description
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Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
class CodeforcesTask108BSolution:
def __init__(self):
self.result = ""
self.n = 0
self.bases = []
def read_input(self):
self.n = int(input())
self.bases = [int(x) for x in input().split(" ")]
def process_task(self):
self.bases = list(set(self.bases))
self.bases.sort(reverse=True)
ending = True
for x in range(len(self.bases) - 1):
if not self.bases[x] >= 2 * self.bases[x + 1]:
ending = False
break
if ending:
self.result = "NO"
else:
self.result = "YES"
def get_result(self):
return self.result
Solution = CodeforcesTask108BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
|
CLASS_DEF FUNC_DEF ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR LIST FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR BIN_OP NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR STRING ASSIGN VAR STRING FUNC_DEF RETURN VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
n = int(input())
l = list(map(int, input().split()))
l.sort()
done = 0
for i in range(n - 1):
if l[i] < l[i + 1] and 2 * l[i] > l[i + 1]:
done = 1
break
if done == 1:
print("YES")
else:
print("NO")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER BIN_OP NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
n = int(input())
flag = True
a = list(map(int, input().split()))
a.sort()
for x in range(1, len(a)):
if a[x] < 2 * a[x - 1] and a[x] != a[x - 1]:
print("YES")
flag = False
break
if flag:
print("NO")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR BIN_OP NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
n = int(input())
a = list(map(int, input().split()))
a = set(a)
a = sorted(a)
for i in range(len(a) - 1):
if a[i] + a[i] > a[i + 1]:
print("YES")
exit(0)
print("NO")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
n = int(input())
l = [int(x) for x in input().split()]
l = list(set(l))
flag = 0
l.sort()
for i in range(len(l) - 1):
if l[i + 1] < 2 * l[i]:
flag = 1
print("YES")
break
if not flag:
print("NO")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER BIN_OP NUMBER VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR EXPR FUNC_CALL VAR STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
input()
b = sorted(list({int(x) for x in input().split()}))
for i in range(1, len(b)):
if 2 * b[i - 1] > b[i]:
print("YES")
break
else:
print("NO")
|
EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP NUMBER VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
n = int(input())
array = list(map(int, input().split(" ")))
array = list(set(array))
array.sort()
flag = False
for i in range(len(array) - 1):
if 2 * array[i] > array[i + 1]:
flag = True
break
if flag:
print("YES")
else:
print("NO")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF BIN_OP NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
x = int(input())
a = list(map(int, input().split()))
a.sort()
f = 0
for i in range(1, x):
if a[i] < 2 * a[i - 1] and a[i] != a[i - 1]:
f = 1
break
j = a[0]
if len(set(a)) == 1 and j == 1:
f = 0
if f:
print("YES")
else:
print("NO")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR BIN_OP NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
for i in range(1, len(a)):
f, s = a[i - 1], a[i]
if f == s:
continue
if f != 1:
f *= 2
if f > s:
print("YES")
exit()
print("NO")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR VAR IF VAR VAR IF VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
n = int(input())
a = sorted(list(map(int, input().split())))
ok = True
for i in range(1, len(a)):
if a[i] != a[i - 1] and a[i - 1] * 2 > a[i]:
ok = False
break
print("NO" if ok else "YES")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR STRING STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
n, t = int(input()) - 1, sorted(list(map(int, input().split())))
print("YNEOS"[all(2 * t[i] <= t[i + 1] for i in range(n) if t[i] < t[i + 1]) :: 2])
|
ASSIGN VAR VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR STRING FUNC_CALL VAR BIN_OP NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER NUMBER
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
I = input
R = range(int(I()) - 1)
a = sorted(map(int, I().split()))
s = "NO"
for i in R:
if a[i] < a[i + 1] and a[i] * 2 > a[i + 1]:
s = "YES"
print(s)
|
ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR STRING FOR VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
n = int(input())
A = sorted(list(set(map(int, input().split()))))
good = True
for i in range(1, len(A)):
if A[i - 1] * 2 > A[i]:
good = False
if not good:
print("YES")
else:
print("NO")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
from sys import *
n = int(stdin.readline())
a = [int(z) for z in stdin.readline().split()]
a.sort()
na = [a[0]]
for i in a:
if i > na[-1]:
na.append(i)
for i in range(1, len(na)):
if na[i] < 2 * na[i - 1]:
stdout.write("YES\n")
exit(0)
stdout.write("NO\n")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST VAR NUMBER FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR BIN_OP NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
n = int(input())
a = [int(x) for x in input().split()]
a.sort()
ans = "NO"
for i in range(1, n):
if a[i] < 2 * a[i - 1] and a[i] != a[i - 1]:
ans = "YES"
break
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR BIN_OP NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
n = int(input())
lst = list(map(int, input().strip().split(" ")))
lst.sort()
f = 0
for j in range(1, n):
if lst[j] < 2 * lst[j - 1] and lst[j] != lst[j - 1]:
f = 1
print("YES")
break
if f == 0:
print("NO")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR BIN_OP NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
from sys import exit
n = int(input())
lst = [*set(map(int, input().split()))]
lst.sort()
for i, x in enumerate(lst[:-1]):
if x * 2 > lst[i + 1]:
from sys import exit
print("YES")
exit()
print("NO")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR VAR FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
def s():
input()
a = list(map(int, input().split()))
a.sort()
for i in range(1, len(a)):
if a[i] != a[i - 1] and a[i] < a[i - 1] * 2:
return "YES"
return "NO"
print(s())
|
FUNC_DEF EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER RETURN STRING RETURN STRING EXPR FUNC_CALL VAR FUNC_CALL VAR
|
Tattah's youngest brother, Tuftuf, is new to programming.
Since his older brother is such a good programmer, his biggest dream is to outshine him. Tuftuf is a student at the German University in Cairo (GUC) where he learns to write programs in Gava.
Today, Tuftuf was introduced to Gava's unsigned integer datatypes. Gava has n unsigned integer datatypes of sizes (in bits) a1, a2, ... an. The i-th datatype have size ai bits, so it can represent every integer between 0 and 2ai - 1 inclusive.
Tuftuf is thinking of learning a better programming language. If there exists an integer x, such that x fits in some type i (in ai bits) and x·x does not fit in some other type j (in aj bits) where ai < aj, then Tuftuf will stop using Gava.
Your task is to determine Tuftuf's destiny.
Input
The first line contains integer n (2 ≤ n ≤ 105) — the number of Gava's unsigned integer datatypes' sizes. The second line contains a single-space-separated list of n integers (1 ≤ ai ≤ 109) — sizes of datatypes in bits. Some datatypes may have equal sizes.
Output
Print "YES" if Tuftuf will stop using Gava, and "NO" otherwise.
Examples
Input
3
64 16 32
Output
NO
Input
4
4 2 1 3
Output
YES
Note
In the second example, x = 7 (1112) fits in 3 bits, but x2 = 49 (1100012) does not fit in 4 bits.
|
n = int(input())
A = list(map(int, input().split()))
A.sort()
per = 0
per2 = True
while per < n - 1:
if A[per] < A[per + 1]:
if A[per + 1] < A[per] * 2:
per2 = False
break
else:
per += 1
else:
per += 1
if per2:
print("NO")
else:
print("YES")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
_ = int(input())
for T in range(_):
h, c, t = [int(x) for x in input().split()]
if t <= (h + c) / 2:
print(2)
continue
x = (t - c) // (2 * t - h - c)
v1 = x * h + (x - 1) * c
v2 = (x + 1) * h + x * c
if abs(v1 - (2 * x - 1) * t) / (2 * x - 1) <= abs(t * (2 * x + 1) - v2) / (
2 * x + 1
):
print(2 * x - 1)
else:
print(2 * x + 1)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR IF BIN_OP FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
eps = 1e-10
t = int(input())
for _ in range(t):
h, c, t = map(int, input().split())
l = 0
r = 100000000
while l + 1 < r:
mid = (l + r) // 2
if mid % 2 == 0:
if mid + 1 < r:
mid += 1
elif mid - 1 > l:
mid -= 1
else:
break
p = (mid // 2 + 1) * h + mid // 2 * c
if p / mid - t > -eps:
l = mid
else:
r = mid
r = l + 2
for_2 = abs(((h + c) / 2 - t) * l * r)
for_l = abs(((l // 2 + 1) * h + l // 2 * c) * r - t * l * r)
for_r = abs(((r // 2 + 1) * h + r // 2 * c) * l - t * l * r)
m = min(for_2, min(for_l, for_r))
if abs(m - for_2) < eps:
print(2)
elif abs(m - for_l) < eps:
print(l)
else:
print(r)
|
ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR NUMBER VAR VAR NUMBER IF BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR IF BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for _ in range(int(input())):
a, b, c = map(int, input().split())
mid = (a + b) / 2
if c == a:
print(1)
elif c <= mid:
print(2)
elif c >= mid + (a - b) / 6 and c < a:
d = mid + (a - b) / 6
if abs(d - c) < abs(a - c):
print(3)
else:
print(1)
else:
c -= mid
d = []
x = (a - b) // c
for i in range(int(x) - 4, int(x) + 5):
if i % 2 == 0 and i // 2 % 2 != 0:
d.append([abs((a - b) / i - c), i])
d.sort()
print(d[0][1] // 2)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR ASSIGN VAR LIST ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR LIST FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def check(h, c, t, n):
return (n + 1) * h + n * c >= (2 * n + 1) * t
def binsearch(h, c, t):
left = 0
right = 1000000000
while right - left > 1:
mid = (right + left) // 2
if check(h, c, t, mid):
left = mid
else:
right = mid
if check(h, c, t, right):
return right
else:
return left
T = int(input())
for z in range(0, T):
h, c, t = [int(i) for i in input().split(" ")]
if 2 * t <= h + c:
print(2)
else:
n = binsearch(h, c, t)
if (2 * n + 3) * ((n + 1) * h + n * c - (2 * n + 1) * t) <= (2 * n + 1) * (
-(n + 2) * h - (n + 1) * c + (2 * n + 3) * t
):
print(2 * n + 1)
else:
print(2 * n + 3)
|
FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR RETURN VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING IF BIN_OP NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for _ in range(int(input())):
h, c, t = map(int, input().split())
if h + c >= t * 2:
print(2)
elif h <= t:
print(1)
else:
k = (h - t) // (t * 2 - h - c)
x = k * k * 2 + k * 5
y = k * k * 2 + k * 3
if (x + 3 + x + 2) * h + (y + y + 1) * c > 2 * (4 * k * k + 8 * k + 3) * t:
k += 1
print(k * 2 + 1)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR NUMBER IF BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP BIN_OP NUMBER BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR VAR BIN_OP NUMBER VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def f(x):
return abs(((x - 1) * (c + h) + h) / (2 * x - 1) - temp)
t = int(input())
for you in range(t):
l = input().split()
h = int(l[0])
c = int(l[1])
temp = int(l[2])
if temp <= (c + h) / 2:
print(2)
elif temp >= h:
print(1)
else:
poss = int((temp - c) / (2 * temp - h - c))
num1 = 2 * temp * (2 * poss - 1) * (2 * poss + 1)
num2 = ((poss - 1) * (c + h) + h) * (2 * poss + 1) + (poss * (c + h) + h) * (
2 * poss - 1
)
if num1 >= num2:
print(2 * poss - 1)
else:
print(2 * poss + 1)
|
FUNC_DEF RETURN FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
def gift():
for _ in range(t):
h, c, tem = list(map(int, input().split()))
if h == tem:
yield 1
elif (h + c) / 2 > tem:
mid = (h + c) / 2
edif = abs(h - c)
ldif = abs(mid - c)
if edif >= ldif:
yield 2
else:
yield 1
elif (h + c) / 2 == tem:
yield 2
else:
dif = tem - (h + c) / 2
posans = (h - c) / 2 // dif
bibi = (h - c) / 2 % dif
if posans % 2 == 0:
posans += 1
oposans = posans + 2
pdif = abs(tem - ((h + c) / 2 + (h - c) / 2 / posans))
edif = abs(tem - ((h + c) / 2 + (h - c) / 2 / oposans))
if oposans >= 3:
kposans = posans - 2
kdif = abs(tem - ((h + c) / 2 + (h - c) / 2 / kposans))
mindif = min(pdif, edif, kdif)
if mindif == pdif:
yield int(posans)
elif mindif == edif:
yield int(oposans)
else:
yield int(kposans)
elif pdif < edif:
yield int(posans)
else:
yield int(oposans)
t = int(input())
ans = gift()
print(*ans, sep="\n")
|
IMPORT ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_DEF FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR EXPR NUMBER EXPR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR STRING
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for _ in range(int(input())):
h, c, t = map(int, input().split())
if h == t:
print(1)
elif t - c <= h - t:
print(2)
else:
temp = (t - h) / (h + c - 2 * t)
if int(temp) == temp:
print(int(2 * temp + 1))
else:
a = int(temp)
b = a + 1
if 2 * t * (2 * a + 1) * (2 * b + 1) >= (2 * b + 1) * (
(a + 1) * h + a * c
) + (2 * a + 1) * ((b + 1) * h + b * c):
print(2 * a + 1)
else:
print(2 * b + 1)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP BIN_OP NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for tt in range(int(input())):
h, c, t = map(int, input().split())
if t <= (h + c) // 2:
print(2)
else:
num = int((t - h) / (h + c - 2 * t))
num = 2 * num + 1
if num % 2:
op1 = num
op2 = num + 2
c1 = abs((op1 // 2 + 1) * h + op1 // 2 * c - t * op1)
c2 = abs((op2 // 2 + 1) * h + op2 // 2 * c - t * op2)
if c1 * op2 <= c2 * op1:
print(op1)
else:
print(op2)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR IF BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
(T,) = map(int, input().split())
def f(h, c, n):
return (h * (n + 1) + c * n) / (2 * n + 1)
def B(h, c, n):
return 2 * n + 1
def C(h, c, n):
return h * (n + 1) + c * n
def H(h, c, t, n):
return (
2 * t * B(h, c, n) * B(h, c, n + 1)
- C(h, c, n + 1) * B(h, c, n)
- C(h, c, n) * B(h, c, n + 1)
>= 0
)
for _ in range(T):
h, c, t = map(int, input().split())
mx = 10**18
if 2 * t <= h + c:
print(2)
else:
n = (t - h) // (h + c - 2 * t)
if H(h, c, t, n):
print(2 * n + 1)
else:
print(2 * (n + 1) + 1)
|
ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER FUNC_DEF RETURN BIN_OP BIN_OP NUMBER VAR NUMBER FUNC_DEF RETURN BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR VAR FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER NUMBER IF BIN_OP NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR IF FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def solve():
h, c, t = [int(x) for x in input().split()]
if h == c:
print(1)
elif t >= h:
print(1)
elif 2 * t <= h + c:
print(2)
else:
ans = (h - c) // (2 * t - h - c)
if ans % 2 == 0:
ans += 1
variants = []
variants.append(ans)
variants.append(2)
variants.append(1)
if ans > 1:
variants.append(ans - 2)
variants.append(ans + 2)
print(
min(
variants,
key=lambda ans: abs(
t - (h * (ans // 2 + ans % 2) + c * (ans // 2)) / ans
),
)
)
t = int(input())
for _ in range(t):
solve()
|
FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF BIN_OP NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for _ in range(int(input())):
a, b, c = map(int, input().split())
mid = (a + b) / 2
if c == a:
print(1)
elif c <= mid:
print(2)
elif c >= mid + (a - b) / 6 and c < a:
d = mid + (a - b) / 6
if abs(d - c) < abs(a - c):
print(3)
else:
print(1)
else:
c -= mid
d = []
x = (a - b) // c
if x % 2 == 0:
xx = x // 2
if xx % 2 == 0:
h = max(6, x - 2)
i = max(6, h - 4)
j = max(6, h + 4)
else:
h = max(6, x)
i = max(6, h - 4)
j = max(6, h + 4)
e = [abs((a - b) / h - c), h]
f = [abs((a - b) / i - c), i]
g = [abs((a - b) / j - c), j]
d.append(e)
d.append(f)
d.append(g)
d.sort()
print(int(d[0][1] // 2))
else:
xx = (x + 1) // 2
if xx % 2 == 0:
h = max(6, x - 1)
i = max(6, h + 4)
else:
h = max(6, x - 3)
i = max(6, h + 4)
e = [abs((a - b) / h - c), h]
f = [abs((a - b) / i - c), i]
d.append(e)
d.append(f)
d.sort()
print(int(d[0][1] // 2))
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR ASSIGN VAR LIST ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR LIST FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR LIST FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR LIST FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR LIST FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR LIST FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def rs():
return input().strip()
def ri():
return int(input())
def ria():
return list(map(int, input().split()))
def ia_to_s(a):
return " ".join([str(s) for s in a])
def temp_at_2k_1(h, c, k):
return k * h + (k - 1) * c, 2 * k - 1
def solve(h, c, t):
if t <= (h + c) // 2:
return 2
k_left = 1
k_right = 10**19
while k_right - k_left > 1:
k_mid = (k_left + k_right) // 2
t_mid_n, t_mid_d = temp_at_2k_1(h, c, k_mid)
if t_mid_n > t * t_mid_d:
k_left = k_mid
else:
k_right = k_mid
abs_min = abs(2 * t - (h + c)), 2
ans = 2
test = [
x for x in [k_left + 2, k_left + 1, k_left, k_left - 1, k_left - 2] if x > 0
]
for at in test:
t_at_n, t_at_d = temp_at_2k_1(h, c, at)
if abs_min[1] * abs(t_at_d * t - t_at_n) <= abs_min[0] * t_at_d:
abs_min = abs(t_at_d * t - t_at_n), t_at_d
ans = 2 * at - 1
return ans
def main():
for _ in range(ri()):
h, c, t = ria()
print(solve(h, c, t))
main()
|
FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR FUNC_DEF RETURN BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER FUNC_DEF IF VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR LIST BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER FOR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER RETURN VAR FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for _ in range(int(input())):
h, c, t = map(int, input().split())
if h <= t:
print(1)
continue
if h + c >= t * 2:
print(2)
continue
lo = 1
hi = 10**10
while lo < hi:
mid = (lo + hi + 1) // 2
totTemp = mid * h + (mid - 1) * c
if totTemp // (mid + mid - 1) >= t:
lo = mid
else:
hi = mid - 1
totTemp = lo * h + (lo - 1) * c
if totTemp // (lo + lo - 1) < t:
print(1 / 0)
a = lo
b = lo - 1
totTemp = a * h + b * c
totTarget = (a + b) * t
totTemp2 = totTemp + c + h
totTarget2 = totTarget + t * 2
if (totTemp - totTarget) * (a + b + 2) <= (totTarget2 - totTemp2) * (a + b):
print(a + b)
else:
print(a + b + 2)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR IF BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR IF BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
def input():
return sys.stdin.readline().rstrip()
def gcd(a, b):
a, b = min(a, b), max(a, b)
if a == 0:
return b
else:
return gcd(b % a, a)
class pos_rational:
def __init__(self, a, b):
self.numerator = a
self.denominator = b
def __eq__(self, other):
val = self.numerator * other.denominator - other.numerator * self.denominator
return val == 0
def __lt__(self, other):
val = self.numerator * other.denominator - other.numerator * self.denominator
return val < 0
def __gt__(self, other):
val = self.numerator * other.denominator - other.numerator * self.denominator
return val > 0
def __ge__(self, other):
return self == other or self > other
def __le__(self, other):
return self == other or self < other
def loss(two_n_plus_one, hot, cold, desired):
n = two_n_plus_one // 2
numerator = int(abs(desired * (2 * n + 1) - ((n + 1) * hot + n * cold)))
denominator = 2 * n + 1
return pos_rational(numerator, denominator)
testcases = int(input())
answers = []
for _ in range(testcases):
hot, cold, desired_temp = [int(i) for i in input().split()]
mid_way = (hot + cold) / 2
if hot == cold:
answers.append(1)
elif desired_temp >= hot:
answers.append(1)
elif desired_temp <= mid_way:
answers.append(2)
else:
frac = (hot - desired_temp) / (desired_temp - mid_way)
frac /= 2
option1 = 2 * int(frac) + 1
option2 = option1 + 2
l1, l2 = loss(option1, hot, cold, desired_temp), loss(
option2, hot, cold, desired_temp
)
if (
min(l1, l2) >= pos_rational(hot - desired_temp, 1)
and hot - desired_temp <= desired_temp - mid_way
):
answers.append(1)
elif min(l1, l2) >= pos_rational(desired_temp - mid_way, 1):
answers.append(2)
elif l1 <= l2:
answers.append(option1)
else:
answers.append(option2)
print(*answers, sep="\n")
|
IMPORT FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR BIN_OP VAR VAR VAR CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR VAR FUNC_DEF ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR RETURN VAR NUMBER FUNC_DEF ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR RETURN VAR NUMBER FUNC_DEF ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR RETURN VAR NUMBER FUNC_DEF RETURN VAR VAR VAR VAR FUNC_DEF RETURN VAR VAR VAR VAR FUNC_DEF ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER RETURN FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR STRING
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
tests = int(input())
for _ in range(tests):
h, c, t = map(int, input().split())
if t == h:
print(1)
continue
if t <= (h + c) / 2:
print(2)
continue
target = t - (h + c) / 2
dif = h - c
tmp = int(dif // (2 * target))
if tmp <= 0:
tmp = 1
if tmp % 2 == 0:
tmp = tmp - 1
val = 500000000
ans = 0
if abs(target - dif / (2 * tmp)) < val:
val = abs(target - dif / (2 * tmp))
ans = tmp
if abs(target - dif / (2 * (tmp + 2))) < val:
val = abs(target - dif / (2 * (tmp + 2)))
ans = tmp + 2
if abs(target - dif / (2 * (tmp + 4))) < val:
ans = tmp + 4
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP NUMBER VAR IF VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR BIN_OP VAR BIN_OP NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR BIN_OP NUMBER VAR ASSIGN VAR VAR IF FUNC_CALL VAR BIN_OP VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
q = int(input())
for _ in range(q):
h, c, t = map(int, input().split())
if 2 * t <= h + c:
print(2)
continue
l = 1
r = 10**20
while r - l > 1:
m = (r + l) // 2
if m * h + (m - 1) * c >= t * (2 * m - 1):
l = m
else:
r = m
a = l
b = l + 1
m = b * h + c * (b - 1)
z = a * h + c * (a - 1)
x = (2 * a - 1) * (2 * b - 1)
if (2 * b - 1) * z - t * x <= t * x - (2 * a - 1) * m:
print(2 * a - 1)
else:
print(2 * b - 1)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER IF BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
t = int(input())
ans = [0] * t
for i in range(t):
h, c, obj = map(int, input().split())
ave = (h + c) / 2
if obj <= ave:
ans[i] = 2
else:
x = obj - ave
y = (h - c) / 2
z = y // x
if z % 2 == 0:
z -= 1
temp0 = abs(obj - ave - y / z)
temp1 = abs(obj - ave - y / (2 + z))
if temp0 <= temp1:
ans[i] = z
else:
ans[i] = z + 2
for i in range(t):
print(int(ans[i]))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP NUMBER VAR IF VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def solver(h, c, t):
if h + c >= 2 * t:
return 2
elif h == t:
return 1
else:
x = (t - c) // (2 * t - h - c)
k1, k2 = abs((2 * x - 1) * t - ((h + c) * x - c)) * (2 * x + 1), abs(
t * (2 * x + 1) - ((h + c) * x + h)
) * (2 * x - 1)
z = 2 * x + 1 if k2 < k1 else 2 * x - 1
return z
for _ in range(int(input())):
H, C, T = [int(i) for i in input().split()]
print(solver(H, C, T))
|
FUNC_DEF IF BIN_OP VAR VAR BIN_OP NUMBER VAR RETURN NUMBER IF VAR VAR RETURN NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for _ in range(int(input())):
p = 0
a, b, c = map(int, input().split())
d = []
h = 0
t = 0
e = []
if a <= c and b <= c:
print(1)
elif 2 * c - a - b <= 0:
print(2)
else:
k = (a - c) // (2 * c - a - b)
if abs(k * (a + b) + a - c * (2 * k + 1)) * (2 * k + 3) <= abs(
(k + 1) * (a + b) + a - c * (2 * k + 3)
) * (2 * k + 1):
print(2 * k + 1)
else:
print(2 * k + 3)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST IF VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER IF BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR IF BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
numcases = int(input())
for _ in range(numcases):
h, c, t = map(int, input().split())
ans = 0
minc = float("inf")
if h + c != t * 2:
diff = t - h
change_diff = t * 2 - (h + c)
if diff > 0 and change_diff < 0 or diff < 0 and change_diff > 0:
mindiff = abs(diff) // abs(change_diff)
val1 = mindiff * (t * 2 - (h + c)) + t - h
val2 = (mindiff + 1) * (t * 2 - (h + c)) + t - h
if abs(val1) / (mindiff * 2 + 1) <= abs(val2) / (mindiff * 2 + 3):
ans = mindiff * 2 + 1
else:
ans = mindiff * 2 + 3
if abs(t * 2 - h + c) < abs(val1) and abs(t * 2 - h + c) < abs(val2):
ans = 2
elif diff == 0:
ans = 1
elif abs(t * 2 - (h + c)) > abs(t - h):
ans = 1
else:
ans = 2
else:
ans = 2
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING IF BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR IF BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def solve(th, tc, t):
ans = 0
if th == t:
ans = 1
elif 2 * t <= th + tc:
ans = 2
else:
k = (th - tc) / (2 * t - th - tc)
if (th - tc) % (2 * t - th - tc) == 0:
if k % 2 == 0:
ans = int(k) + 1
else:
ans = int(k)
else:
k_int = int(k)
if k_int % 2 == 1:
i_candidate = k_int
else:
i_candidate = k_int - 1
if 1 / k - 1 / (i_candidate + 2) + 1e-15 > 1 / i_candidate - 1 / k:
ans = i_candidate
else:
ans = i_candidate + 2
return ans
AA = int(input())
for ___ in range(AA):
th, tc, t = (int(k) for k in input().split(" "))
answer = solve(th, tc, t)
print(answer)
|
FUNC_DEF ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF BIN_OP NUMBER VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR IF BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP BIN_OP NUMBER VAR BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
h, c, temp = map(int, input().split())
ave_temp = (h + c) / 2
if temp <= ave_temp:
print(2)
elif h == temp:
print(1)
else:
u = (h - c) / ((temp - ave_temp) * 2)
u_l = int(u)
if int(u) % 2 == 1:
u1 = u_l
u2 = u_l + 2
else:
u1 = u_l - 1
u2 = u_l + 1
if abs(temp - ave_temp - (h - c) / (u1 * 2)) <= abs(
temp - ave_temp - (h - c) / (u2 * 2)
):
print(u1)
else:
print(u2)
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
from sys import stdin
input = stdin.readline
for _ in range(int(input())):
h, c, t = map(int, input().split())
if t >= h:
print(1)
continue
if 2 * t <= h + c:
print(2)
continue
l, r = 1, 1000000000
while l < r:
m = l + r >> 1
if m * h + (m - 1) * c > (2 * m - 1) * t:
l = m + 1
else:
r = m
k = l
k3 = k - 1
if k3 > 0 and 2 * t * (2 * k - 1) * (2 * k3 - 1) >= (k * h + (k - 1) * c) * (
2 * k3 - 1
) + (k3 * h + (k3 - 1) * c) * (2 * k - 1):
k, x = k3, (k3 * h + (k3 - 1) * c) / (2 * k3 - 1)
else:
x = (k * h + (k - 1) * c) / (2 * k - 1)
mn = min(abs(h - t), abs((h + c) / 2 - t), abs(x - t))
if abs(h - t) == mn:
print(1)
elif abs((h + c) / 2 - t) == mn:
print(2)
else:
print(2 * k - 1)
|
ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF BIN_OP NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER BIN_OP BIN_OP BIN_OP NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR IF FUNC_CALL VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
t = int(input())
for _ in range(t):
h, c, t = map(int, input().split())
avg = (h + c) / 2
if t == h:
print(1)
continue
if avg >= t:
print(2)
continue
else:
n1 = int((t - c) / (2 * t - h - c))
n2 = 2 * t * (2 * n1 - 1) * (2 * n1 + 1)
n3 = ((n1 - 1) * (c + h) + h) * (2 * n1 + 1) + (n1 * (c + h) + h) * (2 * n1 - 1)
if n2 >= n3:
print(2 * n1 - 1)
else:
print(2 * n1 + 1)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
t = int(input())
for i in range(t):
h, c, t = map(int, input().split())
x = t - (h + c) / 2
if x <= 0:
print(2)
else:
n = (h - c) // (2 * t - h - c)
if n % 2 == 0:
if abs(x - (h - c) / (2 * (n + 1))) > abs(x - (h - c) / (2 * (n + 3))):
print(n + 3)
else:
print(n + 1)
elif abs(x - (h - c) / (2 * n)) > abs(x - (h - c) / (2 * (n + 2))):
print(n + 2)
else:
print(n)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
t = int(input())
for _ in range(t):
h, c, t = list(map(int, input().split()))
av = (h + c) / 2
if h == c:
print(1)
elif t <= av:
print(2)
else:
d = (t - av) / (h - av)
k = -(-(h - av) // (t - av))
if k == (h - av) / (t - av):
k += 1
if k % 2 == 0:
k += 1
k = int(k)
av_max_k = (h - av) / k
av_min_k = (h - av) / (k - 2)
d_max_k = abs(t - av - av_max_k)
d_min_k = abs(t - av - av_min_k)
res = k - 2 if d_min_k <= d_max_k else k
print(res)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
from sys import stdin
for _ in range(int(stdin.readline())):
h, c, t = map(int, stdin.readline().split())
if t >= h:
print(1)
continue
if t <= (h + c) / 2:
print(2)
continue
n = 1 / (1 - 2 * ((h - t) / (h - c)))
x = round(n)
if x % 2 == 0:
y = x - 1
ex = (
2 * t * y * (y + 2)
- 2 * ((y // 2 + 1) * h + y // 2 * c) * (y + 1)
- (h + c) * y
)
if ex < 0:
print(y + 2)
else:
print(y)
else:
print(x)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP NUMBER BIN_OP NUMBER BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER BIN_OP BIN_OP NUMBER BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
tt = int(input())
while tt:
h, c, t = map(int, input().split())
if h == t:
print(1)
elif t <= (h + c) / 2:
print(2)
else:
k = (t - c - 1) // (2 * t - h - c)
ans = 2 * k + 1
if (4 * k * k - 1) * (2 * t - h - c) >= 2 * (h - c) * k:
ans -= 2
print(ans)
tt -= 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR VAR NUMBER BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR BIN_OP BIN_OP NUMBER BIN_OP VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def abc(i, j, c):
global a, b
ans = i
while i <= j:
mid = i + (j - i) // 2
if mid * b + (mid + 1) * a <= c * (2 * mid + 1):
ans = mid
i = mid + 1
else:
j = mid - 1
return ans
def bca(i, j, c):
global a, b
ans = i
while i <= j:
mid = i + (j - i) // 2
if mid * b + (mid + 1) * a >= c * (2 * mid + 1):
ans = mid
i = mid + 1
else:
j = mid - 1
return ans
return ans
t = int(input())
for _ in range(t):
a, b, c = map(int, input().split())
if a == b:
print(1)
continue
x = (a + b) / 2
x = abs(c - x)
ans = 2
if b >= a:
y = abc(0, 10**7, c)
m = abs(y * b + (y + 1) * a - (2 * y + 1) * c)
y += 1
v = abs(y * b + (y + 1) * a - (2 * y + 1) * c)
if m == v:
if abs((y * b + (y + 1) * a) / (2 * y + 1) - c) < x:
ans = 2 * y + 1
else:
y -= 1
m = (y * b + (y + 1) * a) / (2 * y + 1)
if abs(m - c) < x:
ans = 2 * y + 1
x = abs(m - c)
y += 1
m = (y * b + (y + 1) * a) / (2 * y + 1)
if abs(m - c) < x:
ans = 2 * y + 1
x = abs(m - c)
else:
y = bca(0, 10**7, c)
m = abs(y * b + (y + 1) * a - (2 * y + 1) * c)
y += 1
v = abs(y * b + (y + 1) * a - (2 * y + 1) * c)
if m == v:
if abs((y * b + (y + 1) * a) / (2 * y + 1) - c) < x:
ans = 2 * y + 1
else:
y -= 1
m = (y * b + (y + 1) * a) / (2 * y + 1)
if abs(m - c) < x:
ans = 2 * y + 1
x = abs(m - c)
y += 1
m = (y * b + (y + 1) * a) / (2 * y + 1)
if abs(m - c) < x:
ans = 2 * y + 1
x = abs(m - c)
print(ans)
|
FUNC_DEF ASSIGN VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP NUMBER NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR IF VAR VAR IF FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP NUMBER NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR IF VAR VAR IF FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def getCups(h, c, t, x):
x = round(x)
minDiff = 100000000000, 1
cups = -100
for i in range(int(x) - 1 - 2, int(x) + 2 + 2):
if i % 2 == 1:
v = h * ((i + 1) / 2) + c * ((i - 1) / 2), i
else:
v = (h + c) / 2, 1
diff = abs(v[0] - t * i), i
if diff[0] * minDiff[1] < minDiff[0] * diff[1]:
minDiff = diff
cups = i
return cups
t = int(input())
for _ in range(t):
h, c, t = [int(i) for i in input().split()]
avg = (h + c) / 2
if t >= h:
print(1)
elif t <= avg:
print(2)
else:
x = (h - c) / (2 * t - h - c)
print(getCups(h, c, t, x))
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR IF BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
from sys import stdin
input = stdin.buffer.readline
for _ in range(int(input())):
h, c, t = map(int, input().split())
d = abs((t << 1) - h - c)
ans = 2
l, r = 0, 1000000
while r - l > 1:
mid = l + r >> 1
if (h + c) * mid + h < t * (mid << 1 | 1):
r = mid
else:
l = mid
tmp = abs(t * (r << 1 | 1) - (h + c) * r - h)
if tmp << 1 < d * (r << 1 | 1) or tmp << 1 == d * (r << 1 | 1) and r << 1 | 1 < ans:
d = tmp
ans = r << 1 | 1
tmp = abs(t * (l << 1 | 1) - (h + c) * l - h)
if (
tmp * ans < d * (l << 1 | 1)
or tmp * ans == d * (l << 1 | 1)
and l << 1 | 1 < ans
):
d = tmp
ans = l << 1 | 1
print(ans)
|
ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP VAR VAR VAR VAR IF BIN_OP VAR NUMBER BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP VAR VAR VAR VAR IF BIN_OP VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
ans = []
for _ in range(int(input())):
h, c, t = map(int, input().split())
if h + c - 2 * t >= 0:
ans.append(2)
continue
a1 = int((t - c) / (2 * t - h - c))
b1 = a1 - 1
a2 = int((t - c) / (2 * t - h - c)) + 1
b2 = a2 - 1
dt1 = abs(t * (a1 + b1) - (a1 * h + b1 * c)) * (a2 + b2)
dt2 = abs(t * (a2 + b2) - (a2 * h + b2 * c)) * (a1 + b1)
if dt1 <= dt2:
ans.append(a1 + b1)
else:
ans.append(a2 + b2)
print("\n".join(map(str, ans)))
|
ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
sys.setrecursionlimit(10**5)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II():
return int(sys.stdin.readline())
def MI():
return map(int, sys.stdin.readline().split())
def LI():
return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number):
return [LI() for _ in range(rows_number)]
def SI():
return sys.stdin.readline()[:-1]
def main():
for _ in range(II()):
h, c, t = MI()
if t >= h:
print(1)
continue
if t * 2 <= c + h:
print(2)
continue
a1 = (-t + h) // (-h - c + 2 * t)
a2 = a1 + 1
if 2 * t * (2 * a2 + 1) * (2 * a1 + 1) - (h * (a2 + 1) + c * a2) * (
2 * a1 + 1
) < (2 * a2 + 1) * (h * (a1 + 1) + c * a1):
ans = a2 * 2 + 1
else:
ans = a1 * 2 + 1
print(ans)
main()
|
IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR STRING FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN FUNC_CALL VAR NUMBER FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF BIN_OP VAR NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
cases = int(input())
for _ in range(cases):
hot, cold, target = map(int, input().split())
if 2 * target <= hot + cold:
print(2)
elif target >= hot:
print(1)
else:
lo, hi = 1, 10**20
while lo < hi:
mid = (lo + 1 + hi) // 2
if mid * hot + (mid - 1) * cold >= (2 * mid - 1) * target:
lo = mid
else:
hi = mid - 1
upper_bound = lo * hot + (lo - 1) * cold, lo + lo - 1
hi = lo + 1
lower_bound = hi * hot + (hi - 1) * cold, hi + hi - 1
ans = lo
if abs((upper_bound[0] - upper_bound[1] * target) * lower_bound[1]) > abs(
(lower_bound[0] - lower_bound[1] * target) * upper_bound[1]
):
ans = hi
print(2 * ans - 1)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP NUMBER NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER IF BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR IF FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def solve(h, c, t):
if t == h:
return 1
if 2 * t <= h + c:
return 2
cands = []
cands.append((h + c, 2, 2))
bx = h - t
by = 2 * t - h - c
b = bx // by
for d in range(2):
bb = b + d
aa = bb + 1
cands.append((aa * h + bb * c, aa + bb, aa + bb))
ans, ansx, ansy = float("inf"), float("inf"), 1
for ox, oy, o in cands:
if ansx * oy > ansy * abs(ox - oy * t):
ans, ansx, ansy = o, abs(ox - oy * t), oy
return ans
tests = int(input())
for _ in range(tests):
h, c, t = map(int, input().split())
print(solve(h, c, t))
|
FUNC_DEF IF VAR VAR RETURN NUMBER IF BIN_OP NUMBER VAR BIN_OP VAR VAR RETURN NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR STRING FUNC_CALL VAR STRING NUMBER FOR VAR VAR VAR VAR IF BIN_OP VAR VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
T = int(input())
for _ in range(T):
h, c, t = map(int, input().split())
s = (h + c) / 2
if s >= t:
print(2)
continue
s = t - s
a = (h - c) / (2 * s)
k = int((a + 1) // 2)
if (h - c) / (4 * k - 2) - s <= s - (h - c) / (4 * k + 2):
print(2 * k - 1)
else:
print(2 * k + 1)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for _ in range(int(input())):
h, c, t = [int(i) for i in input().split()]
if t <= (h + c) / 2:
print(2)
else:
l = 1
r = 10**12
mid = (l + h) / 2
while l < r:
mid = (l + r) // 2
sum = (h * mid + c * (mid - 1)) / (2 * mid - 1)
if sum > t:
l = mid + 1
else:
r = mid
l11 = abs(h * l + c * (l - 1) - t * (2 * l - 1)) * (2 * (l - 1) - 1)
l -= 1
l22 = abs(h * l + c * (l - 1) - t * (2 * l - 1)) * (2 * (l + 1) - 1)
if l11 < l22:
print(2 * (l + 1) - 1)
else:
print(2 * l - 1)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
pprint = lambda s: print(" ".join(map(str, s)))
input = lambda: sys.stdin.readline().strip()
ipnut = input
for i in range(int(input())):
h, c, t = map(int, input().split())
if t <= (h + c) / 2:
print(2)
continue
l = 1
r = 1000000000000000000
def f(m):
return m * (h + c) + h
l = (t - h) // (c + h - 2 * t)
r = l + 1
ansl = abs(t * (2 * l + 1) - f(l)) * (r * 2 + 1), 2 * l + 1
ansr = abs(t * (r * 2 + 1) - f(r)) * (2 * l + 1), r * 2 + 1
print(min(ansl, ansr)[1])
|
IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FUNC_DEF RETURN BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def calc(a, b, t, k):
num1, num2 = abs(k * (a + b) + a - (2 * k + 1) * t), abs(
(k + 1) * (a + b) + a - (2 * k + 3) * t
)
den1, den2 = 2 * k + 1, 2 * k + 3
if num1 * den2 <= num2 * den1:
return 2 * k + 1
return 2 * k + 3
t = int(input())
for _ in range(t):
a, b, t = map(int, input().split())
if t <= (a + b) / 2:
print(2)
continue
k = (a - t) // (2 * t - a - b)
print(calc(a, b, t, k))
|
FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER IF BIN_OP VAR VAR BIN_OP VAR VAR RETURN BIN_OP BIN_OP NUMBER VAR NUMBER RETURN BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
from sys import stdin
input = stdin.readline
INF = 10**9 + 7
MAX = 10**7 + 7
MOD = 10**9 + 7
for Ti in range(int(input().strip())):
h, c, t = [int(x) for x in input().strip().split()]
avg = (h + c) / 2
if t <= avg:
print("2")
continue
ti = (t - h) // (h + c - 2 * t)
l = (2 * ti + 3) * (ti * (h + c) + h - 2 * ti * t - t)
r = (2 * ti + 1) * ((ti + 1) * (h + c) + h - 2 * ti * t - 3 * t)
if abs(l) <= abs(r):
print(2 * ti + 1)
else:
print(2 * ti + 3)
|
ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP BIN_OP NUMBER VAR VAR BIN_OP NUMBER VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
input = sys.stdin.buffer.readline
nTests = int(input())
for _ in range(nTests):
h, c, t = [int(zz) for zz in input().split()]
if t <= (h + c) / 2:
print(2)
elif t == h:
print(1)
else:
nHot = (t - c) // (2 * t - h - c)
x = nHot
den1 = 2 * x - 1
den2 = 2 * x + 1
num1 = abs(x * h + (x - 1) * c - t * den1)
num2 = abs((x + 1) * h + x * c - t * den2)
if num1 * den2 <= num2 * den1:
print(2 * x - 1)
else:
print(2 * x + 1)
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP VAR VAR IF BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
from sys import stdin, stdout
input = stdin.readline
def print(x):
stdout.write(str(x) + "\n")
def solve():
h, c, t = map(int, input().split())
if t <= (h + c) / 2:
print(2)
else:
k = (h - t) // (2 * t - h - c)
k = 2 * k + 1
c1 = (k // 2 + 1) * h + k // 2 * c - t * k
c2 = ((k + 2) // 2 + 1) * h + (k + 2) // 2 * c - t * (k + 2)
if abs(c1) * (k + 2) <= abs(c2) * k:
print(k)
else:
print(k + 2)
t = 1
t = int(input())
for _ in range(t):
solve()
|
ASSIGN VAR VAR FUNC_DEF EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR STRING FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def f(h, c, t, x):
n = x * 2 + 1
return abs(h * (x + 1) + c * x - t * n)
for _ in range(int(input())):
h, c, t = map(int, input().split())
if h + c >= t * 2:
print(2)
continue
if h <= t:
print(1)
continue
ans1 = (h - t) // (2 * t - h - c)
ans2 = ans1 + 1
if f(h, c, t, ans1) * (2 * ans2 + 1) <= f(h, c, t, ans2) * (2 * ans1 + 1):
print(ans1 * 2 + 1)
else:
print(ans2 * 2 + 1)
|
FUNC_DEF ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER RETURN FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
h, c, t = map(int, input().split())
if t * 2 <= h + c:
print(2)
continue
f = lambda x: x * h + (x - 1) * c
ok = 1
ng = 10**6
while abs(ok - ng) > 1:
mid = (ok + ng) // 2
if f(mid) >= t * (mid * 2 - 1):
ok = mid
else:
ng = mid
if abs(t * (ok * 2 - 1) - f(ok)) * (2 * ok + 1) <= abs(
t * (ok * 2 + 1) - f(ok + 1)
) * (2 * ok - 1):
print(ok * 2 - 1)
else:
print(ok * 2 + 1)
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER WHILE FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR VAR IF BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
try:
for _ in range(int(input())):
h, c, t = map(int, input().split())
if h <= t:
print(1)
elif h + c >= 2 * t:
print(2)
else:
x = int((c - t) / (h + c - 2 * t))
m = abs((x * (h + c - 2 * t) + t - c) / (2 * x - 1))
n = abs(((x + 1) * (h + c - 2 * t) + t - c) / (2 * (x + 1) - 1))
if m <= n:
print(2 * x - 1)
else:
print(2 * x + 1)
except:
pass
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF BIN_OP VAR VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR VAR VAR BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
tt = int(input())
for _ in range(tt):
h, c, t = list(map(int, input().split(" ")))
if t >= h:
print(1)
elif t <= (c + h) / 2:
print(2)
else:
n = (t - c) / (2 * t - h - c)
m = int(n) + 1
nn = int(n)
e = nn * h + (nn - 1) * c - t * (2 * nn - 1)
ee = t * (2 * m - 1) - (m * h + (m - 1) * c)
if e * (2 * m - 1) > ee * (2 * nn - 1):
print(2 * m - 1)
else:
print(2 * nn - 1)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR IF BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
q = int(input())
for i in range(q):
h, c, t = [int(j) for j in sys.stdin.readline().split()]
if t == h:
print(1)
elif t <= (h + c) / 2:
print(2)
else:
first = (h + c) / t
first1 = (h + c) / 2
second = (h - t) // (2 * t - h - c)
if second == second + 1:
if second == 0:
print(1)
else:
print(2)
else:
second1 = (h + c) * second + h
second2 = (h + c) * (second + 1) + h
min1 = min(
abs(t - (h + c) / 2),
abs(t * (second * 2 + 1) - second1) / (second * 2 + 1),
abs(t * ((second + 1) * 2 + 1) - second2) / ((second + 1) * 2 + 1),
abs(t - h),
)
if min1 == abs(t - h):
print(1)
elif min1 == abs(t - (h + c) / 2):
print(2)
elif min1 == abs(t * (second * 2 + 1) - second1) / (second * 2 + 1):
print(int(second) * 2 + 1)
elif min1 == abs(t * ((second + 1) * 2 + 1) - second2) / (
(second + 1) * 2 + 1
):
print(int(second + 1) * 2 + 1)
|
IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR IF VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER FUNC_CALL VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
t = int(input())
for i in range(t):
s = input().split()
h = int(s[0])
c = int(s[1])
te = int(s[2])
if te == h:
print("1")
elif h - te == te - c:
print("2")
else:
sum = c + h
if h - te > te - c:
print("2")
else:
a = h - te
b = sum - 2 * te
f = int(a / b)
f = abs(f)
k = abs(2 * f + 1)
l = [k - 2, k, k + 2]
p = [0, 0, 0]
for j in range(3):
p[j] = (a + b * (l[j] - 1) / 2) / l[j]
p[j] = abs(p[j])
print(l[p.index(min(p))])
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING IF BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR VAR IF BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR LIST BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR NUMBER NUMBER VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
T = int(input())
o = []
for i in range(T):
h, c, t = input().split()
h = int(h)
c = int(c)
t = int(t)
s = (h + c) / 2
if t >= h:
o.append(1)
elif t <= s:
o.append(2)
else:
i = (h - t) / (2 * t - (h + c))
if i == int(i):
i = int(i)
o.append(2 * i + 1)
else:
i = int(i)
aprev = abs((i * (h + c - 2 * t) + h - t) / (2 * i + 1))
i += 1
acur = abs((i * (h + c - 2 * t) + h - t) / (2 * i + 1))
if acur >= aprev:
o.append((i - 1) * 2 + 1)
else:
o.append(i * 2 + 1)
for i in range(T):
print(o[i])
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
T = int(input())
for _ in range(T):
h, c, t = map(int, input().split())
if h <= t:
print(1)
elif t <= (h + c) // 2:
print(2)
else:
i1 = (t - h) // (h + c - 2 * t)
i2 = i1 + 1
if ((i1 + 1) * h + i1 * c) * (2 * i2 + 1) + ((i2 + 1) * h + i2 * c) * (
2 * i1 + 1
) <= 2 * t * (2 * i1 + 1) * (2 * i2 + 1):
ans = 2 * i1 + 1
else:
ans = 2 * i2 + 1
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def solve(h, c, t):
if t >= h:
return 1
if 2 * t <= h + c:
return 2
if (h - t) % (2 * t - h - c) == 0:
return 2 * (h - t) // (2 * t - h - c) + 1
nn1 = int((h - t) / (2 * t - h - c))
nn2 = nn1 + 1
if (t * (2 * nn2 + 1) - nn2 * (h + c) - h) * (2 * nn1 + 1) - (
nn1 * (h + c) + h - t * (2 * nn1 + 1)
) * (2 * nn2 + 1) < 0:
return 2 * nn2 + 1
return 2 * nn1 + 1
t = int(input())
for i_t in range(t):
[h, c, tt] = list(map(int, input().split(" ")))
print(solve(h, c, tt))
|
FUNC_DEF IF VAR VAR RETURN NUMBER IF BIN_OP NUMBER VAR BIN_OP VAR VAR RETURN NUMBER IF BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR NUMBER RETURN BIN_OP BIN_OP BIN_OP NUMBER BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER NUMBER RETURN BIN_OP BIN_OP NUMBER VAR NUMBER RETURN BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN LIST VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
t = int(input())
for test in range(t):
h, c, t = map(int, input().split(" "))
t -= c
h -= c
if t <= h / 2:
print(2)
else:
res = int(h / (2 * t - h) / 2)
if abs(t - (res + 1) * h / (2 * res + 1)) > abs(
t - (res + 2) * h / (2 * res + 3)
):
print(2 * res + 3)
else:
print(2 * res + 1)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING VAR VAR VAR VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP NUMBER VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
from sys import stdin
input = stdin.readline
q = int(input())
for _ in range(q):
h, c, t = map(int, input().split())
if 2 * t <= h + c:
print(2)
else:
t *= 2
h *= 2
c *= 2
w = h // 2 - c // 2
pod = (h + c) // 2
goal = t - pod
if goal == 0:
print(1)
else:
x = w // goal
wynik = 2
mini = 23649823847238
for szkl in range(x - 3, x + 4):
if szkl % 2 == 1 and szkl >= 1 and abs(goal - w / szkl) < mini:
mini = abs(goal - w / szkl)
wynik = szkl
print(wynik)
|
ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
t = int(input())
for i in range(0, t):
s = input().split()
h = int(s[0])
c = int(s[1])
t = int(s[2])
if h + c == t * 2:
print(2)
else:
x = (t - h) / (h + c - t * 2)
l = int(x)
r = l + 1
al = abs(h * (l + 1) + c * l - t * (2 * l + 1)) * (2 * r + 1)
ar = abs(h * (r + 1) + c * r - t * (2 * r + 1)) * (2 * l + 1)
am = abs((h + c) / 2 - t) * (2 * l + 1) * (2 * r + 1)
if min(al, ar, am) == al and x >= 0:
print(2 * l + 1)
elif min(al, ar, am) == ar and x >= 0:
print(2 * r + 1)
else:
print(2)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for _ in range(int(input())):
h, c, t = map(int, input().split())
if h == t:
print(1)
else:
avg = (h + c) / 2
if avg == t:
print(2)
elif avg > t:
print(2)
else:
dif = t - avg
j = int(abs((c - h) // (2 * dif)))
if j % 2 == 0:
j += 1
if dif - (h - c) / (2 * j) >= abs(dif - (h - c) / (2 * (j - 2))):
print(j - 2)
else:
print(j)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER IF BIN_OP VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
input = sys.stdin.buffer.readline
def print(val):
sys.stdout.write(str(val) + "\n")
def temp(x, t, h, c):
return abs((t * (2 * x + 1) - ((x + 1) * h + x * c)) / (2 * x + 1))
def prog():
for _ in range(int(input())):
h, c, t = map(int, input().split())
if t <= (h + c) / 2:
print(2)
else:
R = 10**6
L = 0
while R != L:
s = (R - L) // 3
m1 = L + s
m2 = R - s
if temp(m2, t, h, c) >= temp(m1, t, h, c):
R = m2 - 1
else:
L = m1 + 1
print(2 * R + 1)
prog()
|
IMPORT ASSIGN VAR VAR FUNC_DEF EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR STRING FUNC_DEF RETURN FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def f(h: int, c: int, t: int) -> int:
if h + c >= t * 2:
return 2
else:
x = (h - t) // (t * 2 - h - c)
lhs = (h * (x + 1) + x * c) * (x * 2 + 3) - t * (x * 2 + 1) * (x * 2 + 3)
rhs = t * (x * 2 + 1) * (x * 2 + 3) - (h * (x + 2) + (x + 1) * c) * (x * 2 + 1)
if lhs <= rhs:
return x * 2 + 1
else:
return x * 2 + 3
T = int(input())
for _ in range(T):
h, c, t = map(int, input().split())
print(f(h, c, t))
|
FUNC_DEF VAR VAR VAR IF BIN_OP VAR VAR BIN_OP VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR RETURN BIN_OP BIN_OP VAR NUMBER NUMBER RETURN BIN_OP BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for test_ in range(int(input())):
h, c, t = map(int, input().split())
if t <= (h + c) // 2:
print(2)
else:
t -= (h + c) / 2
t1 = (h - c) / 2 / t
x1, x2 = int((t1 - 1) // 2 * 2 + 1), int((t1 + 1) // 2 * 2 + 1)
if (h - c) / (x1 * 2) - t <= t - (h - c) / (x2 * 2):
print(x1)
else:
print(x2)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER NUMBER FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER NUMBER IF BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def temp(h, c, cups, k):
if cups % 2 == 0:
return (h + c) // 2
n = cups // 2
return abs((h + c) * n + h - k * cups), cups
def lessthanore(a, b):
return a[0] * b[1] <= a[1] * b[0]
def check(n, k, h, c):
correct = abs(temp(h, c, 2 * n + 1) - k)
before = abs(temp(h, c, 2 * n - 1) - k)
after = abs(temp(h, c, 2 * n + 3) - k)
print(correct, before, after)
return correct < before and correct < after
def simcups(h, c, n):
i = 0
temp = 0
while i <= n:
if i % 2 == 0:
temp += h
else:
temp += c
i += 1
print(temp / i)
def mincups(h, c, k):
avg = (h + c) / 2
if k <= avg:
return 2
if k >= h:
return 1
n = (h - k) / (2 * k - h - c)
a = temp(h, c, 2 * int(n) + 1, k)
b = temp(h, c, 2 * int(n) + 3, k)
if lessthanore(a, b):
return 2 * int(n) + 1
return 2 * int(n) + 3
t = int(input())
for i in range(t):
h, c, k = map(int, input().split())
print(mincups(h, c, k))
|
FUNC_DEF IF BIN_OP VAR NUMBER NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR FUNC_DEF RETURN BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR RETURN NUMBER IF VAR VAR RETURN NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP NUMBER FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP NUMBER FUNC_CALL VAR VAR NUMBER VAR IF FUNC_CALL VAR VAR VAR RETURN BIN_OP BIN_OP NUMBER FUNC_CALL VAR VAR NUMBER RETURN BIN_OP BIN_OP NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
input = sys.stdin.readline
T = int(input())
def solve(h, c, t, x):
if h * x + c * (x - 1) > t * (2 * x - 1):
return True
else:
return False
for _ in range(T):
h, c, t = [int(x) for x in input().split()]
left = 1
right = 10**20
while right - left > 1:
mid = (right + left) // 2
if solve(h, c, t, mid):
left = mid
else:
right = mid
l = (h * left + c * (left - 1)) * (2 * right - 1) * 2
r = (h * right + c * (right - 1)) * (2 * left - 1) * 2
m = (h * 1 + c * 1) * (2 * left - 1) * (2 * right - 1)
t *= 2 * (2 * left - 1) * (2 * right - 1)
a = [abs(l - t), abs(r - t), abs(m - t)]
if min(a) == abs(t - m):
print(2)
elif min(a) == abs(l - t):
print(2 * left - 1)
elif min(a) == abs(t - r):
print(2 * right - 1)
else:
print(1)
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF IF BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER RETURN NUMBER RETURN NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR LIST FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for _ in range(int(input())):
a, b, c = map(int, input().split())
if a == c:
print(1)
continue
if (a + b) / 2 == c:
print(2)
continue
temp = -1
l = 2
r = c + 1
x = 2
while l < r:
mid = (l + r) // 2
if (mid * a + (mid - 1) * b) / (2 * mid - 1) <= c:
x = mid
r = mid
else:
l = mid + 1
y = x - 1
ans2 = (a + b) / 2
ans1 = (a * x - a + b * y - b) * (x + y)
ans = (a * x + b * y) * (x + y - 2)
pk = (x + y) * (x + y - 2)
if 2 * c * pk >= ans + ans1:
if abs(ans2 - c) <= abs(c - (a * x - a + b * y - b) / (x + y - 2)):
print(2)
else:
print(x + y - 2)
elif abs(ans2 - c) <= abs(c - (a * x + b * y) / (x + y)):
print(2)
else:
print(x + y)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP BIN_OP NUMBER VAR VAR BIN_OP VAR VAR IF FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def solve():
h, c, t = map(int, input().split())
if h == t:
result = 1
elif t <= (h + c) / 2:
result = 2
else:
k = (t - c - 1) // (2 * t - h - c)
result = 2 * k + 1
result -= 2 if (4 * k * k - 1) * (2 * t - h - c) >= 2 * (h - c) * k else 0
print(result)
def main():
for _ in range(int(input())):
solve()
main()
|
FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR VAR NUMBER BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR BIN_OP BIN_OP NUMBER BIN_OP VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for test in range(int(input())):
h, c, t = map(int, input().split())
if h == t:
print(1)
elif t <= (h + c) / 2:
print(2)
elif abs(t - h) <= abs(t - (2 * h + c) / 3):
print(1)
else:
x = (h - t) // (2 * t - h - c)
y = x + 1
dif2 = abs((2 * y + 1) * t - y * h - y * c - h)
dif1 = abs((2 * x + 1) * t - x * h - x * c - h)
if dif2 * (2 * x + 1) < dif1 * (2 * y + 1):
print(2 * y + 1)
else:
print(2 * x + 1)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR IF BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def f(h, c, t):
if t <= (c + h) / 2:
return 2
if t >= h:
return 1
x = (t - c) // (2 * t - h - c)
if (4 * x * x + 2 * x - 1) * h + (4 * x * x - 2 * x - 1) * c <= 2 * (
4 * x * x - 1
) * t:
return 2 * x - 1
else:
return 2 * x + 1
t = int(input())
for i in range(t):
[h, c, t] = input().split(" ")
print(f(int(h), int(c), int(t)))
|
FUNC_DEF IF VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN NUMBER IF VAR VAR RETURN NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR IF BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR VAR BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR VAR BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER BIN_OP BIN_OP BIN_OP NUMBER VAR VAR NUMBER VAR RETURN BIN_OP BIN_OP NUMBER VAR NUMBER RETURN BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN LIST VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
t = int(input())
for i in range(t):
h1, c1, t1 = input().split(" ")
h = int(h1)
c = int(c1)
t = int(t1)
if t <= (h + c) / 2:
print(2)
else:
x = (t - c) // (2 * t - h - c)
m = ((h + c) * x - c) / (2 * x - 1) - t
n = t - ((h + c) * x + h) / (2 * x + 1)
if m < n:
print(2 * x - 1)
elif m == n and m == int(m):
print(2 * x - 1)
elif m == n and m != int(m):
print(2 * x + 1)
else:
print(2 * x + 1)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
input = sys.stdin.buffer.readline
def I():
return list(map(int, input().split()))
def sieve(n):
a = [1] * n
for i in range(2, n):
if a[i]:
for j in range(i * i, n, i):
a[j] = 0
return a
for __ in range(int(input())):
h, c, t = I()
if h == t:
print(1)
continue
if (h + c) / 2 >= t:
print(2)
continue
if t >= (h * 2 + c) / 3:
x1 = (h * 2 + c) / 3
x2 = h
if abs(x2 - t) <= abs(x1 - t):
print(1)
else:
print(3)
continue
l = 1
r = 10**30
ans = 10**100
left = t - (h + c) / 2
while l <= r:
mid = (l + r) // 2
n = mid
delta = (h - c) / (2 * (2 * n + 1))
if delta == left:
ans = n
break
if delta < left:
ans = n
r = mid - 1
else:
l = mid + 1
n = ans
delta1 = (h - c) / (2 * (2 * n + 1))
delta2 = (h - c) / (2 * (2 * n - 1))
if abs(delta1 - left) < abs(delta2 - left):
print(2 * n + 1)
else:
print(2 * n - 1)
|
IMPORT ASSIGN VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR IF FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
from sys import gettrace, stdin
if not gettrace():
def input():
return next(stdin)[:-1]
def main():
def solve():
h, c, t = map(int, input().split())
if t <= (h + c) / 2:
print(2)
return
if t >= h:
print(1)
return
m = (h - t) // (2 * t - h - c)
t1n = (m + 1) * h + m * c
t2n = (m + 2) * h + (m + 1) * c
t1d = 2 * m + 1
t2d = 2 * m + 3
if abs(t1n * t2d - t * t1d * t2d) <= abs(t2n * t1d - t * t1d * t2d):
print(2 * m + 1)
else:
print(2 * m + 3)
q = int(input())
for _ in range(q):
solve()
main()
|
IF FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR NUMBER FUNC_DEF FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN IF VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def read_int():
return int(input())
def read_ints():
return list(map(int, input().split(" ")))
tt = read_int()
for case_num in range(tt):
h, c, t = read_ints()
if 2 * t <= h + c:
print(2)
continue
l = 1
r = int(10000000.0)
def calc(x):
return x * h + (x - 1) * c
while l <= r:
mid = (l + r) // 2
temp = calc(mid)
if temp >= t * (2 * mid - 1):
l = mid + 1
else:
r = mid - 1
a = calc(l - 1)
b = calc(l)
p = 2 * l - 3
q = 2 * l - 1
if abs(a * q - t * p * q) <= abs(b * p - t * p * q):
print(p)
else:
print(q)
|
FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR IF BIN_OP NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER FUNC_DEF RETURN BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def t_on_move(move):
return (move * h + (move - 1) * c) / (2 * move - 1)
for _ in range(int(input())):
h, c, t = list(map(int, input().split()))
even_val = (h + c) // 2
l = 0
r = 2000000000
while r - l > 1:
m = (l + r) // 2
m_val = t_on_move(m)
if m_val > t:
l = m
else:
r = m
delta = abs(t - (h + c) / 2)
ans = 2
if l > 0:
if (r * h + (r - 1) * c) * (2 * l - 1) - t * (2 * r - 1) * (2 * l - 1) > -(
l * h + (l - 1) * c
) * (2 * r - 1) + t * (2 * r - 1) * (2 * l - 1):
if delta > abs(t - t_on_move(r)):
ans = 2 * r - 1
elif delta > abs(t - t_on_move(l)):
ans = 2 * l - 1
elif delta > abs(t - t_on_move(r)):
ans = 2 * r - 1
print(ans)
|
FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER IF BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
t = int(input())
for _ in range(t):
h, c, g = [int(x) for x in input().split()]
av = (h + c) / 2
if g == h:
print(1)
elif g <= av + 0.001:
print(2)
else:
n = int((g - h) / (h + c - g * 2))
best_n = -1
for j in range(max(n - 3, 0), n + 3):
if best_n == -1 or abs(
(best_n + 1) * h + best_n * c - (2 * best_n + 1) * g
) * (2 * j + 1) > abs((j + 1) * h + j * c - g * (2 * j + 1)) * (
2 * best_n + 1
):
best_n = j
print(2 * best_n + 1)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER IF VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
from sys import stdin, stdout
def mixing_water(h, c, t):
r1 = 2
v1 = (h + c) // 2
if h >= c:
rv2 = bs1(h, c, t)
else:
rv2 = bs2(h, c, t)
r2 = rv2[0]
v2 = rv2[1]
dif = abs(t - v1) - abs(t - v2)
if dif == 0:
return min(r1, r2)
elif dif > 0:
return r2
else:
return r1
def bs1(h, c, t):
l = 1
r = 2**31 - 1
while l < r:
m = (l + r) // 2
v1 = ((h + c) * m - c) / (2 * m - 1)
if v1 > t:
l = m + 1
else:
r = m
v1 = (h + c) * r - c
v2 = (h + c) * (r - 1) - c
t1 = 2 * r - 1
t2 = 2 * (r - 1) - 1
if r > 1 and abs(t * t1 * t2 - v2 * t1) <= abs(v1 * t2 - t * t1 * t2):
return [(r - 1) * 2 - 1, v2 / t2]
else:
return [r * 2 - 1, v1 / t1]
def bs2(h, c, t):
l = 1
r = 2**31 - 1
while l < r:
m = (l + r) // 2
v1 = ((h + c) * m - c) / (2 * m - 1)
if v1 < t:
l = m + 1
else:
r = m
v1 = (h + c) * r - c
v2 = (h + c) * (r - 1) - c
t1 = 2 * r - 1
t2 = 2 * (r - 1) - 1
if r > 1 and abs(v2 * t1 - t * t1 * t2) <= abs(t * t1 * t2 - v1 * t2):
return [(r - 1) * 2 - 1, v2 / t2]
else:
return [r * 2, v1 / t1]
T = int(stdin.readline())
for i in range(T):
hct = list(map(int, stdin.readline().split()))
stdout.write(str(mixing_water(hct[0], hct[1], hct[2])) + "\n")
|
FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR NUMBER RETURN FUNC_CALL VAR VAR VAR IF VAR NUMBER RETURN VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER IF VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR RETURN LIST BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER BIN_OP VAR VAR RETURN LIST BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER IF VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR RETURN LIST BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER BIN_OP VAR VAR RETURN LIST BIN_OP VAR NUMBER BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR NUMBER STRING
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
for _ in range(int(input())):
h, c, t = map(int, input().split())
if h + c >= 2 * t:
print(2)
elif t >= h:
print(1)
else:
x = 2 * t - h - c
k = (h - c + x) // (2 * x)
test1 = abs((8 * k**2 - 2) * t - (h * k + c * k - c) * (4 * k + 2))
test2 = abs((8 * k**2 - 2) * t - (h * (k + 1) + c * (k + 1) - c) * (4 * k - 2))
test3 = abs((8 * k**2 - 2) * t - (h + c) * (4 * k**2 - 1))
if k == 1:
if min(test1, test2, test3) == test1:
print(2 * k - 1)
elif min(test1, test2, test3) == test3:
print(2)
else:
print(2 * k + 1)
elif min(test1, test2, test3) == test3:
print(2)
elif min(test1, test2, test3) == test1:
print(2 * k - 1)
else:
print(2 * k + 1)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER IF VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
T = int(input())
for i in range(T):
h, c, t = map(int, input().split())
avg = (h + c) / 2
if avg >= t:
print(2)
else:
n = ((h - c) / (2 * (t - avg)) + 1) / 2
if type(n) == int:
print(2 * n - 1)
else:
n_min = int(n)
n_max = n_min + 1
diff_n_min = abs(t - avg - (h - c) / (2 * (2 * n_min - 1)))
diff_n_max = abs(t - avg - (h - c) / (2 * (2 * n_max - 1)))
if diff_n_min <= diff_n_max:
print(2 * n_min - 1)
else:
print(2 * n_max - 1)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP NUMBER BIN_OP VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
def solve():
[hi, lo, barrel] = map(int, input().split())
if 2 * barrel <= hi + lo:
return 2
items = (hi - barrel) // (2 * barrel - hi - lo)
temp = lambda t: ((t + 1) * hi + t * lo - (2 * t + 1) * barrel) / (2 * t + 1)
if abs(temp(items)) > abs(temp(items + 1)):
items += 1
return 2 * items + 1
for _ in range(int(input())):
print(solve())
|
FUNC_DEF ASSIGN LIST VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP NUMBER VAR BIN_OP VAR VAR RETURN NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER RETURN BIN_OP BIN_OP NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
input = sys.stdin.readline
def solve():
h, c, t = map(int, input().split())
if h <= t:
return 1
if 2 * t <= h + c:
return 2
if 2 * h + c <= t * 3:
if abs(2 * h + c - t * 3) >= (h - t) * 3:
return 1
else:
return 3
l = 0
r = 500000
while r - l > 1:
m = (l + r) // 2
if t * (2 * m + 1) < (m + 1) * h + m * c:
l = m
else:
r = m
if abs((l + 1) * h + l * c - t * (2 * l + 1)) * (2 * r + 1) <= abs(
(r + 1) * h + r * c - t * (2 * r + 1)
) * (2 * l + 1):
return 2 * l + 1
else:
return 2 * r + 1
def main():
T = int(input())
for _ in range(T):
print(solve())
main()
|
IMPORT ASSIGN VAR VAR FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR RETURN NUMBER IF BIN_OP NUMBER VAR BIN_OP VAR VAR RETURN NUMBER IF BIN_OP BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER RETURN NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER RETURN BIN_OP BIN_OP NUMBER VAR NUMBER RETURN BIN_OP BIN_OP NUMBER VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
n = int(input())
for i in range(n):
read = input().split(" ")
a = float(read[0])
b = float(read[1])
c = float(read[2])
if c == a:
print(1)
elif abs(c - (a + b) / 2) == 0:
print(2)
else:
k = abs(c - a) / abs(a + b - 2 * c)
k = int(2 * k + 1)
avr = abs(c - a)
res = 1
if abs(c - (a + b) / 2) < avr:
avr = abs(c - (a + b) / 2)
res = 2
Min = k
Max = min(1000000, k + 2)
for j in range(Min, Max + 1):
if j % 2:
h1 = int(j / 2) + 1
c1 = int(j / 2)
current = abs(c - (a * h1 + b * c1) / j)
if current < avr:
avr = current
res = j
print(res)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
input = sys.stdin.readline
def solve(mid):
return (mid + 1) * h + mid * c >= t * (2 * mid + 1)
t = int(input())
for _ in range(t):
h, c, t = map(int, input().split())
if 2 * t <= h + c:
print(2)
continue
if h <= t:
print(1)
continue
ok = 0
ng = 10**20
while abs(ok - ng) > 1:
mid = (ok + ng) // 2
if solve(mid):
ok = mid
else:
ng = mid
ans = 10**30
t_ = (2 * ok + 1) * (2 * ng + 1) * t
if ((ok + 1) * h + c * ok) * (2 * ng + 1) - t_ > t_ - ((ng + 1) * h + c * ng) * (
2 * ok + 1
):
print(2 * ng + 1)
else:
print(2 * ok + 1)
|
IMPORT ASSIGN VAR VAR FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER WHILE FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER VAR IF BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
t = int(input())
for i in range(t):
a = []
a = list(map(int, input().split()))
h = a[0]
c = a[1]
t = a[2]
ans = 0
m = 2
ans = float(h)
ch = ans
if h <= t:
print(1)
elif (h + c) / 2 >= t:
print(2)
else:
while 0 == 0:
m += 2
ch = ans
ans = ((h + c) * (m // 2) - c) / (m - 1)
if abs(ch - t) <= abs(ans - t):
print(m - 3)
break
if m > 10:
m = (1 - c / t) / (1 - (h + c) / 2 / t)
m = int(m)
if m % 2 == 0:
m += 1
ans1 = ((h + c) * (m // 2) + h) / m
ans2 = ((h + c) * (m // 2 - 1) + h) / (m - 2)
ans3 = ((h + c) * (m // 2 + 1) + h) / (m + 2)
ans1 = abs(t - ans1)
ans2 = abs(t - ans2)
ans3 = abs(t - ans3)
if ans1 <= ans2 and ans1 <= ans3:
print(m)
elif ans2 <= ans3 and ans2 <= ans1:
print(m - 2)
else:
print(m + 2)
break
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER WHILE NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER BIN_OP VAR VAR BIN_OP NUMBER BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
import sys
input = sys.stdin.readline
INF = 10**13
Q = int(input())
Query = [list(map(int, input().split())) for _ in range(Q)]
for h, c, t in Query:
if 2 * t <= h + c:
ans = 2
else:
l = 0
r = INF
while r - l > 1:
m = (l + r) // 2
if (h + c) * m + h > (2 * m + 1) * t:
l = m
else:
r = m
if 2 * t * (2 * r + 1) * (2 * l + 1) - ((h + c) * r + h) * (2 * l + 1) >= (
(h + c) * l + h
) * (2 * r + 1):
ans = 2 * l + 1
else:
ans = 2 * r + 1
print(ans)
|
IMPORT ASSIGN VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR VAR VAR VAR IF BIN_OP NUMBER VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR VAR ASSIGN VAR VAR IF BIN_OP BIN_OP BIN_OP BIN_OP NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
There are two infinite sources of water: hot water of temperature $h$; cold water of temperature $c$ ($c < h$).
You perform the following procedure of alternating moves: take one cup of the hot water and pour it into an infinitely deep barrel; take one cup of the cold water and pour it into an infinitely deep barrel; take one cup of the hot water $\dots$ and so on $\dots$
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to $t$. So if the temperature in the barrel is $t_b$, then the absolute difference of $t_b$ and $t$ ($|t_b - t|$) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to $t$? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
-----Input-----
The first line contains a single integer $T$ ($1 \le T \le 3 \cdot 10^4$) — the number of testcases.
Each of the next $T$ lines contains three integers $h$, $c$ and $t$ ($1 \le c < h \le 10^6$; $c \le t \le h$) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
-----Output-----
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to $t$.
-----Example-----
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
-----Note-----
In the first testcase the temperature after $2$ poured cups: $1$ hot and $1$ cold is exactly $20$. So that is the closest we can achieve.
In the second testcase the temperature after $7$ poured cups: $4$ hot and $3$ cold is about $29.857$. Pouring more water won't get us closer to $t$ than that.
In the third testcase the temperature after $1$ poured cup: $1$ hot is $18$. That's exactly equal to $t$.
|
from sys import stdin
def iinput():
return int(stdin.readline())
def sinput():
return input()
def minput():
return map(int, stdin.readline().split())
def linput():
return list(map(int, stdin.readline().split()))
T = iinput()
while T:
T -= 1
h, c, t = minput()
if h == t:
print(1)
elif (h + c) // 2 >= t:
print(2)
else:
k = (h - t) // (2 * t - h - c)
if abs(k * (h + c) + h - t * (2 * k + 1)) * (2 * k + 3) <= abs(
(k + 1) * (h + c) + h - t * (2 * k + 3)
) * (2 * k + 1):
print(2 * k + 1)
else:
print(2 * k + 3)
|
FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR WHILE VAR VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR IF BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER
|
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