make_factorial_list(self, n):
        # 階乗のリストと階乗のmod逆元のリストを返す O(n)
        # self.make_modinv_list()が先に実行されている必要がある
        fac = [1]
        facinv = [1]
        for i in range(1, n+1):
            fac.append(fac[i-1] * i % self.mod)
            facinv.append(facinv[i-1] * self.modinv[i] % self.mod)
        return fac, facinv

    def make_modinv_list(self, n):
        # 0からnまでのmod逆元のリストを返す O(n)
        modinv = [0] * (n+1)
        modinv[1] = 1
        for i in range(2, n+1):
            modinv[i] = self.mod - self.mod//i * modinv[self.mod%i] % self.mod
        return modinv

if __name__ == "__main__":
	main()

Submission status: Wrong Answer

Solution code:
#!/usr/bin/env python3 ## -*- python -*-
import math

t = input().split()
a = int(t[0])
b = int(t[1])
c = int(t[2])

ans = "No"
if c-a-b>0:
    if 4*a*b<(c-a-b)^2:
        ans = "Yes"

print(ans)

Submission status: Wrong Answer

Solution code:
import math

a,b,c = map(float, input().split())

if math.sqrt(a) + math.sqrt(b) < math.sqrt(c):
  print("Yes")
else:
  print("No")

Submission status: Wrong Answer

Solution code:
import math
a,b,c=map(int,input().split())
A=math.sqrt(a)
B=math.sqrt(b)
C=math.sqrt(c)

if pow(A+B,2)-pow(C,2)<0:
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
# import bisect
# import heapq
# from copy import deepcopy
# from collections import deque
# from collections import Counter
# from itertools import accumulate
# from itertools import permutations
# import numpy as np
import math
a,b,c = map(int, input().split())
# n = int(input())
# p = list(map(int, input().split()))
ans = 'No'


if (math.sqrt(a*b))**2 != float(a*b) and a + b + 2 * int(math.sqrt(a*b)) < c:
    ans = 'Yes'
elif (math.sqrt(a*b))**2 == float(a*b) and a + b + 2 * int(math.sqrt(a*b)) < c:
    ans = 'Yes'
print(ans)

Submission status: Wrong Answer

Solution code:
a, b, c = map(int, input().split())

if a ** (1/2) + b ** (1/2) < c ** (1/2):
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
a,b,c=map(int,input().split())

if (a**0.5)+(b**0.5)<(c**0.5):
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
from math import sqrt
a, b, c = map(int, input().split())

tmp = 2 * (sqrt(a * b))
print("Yes" if a + b + tmp < c else "No")

Submission status: Wrong Answer

Solution code:
#!/usr/bin/env python3

import math

a, b, c = map(int, input().split())

# if math.sqrt(a) + math.sqrt(b) < math.sqrt(c):
#     print("Yes")
# else:
#     print("No")

if a + 2 * math.sqrt(a) * math.sqrt(b) < c:
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):   
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    r_float = lambda: float(r())
    R = lambda: list(map(int, r().split()))
    R_str = lambda: r().split()
    R_map = lambda: map(int, r().split())
    R_float = lambda: list(map(float, r().split()))
    R_tuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main
    
    a, b, c =R()

    if round(c**0.5 - a**0.5 - b**0.5, 20) > 0:
        print('Yes')
    else:
        print('No')
    
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Wrong Answer

Solution code:
import math
a,b,c = map(int, input().split())

if math.sqrt(a) + math.sqrt(b) < math.sqrt(c):
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
from decimal import *
getcontext().prec = 400 
a, b, c = map(int, input().split())
if Decimal(a).sqrt() + Decimal(b).sqrt() < Decimal(c).sqrt():
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
a, b, c = map(int, input().split())
p = 10 ** 9 + 7
if a ** 2 % p + b ** 2 % p + c ** 2 % p - 2 * a * b % p - 2 * b * c % p - 2 * c * a % p > 0:
    print('Yes')
else:
    print('No')

Submission status: Wrong Answer

Solution code:
s = input().split()
a = int(s[0])
b = int(s[1])
c = int(s[2])
if 4 * (a * b) < (c - a - b) * (c - a - b):
    print("Yes")
else:
    print("No")