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Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
t = int(input())
for i in range(t):
a = input()
b = input()
n = len(a)
z, o = 0, 0
for j in a:
if j == "0":
z = 1
else:
o = 1
if z == 0 or o == 0:
print("Unlucky Chef")
else:
n1, n2 = 0, 0
for j in range(n):
e = a[j]
if e == "1" and e != b[j]:
n1 += 1
elif e == "0" and e != b[j]:
n2 += 1
print("Lucky Chef")
ans = max(n1, n2)
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR IF VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR STRING VAR VAR VAR VAR NUMBER IF VAR STRING VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
for _ in range(int(input())):
A = input()
B = input()
cnt0 = 0
cnt1 = 0
if "0" not in A or "1" not in A:
print("Unlucky Chef")
continue
for i in range(len(A)):
if A[i] == "0" and B[i] == "1":
cnt0 += 1
elif A[i] == "1" and B[i] == "0":
cnt1 += 1
print("Lucky Chef")
print(max(cnt0, cnt1))
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF STRING VAR STRING VAR EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR VAR STRING VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
T = int(input())
while T > 0:
T -= 1
A = input()
B = input()
p = 0
q = 0
o = 0
z = 0
for i in range(len(A)):
if A[i] == B[i]:
if A[i] == "1" and B[i] == "1":
o += 1
elif A[i] == "0" and B[i] == "0":
z += 1
elif A[i] == "1" and B[i] == "0":
p += 1
elif A[i] == "0" and B[i] == "1":
q += 1
if p == q:
print("Lucky Chef")
print(p)
elif p != 0 and q != 0:
print("Lucky Chef")
print(max(p, q))
elif p == 0 and o != 0:
print("Lucky Chef")
print(q)
elif q == 0 and z != 0:
print("Lucky Chef")
print(p)
else:
print("Unlucky Chef")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR IF VAR VAR STRING VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR VAR STRING VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
t = int(input())
while t:
t -= 1
a = input()
b = input()
n = len(a)
c1a = 0
c1b = 0
cdiff = 0
for i in range(n):
if a[i] == "1":
c1a += 1
if b[i] == "1":
c1b += 1
if b[i] != a[i]:
cdiff += 1
c1diff = abs(c1a - c1b)
if c1a == n or c1a == 0:
print("Unlucky Chef")
else:
ctot = c1diff + cdiff
ctot = ctot // 2
print("Lucky Chef")
print(ctot)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
t = int(input())
for i in range(t):
a = input()
b = input()
o = 0
z = 0
s = set(a)
if len(s) < 2:
print("Unlucky Chef")
else:
l = len(a)
for j in range(l):
if a[j] != b[j]:
if a[j] == "0":
z += 1
else:
o += 1
count = abs(z - o) + min(z, o)
print("Lucky Chef")
print(count)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
def fun(a, b):
if "1" not in a or "0" not in a:
return "Unlucky Chef"
at = tuple(a)
bt = tuple(b)
a1 = 0
a0 = 0
for i in range(len(at)):
if at[i] != bt[i]:
if at[i] == "1":
a1 += 1
else:
a0 += 1
return min(a0, a1) + abs(a0 - a1)
t = int(input())
for i in range(t):
a = input()
b = input()
ans = fun(a, b)
if ans == "Unlucky Chef":
print("Unlucky Chef")
else:
print("Lucky Chef")
print(ans)
|
FUNC_DEF IF STRING VAR STRING VAR RETURN STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER RETURN BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
for t in range(int(input())):
a = input()
b = input()
f0 = f1 = c10 = c01 = 0
for i in range(len(a)):
if a[i] == "0":
f0 = 1
if b[i] == "1":
c01 += 1
else:
f1 = 1
if b[i] == "0":
c10 += 1
if f0 == 0 or f1 == 0:
print("Unlucky Chef")
else:
print("Lucky Chef")
print(min(c10, c01) + abs(c10 - c01))
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
from sys import stdin, stdout
def main():
from sys import stdin, stdout
for _ in range(int(stdin.readline())):
a = stdin.readline().strip()
b = stdin.readline().strip()
count = 0
ac1 = a.count("1")
ac0 = a.count("0")
a1 = 0
a0 = 0
for i in range(len(a)):
if a[i] != b[i]:
if a[i] == "1":
a1 += 1
else:
a0 += 1
if ac1 and ac0:
stdout.write("Lucky Chef\n" + str(max(a1, a0)) + "\n")
else:
stdout.write("Unlucky Chef\n")
main()
|
FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP STRING FUNC_CALL VAR FUNC_CALL VAR VAR VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
def is_possible(a, b):
return set(a) >= set(b)
t = int(input())
for _ in range(t):
a = input()
b = input()
if not is_possible(a, b):
print("Unlucky Chef")
continue
wrong_ones = sum(1 for x, y in zip(a, b) if x == "1" and y == "0")
wrong_zeroes = sum(1 for x, y in zip(a, b) if x == "0" and y == "1")
swaps = sum(1 for x, y in zip(a, b) if x != y)
print("Lucky Chef")
print(swaps - min(wrong_ones, wrong_zeroes))
|
FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR NUMBER VAR VAR FUNC_CALL VAR VAR VAR VAR STRING VAR STRING ASSIGN VAR FUNC_CALL VAR NUMBER VAR VAR FUNC_CALL VAR VAR VAR VAR STRING VAR STRING ASSIGN VAR FUNC_CALL VAR NUMBER VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
t = int(input())
for i in range(t):
zeroes = 0
ones = 0
zflag = 0
oflag = 0
a = str(input())
b = str(input())
for i in range(0, len(a)):
if a[i] == "0":
zflag = 1
if a[i] == "1":
oflag = 1
if a[i] != b[i]:
if b[i] == "1":
zeroes = zeroes + 1
else:
ones = ones + 1
if oflag == 0 or zflag == 0:
print("Unlucky Chef")
else:
ans = min(zeroes, ones)
ans = ans + max(zeroes, ones) - min(zeroes, ones)
print("Lucky Chef")
print(str(ans))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR NUMBER IF VAR VAR STRING ASSIGN VAR NUMBER IF VAR VAR VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
t = int(input())
for _ in range(t):
a = str(input())
b = str(input())
cnt = 0
cnt1 = 0
for i in range(len(a)):
if a[i] == "0" and b[i] == "1":
cnt += 1
elif a[i] == "1" and b[i] == "0":
cnt1 += 1
x = max(cnt, cnt1)
if a.count("1") == len(a) or a.count("0") == len(a):
print("Unlucky Chef")
else:
print("Lucky Chef")
print(x)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR STRING FUNC_CALL VAR VAR FUNC_CALL VAR STRING FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
t = input()
t = int(t)
while t > 0:
a = list(input())
b = list(input())
if "1" in b and "1" not in a or "0" in b and "0" not in a:
print("Unlucky Chef")
else:
print("Lucky Chef")
ans = 0
zero_to_one = 0
one_to_zero = 0
i = 0
n = len(a)
while i < n:
if a[i] == "1" and b[i] == "0":
one_to_zero += 1
elif a[i] == "0" and b[i] == "1":
zero_to_one += 1
i += 1
ans += min(one_to_zero, zero_to_one)
ans += max(one_to_zero, zero_to_one) - min(one_to_zero, zero_to_one)
print(ans)
t -= 1
|
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF STRING VAR STRING VAR STRING VAR STRING VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR IF VAR VAR STRING VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR VAR STRING VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
t = int(input())
while t > 0:
t = t - 1
a = str(input())
b = str(input())
f = len(a)
c = a.count("1")
d = a.count("0")
e = b.count("1")
g = b.count("0")
z = 0
x = 0
y = 0
k = 0
if f == 1:
k = 1
elif a == b:
k = 2
y = 1
elif c == f or d == f:
k = 1
else:
k = 2
for i in range(0, f):
if a[i] == "0" and b[i] == "1":
z = z + 1
if a[i] == "1" and b[i] == "0":
x = x + 1
if z == x:
y = z
elif z > x:
y = z
else:
y = x
if k == 1:
print("Unlucky Chef")
else:
print("Lucky Chef")
print(y)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
def zero_and_one_count(string):
zero = 0
one = 0
for char in string:
if char == "0":
zero += 1
else:
one += 1
return zero, one
def right_position(string1, string2):
zero = 0
one = 0
for i in range(len(string1)):
zero += string1[i] == string2[i] and string1[i] == "0"
one += string1[i] == string2[i] and string1[i] == "1"
return zero, one
T = int(input())
for _ in range(T):
A = input().strip()
B = input().strip()
zero_a, one_a = zero_and_one_count(A)
zero_b, one_b = zero_and_one_count(B)
zero_rght, one_rght = right_position(A, B)
ans = min(zero_a, zero_b) - zero_rght
one_rght += ans
zero_rght += ans
ans += one_b - one_rght
ans += zero_b - zero_rght
if zero_a == 0 and zero_b != 0 or one_a == 0 and one_b != 0:
print("Unlucky Chef")
else:
print("Lucky Chef")
print(ans)
|
FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER VAR NUMBER RETURN VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR STRING VAR VAR VAR VAR VAR VAR VAR STRING RETURN VAR 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 FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
T = int(input())
for t in range(T):
A = input()
B = input()
N = len(A)
c0, c1, c01, c10 = 0, 0, 0, 0
for i in range(N):
if A[i] == "0":
c0 += 1
c01 += B[i] == "1"
else:
c1 += 1
c10 += B[i] == "0"
if c01 and c0 == N or c10 and c1 == N:
print("Unlucky Chef")
else:
print("Lucky Chef")
print(max(c01, c10))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR NUMBER NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR VAR VAR STRING VAR NUMBER VAR VAR VAR STRING IF VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
for i in range(0, int(input())):
a = input()
b = input()
c = 0
d = 0
if len(a) != len(b):
print("Unlucky Chef")
print(1)
else:
for i in a:
if i == "0":
c = c + 1
else:
d = d + 1
if c == 0 or d == 0:
print("Unlucky Chef")
else:
print("Lucky Chef")
e = []
for i in range(0, len(a)):
if a[i] != b[i]:
e.append(int(a[i]))
l1 = e.count(1)
l0 = e.count(0)
if l1 > l0:
print(len(e) - l0)
else:
print(len(e) - l1)
|
FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER FOR VAR VAR IF VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR
|
Today is Chef's birthday. His mom decided to surprise him with a truly fantastic gift: his favourite binary string B. But, unfortunately, all the stocks of binary string B have been sold out, and only a binary string A (A ≠ B) is available in the market.
She purchases the string A and tries to convert it to string B by applying any of following three operations zero or more times.
AND Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai & Aj
- Ai = result & Ai
- Aj = result & Aj
OR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai | Aj
- Ai = result | Ai
- Aj = result | Aj
XOR Operation:
She will choose a pair of indices i and j such that i != j and perform following sequence of operations.
- result = Ai ^ Aj
- Ai = result ^ Ai
- Aj = result ^ Aj
Chef's mom is eagerly waiting to surprise him with his favourite gift and therefore, she wants to convert string A to string B as fast as possible. Can you please help her by telling her the minimum number of operations she will require? If it is impossible to do so, then let Chef's mom know about it.
-----Input-----
First line of input contains a single integer T denoting the number of test cases. T test cases follow.
First line of each test case, will contain binary string A.
Second line of each test case, will contain binary string B.
-----Output-----
For each test case, Print "Lucky Chef" (without quotes) in first line and minimum number of operations required to convert string A to sting B in second line if conversion is possible. Print "Unlucky Chef" (without quotes) in a new line otherwise.
-----Constraints-----
- 1 ≤ T ≤ 105
- 1 ≤ |A| ≤ 106
- 1 ≤ |B| ≤ 106
- A != B
- |A| = |B|
- sum of |A| over all test cases does not exceed 106
- sum of |B| over all test cases does not exceed 106
-----Subtasks-----
- Subtask #1 (40 points) : Sum of |A| & |B| over all test cases does not exceed 103
- Subtask #2 (60 points) : Sum of |A| & |B| over all test cases does not exceed 106
-----Example-----
Input
2
101
010
1111
1010
Output
Lucky Chef
2
Unlucky Chef
-----Explanation-----
Example case 1.
- Applying XOR operation with indices i = 1 and j = 2. Resulting string will be 011.
- Then, Applying AND operation with indices i = 1 and j = 3. Resulting string will be 010.
Example case 2.
- It is impossible to convert string A to string B.
|
t = int(input())
while t != 0:
zero = 0
one = 0
t = t - 1
A = input()
B = input()
for i in range(len(A)):
if A[i] != B[i]:
if A[i] == "0":
zero = zero + 1
else:
one = one + 1
if "0" not in A or "1" not in A:
print("Unlucky Chef")
else:
print("Lucky Chef")
print(min(zero, one) + abs(zero - one))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF STRING VAR STRING VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR
|
Read problems statements in Russian here
Polo, the Penguin, likes the XOR operation. Please read NOTE if you are not familiar with XOR operation.
XOR-sum of a list of numbers is the result of XOR-ing all of them. XOR-sum of (A[1] XOR A[2] XOR ... XOR A[N]) is defined as A[1] XOR (A[2] XOR (A[3] XOR ( ... XOR A[N]))).
He has an array A consisting of N integers. Index in the array are numbered from 1 to N, inclusive. Let us denote by F(L, R), the XOR-sum of all integers in the array A whose indices lie from L to R, inclusive, i.e. F(L, R) = A[L] XOR A[L+1] XOR ... XOR A[R]. Your task is to find the total sum of XOR-sums F(L, R) over all L and R such that 1 ≤ L ≤ R ≤ N.
------ Input ------
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer N denoting the size of A. The second line contains N space-separated integers A[1], A[2], ..., A[N].
------ Output ------
For each test case, output a single line containing the total sum to the corresponding test case.
------ Constraints ------
$1 ≤ T ≤ 100,000$
$1 ≤ N ≤ 100,000$
$0 ≤ A[i] ≤ 1,000,000,000 (10^{9})$
$The total sum of all N over all test cases will not exceed 100,000.$
------ Example ------
Input:
1
2
1 2
Output:
6
------ Explanation ------
Example case 1. F(1, 1) = A[1] = 1, F(2, 2) = A[2] = 2 and F(1, 2) = A[1] XOR A[2] = 1 XOR 2 = 3. Hence the answer is 1 + 2 + 3 = 6.
------ NOTE ------
XOR operation is a bitwise "Exclusive OR" operation performed on two integers in binary representation. First, the shorter number is prepended with leading zeroes until the numbers have equal size in binary. Then the resulting number (also in binary) contains 0 in all positions where the corresponding bits coincide, and 1 on the rest of the positions.
For example, 3 XOR 5 = 011_{2} XOR 101_{2} = 110_{2} = 6.
|
t = int(input())
while t:
t -= 1
n = int(input())
s = input()
s = s.split()
li = []
for i in s:
li += [int(i)]
res = 0
for i in range(0, 31):
s = 0
ones = 0
e = 0
o = 0
zero = 0
for j in li:
if 1 << i & j == 0:
if ones % 2 == 0:
s += e
else:
s += o
zero += 1
else:
ones += 1
if ones % 2 == 0:
e += zero + 1
s += e - 1
else:
o += zero + 1
s += o - 1
zero = 0
res += 2**i * s
res += sum(li)
print(res)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR VAR LIST FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP BIN_OP NUMBER VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
Read problems statements in Russian here
Polo, the Penguin, likes the XOR operation. Please read NOTE if you are not familiar with XOR operation.
XOR-sum of a list of numbers is the result of XOR-ing all of them. XOR-sum of (A[1] XOR A[2] XOR ... XOR A[N]) is defined as A[1] XOR (A[2] XOR (A[3] XOR ( ... XOR A[N]))).
He has an array A consisting of N integers. Index in the array are numbered from 1 to N, inclusive. Let us denote by F(L, R), the XOR-sum of all integers in the array A whose indices lie from L to R, inclusive, i.e. F(L, R) = A[L] XOR A[L+1] XOR ... XOR A[R]. Your task is to find the total sum of XOR-sums F(L, R) over all L and R such that 1 ≤ L ≤ R ≤ N.
------ Input ------
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer N denoting the size of A. The second line contains N space-separated integers A[1], A[2], ..., A[N].
------ Output ------
For each test case, output a single line containing the total sum to the corresponding test case.
------ Constraints ------
$1 ≤ T ≤ 100,000$
$1 ≤ N ≤ 100,000$
$0 ≤ A[i] ≤ 1,000,000,000 (10^{9})$
$The total sum of all N over all test cases will not exceed 100,000.$
------ Example ------
Input:
1
2
1 2
Output:
6
------ Explanation ------
Example case 1. F(1, 1) = A[1] = 1, F(2, 2) = A[2] = 2 and F(1, 2) = A[1] XOR A[2] = 1 XOR 2 = 3. Hence the answer is 1 + 2 + 3 = 6.
------ NOTE ------
XOR operation is a bitwise "Exclusive OR" operation performed on two integers in binary representation. First, the shorter number is prepended with leading zeroes until the numbers have equal size in binary. Then the resulting number (also in binary) contains 0 in all positions where the corresponding bits coincide, and 1 on the rest of the positions.
For example, 3 XOR 5 = 011_{2} XOR 101_{2} = 110_{2} = 6.
|
import sys
data = sys.stdin.read().split()
data.reverse()
def read():
return int(data.pop())
def readStr():
return data.pop()
T = read()
res = []
for test in range(T):
n = read()
A = [read() for i in range(n)]
ans = 0
for b in range(30):
cnt = 0
V = []
sum = [0, 0]
for a in A:
if a & 1 << b:
V.append(cnt + 1)
sum[len(V) % 2] += cnt + 1
cnt = 0
else:
cnt += 1
V.append(cnt + 1)
sum[len(V) % 2] += cnt + 1
cnt = 0
for i in range(len(V)):
s = V[i]
cnt += s * sum[i % 2]
sum[(i + 1) % 2] -= s
ans += cnt * (1 << b)
res.append(str(ans))
print("\n".join(res))
|
IMPORT ASSIGN VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST NUMBER NUMBER FOR VAR VAR IF BIN_OP VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Read problems statements in Russian here
Polo, the Penguin, likes the XOR operation. Please read NOTE if you are not familiar with XOR operation.
XOR-sum of a list of numbers is the result of XOR-ing all of them. XOR-sum of (A[1] XOR A[2] XOR ... XOR A[N]) is defined as A[1] XOR (A[2] XOR (A[3] XOR ( ... XOR A[N]))).
He has an array A consisting of N integers. Index in the array are numbered from 1 to N, inclusive. Let us denote by F(L, R), the XOR-sum of all integers in the array A whose indices lie from L to R, inclusive, i.e. F(L, R) = A[L] XOR A[L+1] XOR ... XOR A[R]. Your task is to find the total sum of XOR-sums F(L, R) over all L and R such that 1 ≤ L ≤ R ≤ N.
------ Input ------
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer N denoting the size of A. The second line contains N space-separated integers A[1], A[2], ..., A[N].
------ Output ------
For each test case, output a single line containing the total sum to the corresponding test case.
------ Constraints ------
$1 ≤ T ≤ 100,000$
$1 ≤ N ≤ 100,000$
$0 ≤ A[i] ≤ 1,000,000,000 (10^{9})$
$The total sum of all N over all test cases will not exceed 100,000.$
------ Example ------
Input:
1
2
1 2
Output:
6
------ Explanation ------
Example case 1. F(1, 1) = A[1] = 1, F(2, 2) = A[2] = 2 and F(1, 2) = A[1] XOR A[2] = 1 XOR 2 = 3. Hence the answer is 1 + 2 + 3 = 6.
------ NOTE ------
XOR operation is a bitwise "Exclusive OR" operation performed on two integers in binary representation. First, the shorter number is prepended with leading zeroes until the numbers have equal size in binary. Then the resulting number (also in binary) contains 0 in all positions where the corresponding bits coincide, and 1 on the rest of the positions.
For example, 3 XOR 5 = 011_{2} XOR 101_{2} = 110_{2} = 6.
|
for _ in range(int(input())):
n = int(input())
arr = [int(c) for c in input().split()]
s = []
xorval = 0
for i in arr:
xorval ^= i
s.append(xorval)
count = [(0) for i in range(32)]
ans = sum(s)
for i in range(32):
if s[0] & 1 << i:
count[i] += 1
for j in range(1, len(s)):
for i in range(32):
if s[j] & 1 << i:
ans += (j - count[i]) * 1 << i
else:
ans += count[i] * (1 << i)
for i in range(32):
if s[j] & 1 << i:
count[i] += 1
print(ans)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR NUMBER BIN_OP NUMBER VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR VAR BIN_OP NUMBER VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER VAR VAR BIN_OP VAR VAR BIN_OP NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR VAR BIN_OP NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Read problems statements in Russian here
Polo, the Penguin, likes the XOR operation. Please read NOTE if you are not familiar with XOR operation.
XOR-sum of a list of numbers is the result of XOR-ing all of them. XOR-sum of (A[1] XOR A[2] XOR ... XOR A[N]) is defined as A[1] XOR (A[2] XOR (A[3] XOR ( ... XOR A[N]))).
He has an array A consisting of N integers. Index in the array are numbered from 1 to N, inclusive. Let us denote by F(L, R), the XOR-sum of all integers in the array A whose indices lie from L to R, inclusive, i.e. F(L, R) = A[L] XOR A[L+1] XOR ... XOR A[R]. Your task is to find the total sum of XOR-sums F(L, R) over all L and R such that 1 ≤ L ≤ R ≤ N.
------ Input ------
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer N denoting the size of A. The second line contains N space-separated integers A[1], A[2], ..., A[N].
------ Output ------
For each test case, output a single line containing the total sum to the corresponding test case.
------ Constraints ------
$1 ≤ T ≤ 100,000$
$1 ≤ N ≤ 100,000$
$0 ≤ A[i] ≤ 1,000,000,000 (10^{9})$
$The total sum of all N over all test cases will not exceed 100,000.$
------ Example ------
Input:
1
2
1 2
Output:
6
------ Explanation ------
Example case 1. F(1, 1) = A[1] = 1, F(2, 2) = A[2] = 2 and F(1, 2) = A[1] XOR A[2] = 1 XOR 2 = 3. Hence the answer is 1 + 2 + 3 = 6.
------ NOTE ------
XOR operation is a bitwise "Exclusive OR" operation performed on two integers in binary representation. First, the shorter number is prepended with leading zeroes until the numbers have equal size in binary. Then the resulting number (also in binary) contains 0 in all positions where the corresponding bits coincide, and 1 on the rest of the positions.
For example, 3 XOR 5 = 011_{2} XOR 101_{2} = 110_{2} = 6.
|
import sys
def bin_one(num):
s = bin(num)[2:]
return set([(len(s) - i - 1) for i in range(len(s)) if s[i] == "1"])
t = int(sys.stdin.readline())
for _ in range(t):
n = int(sys.stdin.readline())
A = [int(i) for i in sys.stdin.readline().rstrip().split()]
max_bin_len = len(bin(max(A))[2:])
C = [0] * max_bin_len
S = 0
total_sum = 0
for i in range(0, n):
S ^= A[i]
index_ones = bin_one(S)
for j in range(len(C)):
if j in index_ones:
total_sum += (i + 1 - C[j]) * pow(2, j)
C[j] += 1
else:
total_sum += C[j] * pow(2, j)
print(total_sum)
|
IMPORT FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR NUMBER VAR VAR VAR NUMBER VAR BIN_OP VAR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR VAR
|
Read problems statements in Russian here
Polo, the Penguin, likes the XOR operation. Please read NOTE if you are not familiar with XOR operation.
XOR-sum of a list of numbers is the result of XOR-ing all of them. XOR-sum of (A[1] XOR A[2] XOR ... XOR A[N]) is defined as A[1] XOR (A[2] XOR (A[3] XOR ( ... XOR A[N]))).
He has an array A consisting of N integers. Index in the array are numbered from 1 to N, inclusive. Let us denote by F(L, R), the XOR-sum of all integers in the array A whose indices lie from L to R, inclusive, i.e. F(L, R) = A[L] XOR A[L+1] XOR ... XOR A[R]. Your task is to find the total sum of XOR-sums F(L, R) over all L and R such that 1 ≤ L ≤ R ≤ N.
------ Input ------
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer N denoting the size of A. The second line contains N space-separated integers A[1], A[2], ..., A[N].
------ Output ------
For each test case, output a single line containing the total sum to the corresponding test case.
------ Constraints ------
$1 ≤ T ≤ 100,000$
$1 ≤ N ≤ 100,000$
$0 ≤ A[i] ≤ 1,000,000,000 (10^{9})$
$The total sum of all N over all test cases will not exceed 100,000.$
------ Example ------
Input:
1
2
1 2
Output:
6
------ Explanation ------
Example case 1. F(1, 1) = A[1] = 1, F(2, 2) = A[2] = 2 and F(1, 2) = A[1] XOR A[2] = 1 XOR 2 = 3. Hence the answer is 1 + 2 + 3 = 6.
------ NOTE ------
XOR operation is a bitwise "Exclusive OR" operation performed on two integers in binary representation. First, the shorter number is prepended with leading zeroes until the numbers have equal size in binary. Then the resulting number (also in binary) contains 0 in all positions where the corresponding bits coincide, and 1 on the rest of the positions.
For example, 3 XOR 5 = 011_{2} XOR 101_{2} = 110_{2} = 6.
|
def solver():
T = int(input())
for _ in range(T):
ans = 0
n = int(input())
arr = [int(x) for x in input().split()]
for i in range(1, len(arr)):
arr[i] = arr[i] ^ arr[i - 1]
for i in range(32, -1, -1):
t = 1 << i
cnt = 0
for e in arr:
if e & t != 0:
cnt += 1
ans += cnt * (n - cnt + 1) * t
print(ans)
solver()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
You are given arrays A and B with N non-negative integers each.
An array X of length N is called *good*, if:
All elements of the array X are non-negative;
X_{1} | X_{2} | \ldots | X_{i} = A_{i} for all (1≤ i≤ N);
X_{i} \& X_{(i+1)} \& \ldots \& X_{N} = B_{i} for all (1≤ i≤ N).
Find the maximum bitwise XOR of all elements over all good arrays X.
More formally, find the maximum value of X_{1}\oplus X_{2}\oplus \ldots X_{N}, over all good arrays X.
It is guaranteed that at least one such array X exists.
Note that |, \&, and \oplus denote the bitwise [or], [and], and [xor] operations respectively.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of multiple lines of input.
- The first line of each test case contains one integer N — the size of the array.
- The next line contains N space-separated integers describing the elements of the array A.
- The next line contains N space-separated integers describing the elements of the array B.
------ Output Format ------
For each test case, output on a new line, the maximum bitwise XOR of all elements over all good arrays X.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} < 2^{30}$
$0 ≤ B_{i} < 2^{30}$
- It is guaranteed that at least one such array $X$ exists.
- Sum of $N$ over all test cases is less than $3\cdot 10^{5}$.
----- Sample Input 1 ------
2
3
0 3 3
0 2 2
2
2 3
0 1
----- Sample Output 1 ------
1
3
----- explanation 1 ------
Test case $1$: An optimal *good* array is $X = [0, 3, 2]$.
- For $i = 1$: $A_{1} = X_{1} = 0$ and $B_{1} = X_{1}$ $\&$ $X_{2}$ $\&$ $X_{3} = 0$.
- For $i = 2$: $A_{2} = X_{1}$ $|$ $X_{2} = 3$ and $B_{2} = X_{2}$ $\&$ $X_{3} = 2$.
- For $i = 3$: $A_{3} = X_{1}$ $|$ $X_{2}$ $|$ $X_{3} = 3$ and $B_{3} = X_{3} = 2$.
The XOR of all elements of $X$ is $0\oplus 3\oplus 2 = 1$. It can be proven that this is the maximum XOR value for any $X$.
Test case $2$: An optimal *good* array is $X = [2, 1]$.
- For $i = 1$: $A_{1} = X_{1} = 2$ and $B_{1} = X_{1}$ $\&$ $X_{2} = 0$.
- For $i = 2$: $A_{2} = X_{1}$ $|$ $X_{2} = 3$ and $B_{2} = X_{2} = 1$.
The XOR of all elements of $X$ is $2\oplus 1 = 3$. It can be proven that this is the maximum XOR value for any $X$.
|
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
first_1 = [-1] * 31
last_1 = [n] * 31
for i in range(n):
for bit in range(30):
if a[i] & 1 << bit and first_1[bit] == -1:
first_1[bit] = i
if b[i] & 1 << bit and last_1[bit] == n:
last_1[bit] = i
ans = 0
for i in range(31):
L, R = first_1[i], last_1[i]
if L == -1:
continue
if L == R:
if (n - R) % 2:
ans += 2**i
elif L + 2 == R:
if (n - R) % 2 == 0:
ans += 2**i
else:
ans += 2**i
print(ans)
|
IMPORT ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR 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 FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR VAR BIN_OP NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR IF BIN_OP VAR VAR BIN_OP NUMBER VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR IF VAR NUMBER IF VAR VAR IF BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP NUMBER VAR IF BIN_OP VAR NUMBER VAR IF BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR BIN_OP NUMBER VAR VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given arrays A and B with N non-negative integers each.
An array X of length N is called *good*, if:
All elements of the array X are non-negative;
X_{1} | X_{2} | \ldots | X_{i} = A_{i} for all (1≤ i≤ N);
X_{i} \& X_{(i+1)} \& \ldots \& X_{N} = B_{i} for all (1≤ i≤ N).
Find the maximum bitwise XOR of all elements over all good arrays X.
More formally, find the maximum value of X_{1}\oplus X_{2}\oplus \ldots X_{N}, over all good arrays X.
It is guaranteed that at least one such array X exists.
Note that |, \&, and \oplus denote the bitwise [or], [and], and [xor] operations respectively.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of multiple lines of input.
- The first line of each test case contains one integer N — the size of the array.
- The next line contains N space-separated integers describing the elements of the array A.
- The next line contains N space-separated integers describing the elements of the array B.
------ Output Format ------
For each test case, output on a new line, the maximum bitwise XOR of all elements over all good arrays X.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} < 2^{30}$
$0 ≤ B_{i} < 2^{30}$
- It is guaranteed that at least one such array $X$ exists.
- Sum of $N$ over all test cases is less than $3\cdot 10^{5}$.
----- Sample Input 1 ------
2
3
0 3 3
0 2 2
2
2 3
0 1
----- Sample Output 1 ------
1
3
----- explanation 1 ------
Test case $1$: An optimal *good* array is $X = [0, 3, 2]$.
- For $i = 1$: $A_{1} = X_{1} = 0$ and $B_{1} = X_{1}$ $\&$ $X_{2}$ $\&$ $X_{3} = 0$.
- For $i = 2$: $A_{2} = X_{1}$ $|$ $X_{2} = 3$ and $B_{2} = X_{2}$ $\&$ $X_{3} = 2$.
- For $i = 3$: $A_{3} = X_{1}$ $|$ $X_{2}$ $|$ $X_{3} = 3$ and $B_{3} = X_{3} = 2$.
The XOR of all elements of $X$ is $0\oplus 3\oplus 2 = 1$. It can be proven that this is the maximum XOR value for any $X$.
Test case $2$: An optimal *good* array is $X = [2, 1]$.
- For $i = 1$: $A_{1} = X_{1} = 2$ and $B_{1} = X_{1}$ $\&$ $X_{2} = 0$.
- For $i = 2$: $A_{2} = X_{1}$ $|$ $X_{2} = 3$ and $B_{2} = X_{2} = 1$.
The XOR of all elements of $X$ is $2\oplus 1 = 3$. It can be proven that this is the maximum XOR value for any $X$.
|
import sys
def mi():
return map(int, input().split())
def li():
return list(mi())
def si():
return str(input())
def lsi():
return si().strip().split(" ")
def ni():
return int(input())
input = sys.stdin.readline
for _ in range(ni()):
n = ni()
a = li()
b = li()
first_1 = [-1] * 31
last_1 = [n] * 31
for i in range(n):
for bit in range(30):
if a[i] & 1 << bit and first_1[bit] == -1:
first_1[bit] = i
if b[i] & 1 << bit and last_1[bit] == n:
last_1[bit] = i
ans = 0
for i in range(31):
L, R = first_1[i], last_1[i]
if L == -1:
continue
if L == R:
if (n - R) % 2:
ans += 2**i
elif L + 2 == R:
if (n - R) % 2 == 0:
ans += 2**i
else:
ans += 2**i
print(ans)
|
IMPORT FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR VAR BIN_OP NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR IF BIN_OP VAR VAR BIN_OP NUMBER VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR IF VAR NUMBER IF VAR VAR IF BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP NUMBER VAR IF BIN_OP VAR NUMBER VAR IF BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR BIN_OP NUMBER VAR VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given arrays A and B with N non-negative integers each.
An array X of length N is called *good*, if:
All elements of the array X are non-negative;
X_{1} | X_{2} | \ldots | X_{i} = A_{i} for all (1≤ i≤ N);
X_{i} \& X_{(i+1)} \& \ldots \& X_{N} = B_{i} for all (1≤ i≤ N).
Find the maximum bitwise XOR of all elements over all good arrays X.
More formally, find the maximum value of X_{1}\oplus X_{2}\oplus \ldots X_{N}, over all good arrays X.
It is guaranteed that at least one such array X exists.
Note that |, \&, and \oplus denote the bitwise [or], [and], and [xor] operations respectively.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of multiple lines of input.
- The first line of each test case contains one integer N — the size of the array.
- The next line contains N space-separated integers describing the elements of the array A.
- The next line contains N space-separated integers describing the elements of the array B.
------ Output Format ------
For each test case, output on a new line, the maximum bitwise XOR of all elements over all good arrays X.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} < 2^{30}$
$0 ≤ B_{i} < 2^{30}$
- It is guaranteed that at least one such array $X$ exists.
- Sum of $N$ over all test cases is less than $3\cdot 10^{5}$.
----- Sample Input 1 ------
2
3
0 3 3
0 2 2
2
2 3
0 1
----- Sample Output 1 ------
1
3
----- explanation 1 ------
Test case $1$: An optimal *good* array is $X = [0, 3, 2]$.
- For $i = 1$: $A_{1} = X_{1} = 0$ and $B_{1} = X_{1}$ $\&$ $X_{2}$ $\&$ $X_{3} = 0$.
- For $i = 2$: $A_{2} = X_{1}$ $|$ $X_{2} = 3$ and $B_{2} = X_{2}$ $\&$ $X_{3} = 2$.
- For $i = 3$: $A_{3} = X_{1}$ $|$ $X_{2}$ $|$ $X_{3} = 3$ and $B_{3} = X_{3} = 2$.
The XOR of all elements of $X$ is $0\oplus 3\oplus 2 = 1$. It can be proven that this is the maximum XOR value for any $X$.
Test case $2$: An optimal *good* array is $X = [2, 1]$.
- For $i = 1$: $A_{1} = X_{1} = 2$ and $B_{1} = X_{1}$ $\&$ $X_{2} = 0$.
- For $i = 2$: $A_{2} = X_{1}$ $|$ $X_{2} = 3$ and $B_{2} = X_{2} = 1$.
The XOR of all elements of $X$ is $2\oplus 1 = 3$. It can be proven that this is the maximum XOR value for any $X$.
|
def read_int():
return int(input())
def read_ints():
return [int(x) for x in input().split()]
def read_case():
n = read_int()
a = read_ints()
b = read_ints()
return n, a, b
def solve(n, a, b):
x = 0
bit = 1
while bit <= 2**30:
first_1 = 0
while first_1 < len(a) and a[first_1] & bit == 0:
first_1 += 1
if first_1 == len(a):
bit *= 2
continue
last_0 = len(b) - 1
while last_0 >= 0 and b[last_0] & bit != 0:
last_0 -= 1
if last_0 > first_1 + 1:
x |= bit
elif last_0 == first_1 + 1:
if (len(b) - last_0) % 2 == 1:
x |= bit
elif (len(b) - first_1) % 2 == 1:
x |= bit
bit *= 2
return x
cases = read_int()
for case in range(cases):
print(solve(*read_case()))
|
FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR RETURN VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR NUMBER BIN_OP VAR VAR VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR IF VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR VAR IF BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR
|
You are given arrays A and B with N non-negative integers each.
An array X of length N is called *good*, if:
All elements of the array X are non-negative;
X_{1} | X_{2} | \ldots | X_{i} = A_{i} for all (1≤ i≤ N);
X_{i} \& X_{(i+1)} \& \ldots \& X_{N} = B_{i} for all (1≤ i≤ N).
Find the maximum bitwise XOR of all elements over all good arrays X.
More formally, find the maximum value of X_{1}\oplus X_{2}\oplus \ldots X_{N}, over all good arrays X.
It is guaranteed that at least one such array X exists.
Note that |, \&, and \oplus denote the bitwise [or], [and], and [xor] operations respectively.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of multiple lines of input.
- The first line of each test case contains one integer N — the size of the array.
- The next line contains N space-separated integers describing the elements of the array A.
- The next line contains N space-separated integers describing the elements of the array B.
------ Output Format ------
For each test case, output on a new line, the maximum bitwise XOR of all elements over all good arrays X.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} < 2^{30}$
$0 ≤ B_{i} < 2^{30}$
- It is guaranteed that at least one such array $X$ exists.
- Sum of $N$ over all test cases is less than $3\cdot 10^{5}$.
----- Sample Input 1 ------
2
3
0 3 3
0 2 2
2
2 3
0 1
----- Sample Output 1 ------
1
3
----- explanation 1 ------
Test case $1$: An optimal *good* array is $X = [0, 3, 2]$.
- For $i = 1$: $A_{1} = X_{1} = 0$ and $B_{1} = X_{1}$ $\&$ $X_{2}$ $\&$ $X_{3} = 0$.
- For $i = 2$: $A_{2} = X_{1}$ $|$ $X_{2} = 3$ and $B_{2} = X_{2}$ $\&$ $X_{3} = 2$.
- For $i = 3$: $A_{3} = X_{1}$ $|$ $X_{2}$ $|$ $X_{3} = 3$ and $B_{3} = X_{3} = 2$.
The XOR of all elements of $X$ is $0\oplus 3\oplus 2 = 1$. It can be proven that this is the maximum XOR value for any $X$.
Test case $2$: An optimal *good* array is $X = [2, 1]$.
- For $i = 1$: $A_{1} = X_{1} = 2$ and $B_{1} = X_{1}$ $\&$ $X_{2} = 0$.
- For $i = 2$: $A_{2} = X_{1}$ $|$ $X_{2} = 3$ and $B_{2} = X_{2} = 1$.
The XOR of all elements of $X$ is $2\oplus 1 = 3$. It can be proven that this is the maximum XOR value for any $X$.
|
def check_bit(n, a, b, bit):
required = 0
possible = False
for i in range(n):
must_be_0 = must_be_1 = False
ai = a[i] >> bit & 1
must_be_0 = must_be_0 or ai == 0
if i > 0:
aim1 = a[i - 1] >> bit & 1
must_be_1 = must_be_1 or aim1 == 0 and ai == 1
else:
must_be_1 = must_be_1 or ai == 1
bi = b[i] >> bit & 1
must_be_1 = must_be_1 or bi == 1
if i + 1 < n:
bip1 = b[i + 1] >> bit & 1
must_be_0 = must_be_0 or bip1 == 1 and bi == 0
else:
must_be_0 = must_be_0 or bi == 0
if must_be_1:
required += 1
if not must_be_1 and not must_be_0:
possible = True
return required & 1 == 1 or possible
def solve(n, a, b):
answer = 0
for bit in range(30):
if check_bit(n, a, b, bit):
answer |= 1 << bit
return answer
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
print(solve(n, a, b))
|
FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER RETURN BIN_OP VAR NUMBER NUMBER VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR VAR BIN_OP NUMBER VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR 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 FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
for t1 in range(t):
n = int(input())
a = list(map(int, input().split()))
x = [0] * 10000
k = 0
for i in a:
j = i
z = 0
while j > 0:
if j % 2 == 1:
x[z] = x[z] + 1
k = max(k, z)
z = z + 1
j = j // 2
for i in range(k + 1):
if x[i] >= 2:
x[i] = 1
else:
x[i] = 0
ans = 0
for i in range(k + 1):
if x[i] == 1:
ans = ans + 2**i
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
T = int(input())
while T > 0:
n = int(input())
arr = list(map(int, input().strip().split()))
setcount = []
for i in range(30):
temp = 0
setcount.append(temp)
for i in range(n):
for j in range(30):
if arr[i] & 1 * 2**j:
setcount[j] += 1
answer = 0
for i in range(30):
if setcount[i] > 1:
answer |= 1 * 2**i
print(answer)
T = T - 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR VAR BIN_OP NUMBER BIN_OP NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER VAR BIN_OP NUMBER BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
while t:
n = int(input())
a = list(map(int, input().strip().split()))
b = [(0) for i in range(30)]
for i in range(n):
for j in range(30):
if a[i] & 1 << j:
b[j] += 1
ans = 0
for i in range(30):
if b[i] > 1:
ans |= 1 << i
print(ans)
t -= 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR VAR BIN_OP NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
T = int(input())
for i in range(T):
N = int(input())
arrA = [int(x) for x in input().split()]
BinArr = [0] * 32
for num in arrA:
i = -1
while i >= -32:
if num % 2 == 1:
BinArr[i] += 1
num = num // 2
i -= 1
sum = 0
for j in range(len(BinArr)):
if BinArr[j] >= 2:
sum += 2 ** (len(BinArr) - j - 1)
print(sum)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR BIN_OP NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
cn = 1
ans = 0
for j in range(32):
c = 0
for k in range(n):
if arr[k] % 2:
c += 1
arr[k] = arr[k] // 2
if c > 1:
ans += cn
cn *= 2
print(ans)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
for l in range(t):
n = int(input())
a = [k for k in map(int, input().split())]
b = [(0) for i in range(32)]
j = 0
for k in range(0, n):
z = 0
for m in range(0, 31):
if a[k] % 2 == 1:
b[z] += 1
z += 1
a[k] = a[k] // 2
else:
a[k] = a[k] // 2
z += 1
ans = 0
for e in b:
if e > 1:
ans = ans + 2**j
j += 1
else:
j += 1
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER IF BIN_OP VAR VAR NUMBER NUMBER VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
while t:
n = int(input())
li = list(map(int, input().strip().split()))
l = []
di = {}
ml = 0
c = 0
for i in range(32):
di[i] = "0"
for i in li:
bs = bin(i)
l.append(bs[::-1])
lon = len(l[len(l) - 1])
if ml < lon:
ml = lon
for i in range(ml):
c = 0
for j in range(len(l)):
if i < len(l[j]) and l[j][i] == "1":
c += 1
if c > 1:
di[i] = "1"
ans = ""
for i in range(32):
ans += di[i]
print(int(ans[::-1], 2))
t -= 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR DICT ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR STRING FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR STRING FOR VAR FUNC_CALL VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
for i in range(t):
n = int(input())
li = input().split()
for j in range(n):
li[j] = int(li[j])
m = [0] * 40
for j in range(n):
temp = bin(li[j])[2:]
for k in range(len(temp)):
m[-k - 1] = m[-k - 1] + int(temp[-k - 1])
ans = 0
for j in range(40):
if m[j] > 1:
ans = ans + 2 ** (40 - j - 1)
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
T = int(input())
for i in range(T):
N = int(input())
Arr = [int(i) for i in input().split()]
pos = 1
ans = 0
for i in range(32):
count = 0
for j in range(len(Arr)):
if Arr[j] % 2 == 1:
count += 1
Arr[j] = Arr[j] // 2
if count > 1:
ans = ans + pos
pos = pos * 2
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
case = int(input())
for i in range(case):
n = int(input())
arr = list(map(int, input().split(" ")))
i = 30
onebits = []
while i > 0:
count = 0
for j in range(n):
count += arr[j] & 1
arr[j] = arr[j] >> 1
onebits.append(count)
i -= 1
res = 0
power = 0
for counts in onebits:
if counts >= 2:
res += pow(2, power)
power += 1
print(res)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR 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 NUMBER ASSIGN VAR LIST WHILE VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
def solve(ar):
n = len(ar)
out = [0] * 32
for i in range(n):
idx = 0
tmp = ar[i]
while tmp != 0:
if tmp & 1 == 1:
out[idx] += 1
tmp = tmp >> 1
idx += 1
res = 0
p = 0
for i in out:
if i > 1:
res += 2**p
p += 1
return res
for i in range(t):
n = int(input())
x = list(map(int, input().split()))
print(solve(x))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR WHILE VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR BIN_OP NUMBER VAR VAR NUMBER RETURN VAR FOR VAR FUNC_CALL VAR VAR 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 FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
while t:
n = int(input())
l = [int(i) for i in input().split()]
if n == 1:
print(l[0])
x = ans = 0
for i in range(n - 1, 0, -1):
x |= l[i]
ans |= l[i - 1] & x
print(ans)
t -= 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
nl = [0] * 32
for i in range(n):
for j in range(32):
nl[j] += l[i] & 1 << j > 0
ans = 0
for i in range(32):
if nl[i] >= 2:
ans |= 1 << i
print(ans)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR VAR BIN_OP VAR VAR BIN_OP NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
num = int(input())
while num:
num -= 1
length = int(input())
a1 = input().split()
if length == 2:
print(int(a1[0]) & int(a1[1]))
continue
length = 0
answer = ""
for i in a1:
length = max(length, len(bin(int(i))))
for j in range(0, length):
gotans = False
is1present = False
for i in range(0, len(a1)):
temp = 1 << j
if temp & int(a1[i]) != 0:
if is1present:
gotans = True
break
else:
is1present = True
if gotans:
answer = "1" + answer
else:
answer = "0" + answer
print(int(answer, 2))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP NUMBER VAR IF BIN_OP VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR BIN_OP STRING VAR ASSIGN VAR BIN_OP STRING VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
while t:
n = int(input())
l = list(map(int, input().split()))
b = [0] * 30
ct = 0
for c in l:
bi = format(c, "030b")
for i in range(29, -1, -1):
if bi[i] == "1":
b[i] += 1
op = "0b"
for k in range(30):
if b[k] > 1:
op += "1"
else:
op += "0"
print(int(op, 2))
t -= 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR STRING FOR VAR FUNC_CALL VAR NUMBER NUMBER NUMBER IF VAR VAR STRING VAR VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER VAR STRING VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for i in range(int(input())):
n = int(input())
setbits = [0] * 32
num = list(map(int, input().split()))[:n]
for j in range(n):
for k in range(32):
if num[j] & 1 << k:
setbits[k] += 1
ans = 0
for j in range(32):
if setbits[j] > 1:
ans |= 1 << j
print(ans)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR VAR BIN_OP NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
def main():
for _ in range(int(input())):
n = int(input())
arr = list(map(lambda x: "{0:032b}".format(int(x)), input().split()))
ans = ["0" for _ in range(32)]
for i in range(32):
counter = 0
for element in arr:
if element[i] == "1":
if counter == 0:
counter += 1
elif counter == 1:
ans[i] = "1"
break
print(int("".join(ans), 2))
main()
|
FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR STRING VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR STRING IF VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL STRING VAR NUMBER EXPR FUNC_CALL VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for _ in range(int(input())):
n = int(input())
X = [int(x) for x in input().split()]
l = []
for i in X:
l.append(format(i, "032b"))
ans = ""
for i in range(32):
count = 0
for s in l:
if s[i] == "1":
count += 1
if count >= 2:
break
if count >= 2:
ans += "1"
else:
ans += "0"
print(int(ans, 2))
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR STRING ASSIGN VAR STRING FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR NUMBER IF VAR NUMBER VAR STRING VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
or_Array = [0] * len(a)
or_Array[n - 1] = a[n - 1]
for i in range(n - 2, 0, -1):
or_Array[i] = a[i] | or_Array[i + 1]
res = 0
for j in range(n - 2):
res = res | a[j] & or_Array[j + 1]
res = res | a[n - 1] & a[n - 2]
print(res)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for _ in range(int(input())):
n = int(input())
res = list(map(int, input().split()))
vect = [0] * 30
for i in res:
s = bin(i)[2:]
k = -1
for j in range(len(s) - 1, -1, -1):
if s[j] == "1":
vect[k] += 1
k -= 1
final_result = ["0"] * 30
for i in range(29, -1, -1):
if vect[i] > 1:
final_result[i] = "1"
print(int("".join(final_result), 2))
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR STRING VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP LIST STRING NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL STRING VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
mod = 1000000007
read_int = lambda: int(input().strip())
read_str = lambda: input().strip()
read_str_arr = lambda: input().strip().split()
read_int_arr = lambda: [int(x) for x in input().strip().split()]
def solve():
N = read_int()
A = read_int_arr()
ans = 0
or_acc = A[N - 1]
for i in range(N - 2, -1, -1):
ans |= A[i] & or_acc
or_acc |= A[i]
print(ans)
for _ in range(int(input())):
solve()
|
ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR BIN_OP VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
for i in range(t):
n = int(input())
l = list(map(int, list(input().split())))
arr = [0] * 35
for j in range(n):
count = 0
while l[j] > 0:
x = l[j] & 1
if x == 1:
arr[count] = arr[count] + 1
count = count + 1
l[j] = l[j] >> 1
ans = 0
for i in range(len(arr)):
if arr[i] > 1:
ans = ans + pow(2, i)
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for t in range(int(input())):
n = int(input())
a = [int(i) for i in input().split()]
byte_list = [0] * 32
ans = 0
for num in a:
i = 0
bit = num >> i
while bit > 0:
byte_list[i] = byte_list[i] + (bit & 1)
i += 1
bit = num >> i
for i, bit_count in enumerate(byte_list):
if bit_count > 1:
ans = ans | 1 << i
print(ans)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR WHILE VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR FOR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
def binary(l, num):
j = 0
while num != 0:
rem = num % 2
l[j] = l[j] + rem
j += 1
num //= 2
return l, num
n = int(input())
for _ in range(n):
l = [0] * 30
x = int(input())
y = list(map(int, input().split()))
for i in range(x):
binary(l, y[i])
for i in range(30):
if l[i] > 1:
l[i] = 1
else:
l[i] = 0
l.reverse()
l2 = map(str, l)
st = "".join(map(str, l2))
print(int(st, 2))
|
FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR VAR NUMBER VAR NUMBER RETURN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
T = int(input())
for t in range(T):
n = int(input())
a = list(map(int, input().split()))
lenA = len(a)
if lenA == 2:
print(a[0] & a[1])
continue
acc = 0
x = a[lenA - 1]
for i in range(lenA - 2, -1, -1):
acc = acc | a[i] & x
x = x | a[i]
print(acc)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR 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 IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int, input().split()))
x = arr[0]
left = [arr[0]]
for i in range(1, len(arr)):
x = x | arr[i]
left.append(x)
right = [arr[-1]]
x = arr[-1]
for i in range(len(arr) - 2, -1, -1):
x = x | arr[i]
right.append(x)
right = list(reversed(right))
ans = 0
for i in range(len(arr)):
if i == 0:
z = 0
else:
z = left[i - 1]
if i + 1 == len(arr):
z1 = 0
else:
z1 = right[i + 1]
z = z | z1
ans = ans | z & arr[i]
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR LIST VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
T = int(input())
for _ in range(T):
N = int(input())
A = list(map(int, input().split()))
B = []
maxlen = 0
for each in A:
binary = bin(each)[2:]
B.append(binary)
if len(binary) > maxlen:
maxlen = len(binary)
for i in range(len(B)):
B[i] = "0" * (maxlen - len(B[i])) + B[i]
d = dict()
for i in range(maxlen):
d[i] = 0
final = ""
for each in B:
for i in range(maxlen):
if each[i] == "1":
d[i] += 1
for i in d.values():
if i > 1:
final += "1"
else:
final += "0"
print(int(final, 2))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP STRING BIN_OP VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR STRING FOR VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR VAR NUMBER FOR VAR FUNC_CALL VAR IF VAR NUMBER VAR STRING VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
d = {}
for i in range(30, -1, -1):
d[i] = 0
for bit in range(30, -1, -1):
for j in arr:
if 1 << bit & j:
d[bit] += 1
ans = 0
for i in range(30, -1, -1):
if d[i] > 1:
ans += 1 << i
print(ans)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR DICT FOR VAR FUNC_CALL VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER NUMBER FOR VAR VAR IF BIN_OP BIN_OP NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
T = int(input())
for _ in range(T):
n = int(input())
A = list(map(int, input().split(" ")))
B = []
if n == 2:
print(A[1] & A[0])
continue
for i in range(n - 1, 0, -1):
if i == n - 1:
B.append(A[i])
else:
B.append(B[-1] | A[i])
ans = 0
for i in range(n - 1):
ans = ans | A[i] & B[-1]
B.pop()
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR 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 LIST IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
for _ in range(t):
n = int(input())
p = [int(i) for i in input().split()]
l = []
temp = 0
for i in range(n - 1):
temp = temp | p[n - i - 1]
l.append(temp)
temp = 0
for i in range(n - 1):
temp = temp | p[i] & l[n - i - 2]
print(temp)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
while t > 0:
n = int(input())
a = input().split()
for i in range(n):
a[i] = int(a[i])
b = [(0) for i in range(32)]
for i in range(n):
r = 0
while a[i] > 0:
if a[i] % 2:
b[r] += 1
a[i] = a[i] // 2
r += 1
ans = 0
m = 0
for i in range(32):
if b[i] > 1:
m = i
for i in range(m, -1, -1):
if b[i] > 1:
ans = 2 * ans + 1
else:
ans = 2 * ans
print(ans)
t -= 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
for i in range(t):
n = int(input())
A = list(map(int, input().split(" ")))
A = A[:n]
li = [
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
]
for k in A:
for j in range(31):
if k & 1 << j:
li[j] += 1
ans = 0
for j in range(31):
if li[j] > 1:
ans = ans | 1 << j
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR 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 VAR VAR ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER FOR VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR BIN_OP NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
a.sort()
res = 0
for i in range(32, -1, -1):
cnt = 0
for j in range(n):
if a[j] & 1 << i:
cnt += 1
if cnt > 1:
res += pow(2, i)
print(res)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR 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 NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR BIN_OP NUMBER VAR VAR NUMBER IF VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
ma = format(max(l), "b")
l1 = [0] * len(ma)
i = 0
for x in l:
a = format(x, "b")
z = 0
for y in a[::-1]:
if y == "1":
if l1[z] == 2:
pass
else:
l1[z] += 1
z += 1
s = ""
for a in l1[::-1]:
if a <= 1:
s += "0"
else:
s += "1"
print(int(s, 2))
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR 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 FUNC_CALL VAR VAR STRING ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR STRING ASSIGN VAR NUMBER FOR VAR VAR NUMBER IF VAR STRING IF VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER ASSIGN VAR STRING FOR VAR VAR NUMBER IF VAR NUMBER VAR STRING VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
def solve(ar):
t = 0
ans = 0
for i in range(32):
t = 0
for x in ar:
if x & 1 << i:
t += 1
if t > 1:
ans |= 1 << i
return ans
for i in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
print(solve(l))
|
FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR BIN_OP NUMBER VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP NUMBER VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR 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 FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
while t > 0:
n = int(input())
lst = list(map(int, input().split()))
or_lst = []
initial_or = 0
for i in range(n - 1, -1, -1):
or_lst.insert(0, initial_or)
initial_or |= lst[i]
or_component = 0
for i in range(n):
temp = or_lst[i] & lst[i]
or_component |= temp
print(or_component)
t -= 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
for tc in range(0, t):
n = int(input())
arr = list(map(int, input().split()))
out = [(0) for i in range(32)]
for j in range(len(arr)):
for i in range(32):
k = arr[j] >> i
if k & 1:
out[i] += 1
ans = 0
for i in range(32):
if out[i] > 1:
ans += pow(2, i)
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR IF BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
try:
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
matrix = []
for i in arr:
s = "0" * 30
temp = bin(i).replace("0b", "")
matrix.append(s[len(temp) :] + temp)
result = ["0"] * 30
for i in range(30):
cnt = 0
for j in range(n):
if matrix[j][i] == "1":
cnt += 1
if cnt > 1:
result[i] = "1"
print(int("".join(result), 2))
except:
pass
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR BIN_OP STRING NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR VAR STRING STRING EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP LIST STRING NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL STRING VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
def main():
n = int(input())
a = list(map(int, input().split()))
ai = [(0) for j in range(30)]
for j in a:
n = 0
while n < 31:
if j & 1 << 0 + n >= 1:
ai[n] += 1
n += 1
for k in range(30):
if ai[k] < 2:
ai[k] = 0
else:
ai[k] = 1
print(int("".join(map(str, ai[::-1])), 2))
for _ in range(int(input())):
main()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER IF BIN_OP VAR BIN_OP NUMBER BIN_OP NUMBER VAR NUMBER VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
while t > 0:
n = int(input())
a = list(map(int, input().split()))
ans = 0
c = a[0]
for i in range(1, n):
ans = ans | c & a[i]
c |= a[i]
print(ans)
t -= 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for i in range(int(input())):
n = int(input())
ar = [int(x) for x in input().split()]
l = []
m = 1
for i in range(0, 32):
count = 0
for j in ar:
if j >> i & m == 1:
count += 1
if count == 2:
l.append(2)
break
else:
l.append(1)
s = 0
for i in range(0, 32):
if l[i] == 2:
s = s + pow(2, i)
print(s)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP BIN_OP VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER IF VAR VAR NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
while t:
t = t - 1
n = int(input())
arr = list(map(int, input().split()))
res = []
for i in range(0, 32, 1):
res.append(0)
ans = 0
for i in range(0, 32, 1):
for j in range(n):
if arr[j] % 2 == 1:
res[i] = res[i] + 1
else:
res[i] = res[i]
arr[j] = arr[j] // 2
for i in range(0, len(res)):
if res[i] > 1:
ans = ans + 2**i
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for t in range(int(input())):
N = int(input())
A = list(map(int, input().split()))
suffixOR = [0]
for i in range(len(A) - 1, 0, -1):
suffixOR.append(suffixOR[-1] | A[i])
suffixOR.reverse()
ans = 0
for i in range(N - 1):
ans = ans | A[i] & suffixOR[i]
print(ans)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = int(input())
for _ in range(t):
n = int(input())
a = [int(x) for x in input().split()]
ans = 0
for i in range(32):
c = 0
for x in a:
if x & 1 << i:
c += 1
if c > 1:
ans = ans | 1 << i
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR BIN_OP NUMBER VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for i in range(int(input())):
N = int(input())
arr = [int(x) for x in input().split()]
brr = []
dict = set()
while len(arr) > 0:
count = 0
dict = sorted(dict)
dict = reversed(dict)
for i in dict:
arr.pop(i)
n = len(arr)
dict = set()
for i in range(len(arr)):
if arr[i] == 0:
dict.add(i)
if arr[i] & 1 == 1:
count += 1
arr[i] = arr[i] >> 1
if arr[i] == 0:
dict.add(i)
if count > 1:
brr.append(1)
else:
brr.append(0)
sum = 0
k = 0
for i in brr:
sum += i * 2**k
k += 1
print(sum)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR WHILE FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP VAR BIN_OP NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
t = input()
t = int(t)
while t:
n = input()
n = int(n)
a = [int(a) for a in input().split()]
setBitCount = [0] * 30
for num in a:
for i in range(0, 30):
if num & 1 << i:
setBitCount[i] += 1
ans = 0
for i in range(30):
if setBitCount[i] > 1:
ans |= 1 << i
print(ans)
t -= 1
|
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR VAR FOR VAR FUNC_CALL VAR NUMBER NUMBER IF BIN_OP VAR BIN_OP NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
try:
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
bits = []
ans = ["0"] * 32
for i in arr:
temp1 = bin(i).replace("0b", "")
temp2 = "0" * (32 - len(temp1))
bits.append(temp2 + temp1)
for i in range(32):
cnt = 0
for j in bits:
if j[i] == "1":
cnt += 1
if cnt == 2:
ans[i] = "1"
break
print(int("".join(ans), 2))
except:
pass
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR BIN_OP LIST STRING NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR VAR STRING STRING ASSIGN VAR BIN_OP STRING BIN_OP NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL STRING VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for _ in range(int(input())):
n = int(input())
q = [0] * 32
a = list(map(int, input().split()))
for i in a:
p = 0
while i:
q[p] += i & 1
i >>= 1
p += 1
for i in range(32):
if q[i] > 1:
q[i] = "1"
else:
q[i] = "0"
print(int("".join(q[::-1]), 2))
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL STRING VAR NUMBER NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
def to_find(n, ll):
b = [0] * 30
for i in range(n):
for j in range(30):
if ll[i] & 1 << j > 0:
b[j] += 1
res = 0
for i in range(30):
if b[i] >= 2:
res = res | 1 << i
return res
for i in range(int(input())):
n = int(input())
ll = list(map(int, input().split()))
res = to_find(n, ll)
print(res)
|
FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR VAR BIN_OP NUMBER VAR NUMBER VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR 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 VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for _ in range(int(input())):
n = int(input())
a = [int(i) for i in input().split()]
for i in range(n):
a[i] = (30 - len(bin(a[i])[2:])) * "0" + bin(a[i])[2:]
y = [0] * 30
for i in range(30):
for j in range(n):
if a[j][i] == "1":
y[i] += 1
s = ""
for i in range(30):
if y[i] > 1:
s += "1"
else:
s += "0"
print(int(s, 2))
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP BIN_OP NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER STRING FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR STRING VAR VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER VAR STRING VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
for _ in range(int(input())):
n = int(input())
num_bits = [0] * 30
res = 0
for x in input().split():
a = int(x)
for j in range(len(num_bits)):
if a & 1 << j:
num_bits[j] += 1
if num_bits[j] == 2:
res += 2**j
print(res)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP NUMBER VAR VAR VAR NUMBER IF VAR VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR
|
You are given an array A consisting of N integers.
From array A, we create a new array B containing all pairwise bitwise ANDs of elements from A. That is, B consists of N\cdot(N-1)/2 elements, each of the form A_{i} \mathbin{\&} A_{j} for some 1 ≤ i < j ≤ N.
In array B, we repeat the following process till there is only one element remaining:
Let the current maximum and minimum element of array B be X and Y respectively.
Remove X and Y from array B and insert X \mathbin{|} Y into it.
Determine the last remaining element in array B.
Here, \mathbin{\&} denotes the [Bitwise AND operation] and \mathbin{|} denotes the [Bitwise OR operation].
------ Input Format ------
- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N, denoting the number of elements in A.
- The second line of each test case contains N space-separated integers A_{1}, A_{2}, \ldots, A_{N}.
------ Output Format ------
For each test case, print a single line containing one integer — the last remaining element in array B.
------ Constraints ------
$1 ≤ T ≤ 10^{4}$
$2 ≤ N ≤ 10^{5}$
$0 ≤ A_{i} ≤ 10^{9}$
- The sum of $N$ over all test cases does not exceed $2\cdot 10^{5}$.
----- Sample Input 1 ------
3
2
1 3
3
2 7 1
4
4 6 7 2
----- Sample Output 1 ------
1
3
6
----- explanation 1 ------
Test Case $1$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}] = [1 \mathbin{\&} 3] = [1]$. There is only one element in $B$ so the answer is $1$.
Test Case $2$: Array $B$ will be $[A_{1} \mathbin{\&} A_{2}, A_{1} \mathbin{\&} A_{3}, A_{2} \mathbin{\&} A_{3}] = [2 \mathbin{\&} 7, 2 \mathbin{\&} 1, 7 \mathbin{\&} 1] = [2, 0, 1]$.
Then, we do the following operations on $B$:
1. Remove $2$ and $0$ from $B$ and insert $2\mathbin{|} 0 = 2$ into it. Now, $B = [1, 2]$.
2. Remove $2$ and $1$ from $B$ and insert $2\mathbin{|} 1 = 3$ into it. Now, $B = [3]$.
The last remaining element is thus $3$.
|
def Binary(n):
s = bin(n)
s1 = s[2:]
return s1
def binaryToDecimal(n):
return int(n, 2)
for i in range(int(input())):
N = int(input())
elements = list(map(int, input().split()))
newelements = [Binary(i) for i in elements]
a = max(elements)
a = Binary(a)
a = len(a)
for i in range(len(newelements)):
while len(newelements[i]) != a:
newelements[i] = "0" + newelements[i]
def yes(i):
count = 0
for j in range(len(newelements)):
if newelements[j][i] == "1":
count += 1
if count >= 2:
return True
else:
return False
ans = ""
for i in range(a):
if yes(i):
ans = ans + "1"
else:
ans = ans + "0"
print(binaryToDecimal(ans))
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER RETURN VAR FUNC_DEF RETURN FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR 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 VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR WHILE FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP STRING VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR STRING VAR NUMBER IF VAR NUMBER RETURN NUMBER RETURN NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR STRING ASSIGN VAR BIN_OP VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR
|
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.
For a given $N$, find the number of ways to choose an integer $x$ from the range $[0, 2^{N} - 1]$ such that $x \oplus (x+1) = (x+2) \oplus (x+3)$, where $\oplus$ denotes the bitwise XOR operator.
Since the number of valid $x$ can be large, output it modulo $10^{9}+7$.
------ Input ------
The first line contains an integer $T$, the number of test cases. Then the test cases follow.
The only line of each test case contains a single integer $N$.
------ Output ------
For each test case, output in a single line the answer to the problem modulo $10^{9} + 7$.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
------ Subtasks ------
Subtask #1 (100 points): Original Constraints
----- Sample Input 1 ------
2
1
2
----- Sample Output 1 ------
1
2
----- explanation 1 ------
Test Case $1$: The possible values of $x$ are $\{0\}$.
Test Case $2$: The possible values of $x$ are $\{0, 2\}$.
|
modNo = int(1000000000.0 + 7)
t = int(input())
while t > 0:
n = int(input())
print(pow(2, n - 1, modNo))
t = t - 1
|
ASSIGN VAR FUNC_CALL VAR BIN_OP NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER
|
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.
For a given $N$, find the number of ways to choose an integer $x$ from the range $[0, 2^{N} - 1]$ such that $x \oplus (x+1) = (x+2) \oplus (x+3)$, where $\oplus$ denotes the bitwise XOR operator.
Since the number of valid $x$ can be large, output it modulo $10^{9}+7$.
------ Input ------
The first line contains an integer $T$, the number of test cases. Then the test cases follow.
The only line of each test case contains a single integer $N$.
------ Output ------
For each test case, output in a single line the answer to the problem modulo $10^{9} + 7$.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
------ Subtasks ------
Subtask #1 (100 points): Original Constraints
----- Sample Input 1 ------
2
1
2
----- Sample Output 1 ------
1
2
----- explanation 1 ------
Test Case $1$: The possible values of $x$ are $\{0\}$.
Test Case $2$: The possible values of $x$ are $\{0, 2\}$.
|
t = int(input())
for i in range(t):
n = int(input())
c = 1
n = n - 1
b = 2
while n:
if n % 2 != 0:
c = c * b % 1000000007
b = b * b % 1000000007
n //= 2
print(c % 1000000007)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER
|
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.
For a given $N$, find the number of ways to choose an integer $x$ from the range $[0, 2^{N} - 1]$ such that $x \oplus (x+1) = (x+2) \oplus (x+3)$, where $\oplus$ denotes the bitwise XOR operator.
Since the number of valid $x$ can be large, output it modulo $10^{9}+7$.
------ Input ------
The first line contains an integer $T$, the number of test cases. Then the test cases follow.
The only line of each test case contains a single integer $N$.
------ Output ------
For each test case, output in a single line the answer to the problem modulo $10^{9} + 7$.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
------ Subtasks ------
Subtask #1 (100 points): Original Constraints
----- Sample Input 1 ------
2
1
2
----- Sample Output 1 ------
1
2
----- explanation 1 ------
Test Case $1$: The possible values of $x$ are $\{0\}$.
Test Case $2$: The possible values of $x$ are $\{0, 2\}$.
|
poww = [0]
res = 2
for i in range(1, 100001):
poww.append(res)
res *= 2
for _ in range(int(input())):
n = int(input())
print(poww[n] // 2 % 1000000007)
|
ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER
|
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.
For a given $N$, find the number of ways to choose an integer $x$ from the range $[0, 2^{N} - 1]$ such that $x \oplus (x+1) = (x+2) \oplus (x+3)$, where $\oplus$ denotes the bitwise XOR operator.
Since the number of valid $x$ can be large, output it modulo $10^{9}+7$.
------ Input ------
The first line contains an integer $T$, the number of test cases. Then the test cases follow.
The only line of each test case contains a single integer $N$.
------ Output ------
For each test case, output in a single line the answer to the problem modulo $10^{9} + 7$.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
------ Subtasks ------
Subtask #1 (100 points): Original Constraints
----- Sample Input 1 ------
2
1
2
----- Sample Output 1 ------
1
2
----- explanation 1 ------
Test Case $1$: The possible values of $x$ are $\{0\}$.
Test Case $2$: The possible values of $x$ are $\{0, 2\}$.
|
for i in range(int(input())):
n = int(input())
if n == 1:
print(1)
elif n == 2:
print(2)
else:
print(pow(2, n - 1, 10**9 + 7))
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER BIN_OP BIN_OP NUMBER NUMBER NUMBER
|
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.
For a given $N$, find the number of ways to choose an integer $x$ from the range $[0, 2^{N} - 1]$ such that $x \oplus (x+1) = (x+2) \oplus (x+3)$, where $\oplus$ denotes the bitwise XOR operator.
Since the number of valid $x$ can be large, output it modulo $10^{9}+7$.
------ Input ------
The first line contains an integer $T$, the number of test cases. Then the test cases follow.
The only line of each test case contains a single integer $N$.
------ Output ------
For each test case, output in a single line the answer to the problem modulo $10^{9} + 7$.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
------ Subtasks ------
Subtask #1 (100 points): Original Constraints
----- Sample Input 1 ------
2
1
2
----- Sample Output 1 ------
1
2
----- explanation 1 ------
Test Case $1$: The possible values of $x$ are $\{0\}$.
Test Case $2$: The possible values of $x$ are $\{0, 2\}$.
|
maxn = 10**5
mod = 10**9 + 7
dp = [1]
for i in range(1, maxn + 1):
dp.append(dp[i - 1] * 2 % mod)
for case in range(int(input())):
n = int(input())
print(dp[n - 1])
|
ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER
|
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.
For a given $N$, find the number of ways to choose an integer $x$ from the range $[0, 2^{N} - 1]$ such that $x \oplus (x+1) = (x+2) \oplus (x+3)$, where $\oplus$ denotes the bitwise XOR operator.
Since the number of valid $x$ can be large, output it modulo $10^{9}+7$.
------ Input ------
The first line contains an integer $T$, the number of test cases. Then the test cases follow.
The only line of each test case contains a single integer $N$.
------ Output ------
For each test case, output in a single line the answer to the problem modulo $10^{9} + 7$.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
------ Subtasks ------
Subtask #1 (100 points): Original Constraints
----- Sample Input 1 ------
2
1
2
----- Sample Output 1 ------
1
2
----- explanation 1 ------
Test Case $1$: The possible values of $x$ are $\{0\}$.
Test Case $2$: The possible values of $x$ are $\{0, 2\}$.
|
factDp = [(-1) for i in range(10**5)]
curr = 2
factDp[0] = 1
factDp[1] = 2
MOD = 10**9 + 7
for i in range(2, 10**5):
curr = curr * 2 % MOD
factDp[i] = curr
for _ in range(int(input())):
n = int(input())
res = factDp[n - 1]
print(res)
|
ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.
For a given $N$, find the number of ways to choose an integer $x$ from the range $[0, 2^{N} - 1]$ such that $x \oplus (x+1) = (x+2) \oplus (x+3)$, where $\oplus$ denotes the bitwise XOR operator.
Since the number of valid $x$ can be large, output it modulo $10^{9}+7$.
------ Input ------
The first line contains an integer $T$, the number of test cases. Then the test cases follow.
The only line of each test case contains a single integer $N$.
------ Output ------
For each test case, output in a single line the answer to the problem modulo $10^{9} + 7$.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
------ Subtasks ------
Subtask #1 (100 points): Original Constraints
----- Sample Input 1 ------
2
1
2
----- Sample Output 1 ------
1
2
----- explanation 1 ------
Test Case $1$: The possible values of $x$ are $\{0\}$.
Test Case $2$: The possible values of $x$ are $\{0, 2\}$.
|
from sys import stdin
l = []
lim = 10000000000.0 + 7
for _ in range(int(stdin.readline())):
l.append(pow(2, int(stdin.readline()) - 1, 1000000007))
for i in l:
print(i)
|
ASSIGN VAR LIST ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR FUNC_CALL VAR NUMBER NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR
|
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.
For a given $N$, find the number of ways to choose an integer $x$ from the range $[0, 2^{N} - 1]$ such that $x \oplus (x+1) = (x+2) \oplus (x+3)$, where $\oplus$ denotes the bitwise XOR operator.
Since the number of valid $x$ can be large, output it modulo $10^{9}+7$.
------ Input ------
The first line contains an integer $T$, the number of test cases. Then the test cases follow.
The only line of each test case contains a single integer $N$.
------ Output ------
For each test case, output in a single line the answer to the problem modulo $10^{9} + 7$.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
------ Subtasks ------
Subtask #1 (100 points): Original Constraints
----- Sample Input 1 ------
2
1
2
----- Sample Output 1 ------
1
2
----- explanation 1 ------
Test Case $1$: The possible values of $x$ are $\{0\}$.
Test Case $2$: The possible values of $x$ are $\{0, 2\}$.
|
t = int(input())
while t:
count1 = 0
n = int(input())
if n == 0:
print(1)
else:
print(pow(2, n - 1, 1000000007))
t = t - 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER
|
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.
For a given $N$, find the number of ways to choose an integer $x$ from the range $[0, 2^{N} - 1]$ such that $x \oplus (x+1) = (x+2) \oplus (x+3)$, where $\oplus$ denotes the bitwise XOR operator.
Since the number of valid $x$ can be large, output it modulo $10^{9}+7$.
------ Input ------
The first line contains an integer $T$, the number of test cases. Then the test cases follow.
The only line of each test case contains a single integer $N$.
------ Output ------
For each test case, output in a single line the answer to the problem modulo $10^{9} + 7$.
------ Constraints ------
$1 ≤ T ≤ 10^{5}$
$1 ≤ N ≤ 10^{5}$
------ Subtasks ------
Subtask #1 (100 points): Original Constraints
----- Sample Input 1 ------
2
1
2
----- Sample Output 1 ------
1
2
----- explanation 1 ------
Test Case $1$: The possible values of $x$ are $\{0\}$.
Test Case $2$: The possible values of $x$ are $\{0, 2\}$.
|
tum = int(input())
pum = int(1000000000.0 + 7)
for i in range(tum):
num = int(input())
if num == 1:
print(1)
else:
print((2 << num - 2) % pum)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER VAR
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
def mi():
return map(int, input().split())
def li():
return list(mi())
def si():
return str(input())
def ni():
return int(input())
mod = 998244353
for _ in range(ni()):
n, m = mi()
res = 0
for bit in range(31):
b2 = 2 << bit
count = m // b2 * (b2 >> 1) + max(0, m % b2 - (b2 >> 1) + 1)
res += pow(count, n, mod) << bit
print(res % mod)
|
FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR NUMBER BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
mod = 998244353
def power(x, y, p):
res = 1
x = x % p
if x == 0:
return 0
while y > 0:
if y & 1 == 1:
res = res * x % p
y //= 2
x = x * x % p
return res
def power_ka_function(a, b):
return a // 2 ** (b + 1) * 2**b
def rem_ka_function(a, b):
res = a % 2 ** (b + 1) - 2**b + 1
if res < 0:
return 0
return res
def solve():
size, limit = map(int, input().split())
ans = 0
for i in range(32):
var = power_ka_function(limit, i) + rem_ka_function(limit, i)
var = power(var, size, mod)
var %= mod
var *= 2**i
ans += var
ans %= mod
print(ans % mod)
t = int(input())
for _ in range(t):
solve()
|
ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER RETURN NUMBER WHILE VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR RETURN VAR FUNC_DEF RETURN BIN_OP BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP NUMBER VAR FUNC_DEF ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP NUMBER VAR NUMBER IF VAR NUMBER RETURN NUMBER RETURN VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR BIN_OP NUMBER VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
def pow(bg, num):
count = 1
while num:
if num % 2 == 1:
count = count * bg % 998244353
bg = bg * bg % 998244353
num //= 2
return count
def coun(arg1, arg2):
return (arg1 // (1 << arg2 + 1) << arg2) + max(
0, arg1 % (1 << arg2 + 1) - (1 << arg2) + 1
)
for trie in range(int(input())):
n, m = map(int, input().split())
out_put = 0
for i in range(0, 31):
out_put = (out_put + (pow(coun(m, i), n) << i) % 998244353) % 998244353
print(out_put)
|
FUNC_DEF ASSIGN VAR NUMBER WHILE VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER RETURN VAR FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
def fast_mod_exp(n, exp, mod):
if exp == 0:
return 1 % mod
b = binary(exp, 0)
res = 1
current = n
for i in range(len(b)):
if b[len(b) - i - 1] == 1:
res *= current
res %= mod
current *= current
current %= mod
return res
def binary(n, min_size):
if n == 0:
return [0]
res = []
while n > 0 or len(res) < min_size:
res.append(n % 2)
n //= 2
return list(reversed(res))
def get_count(p, limit, mod):
bl = binary(limit, 32)
res = 0
e = 1
for i, el in enumerate(bl):
current_p = len(bl) - i - 1
if p == current_p:
if el == 0:
break
else:
e = 0
continue
if len(bl) - i - (1 + e) < 0:
break
if el == 0:
continue
res += 2 ** (len(bl) - i - (1 + e))
res %= mod
return res
def f(n, limit, mod):
res = 0
p = 0
while 2**p <= limit:
count = get_count(p, limit, mod)
count = fast_mod_exp(count, n, mod)
res += count * 2**p
res %= mod
p += 1
return res
t = int(input())
for i in range(t):
n, m = [int(el) for el in input().split(" ")]
print(f(n, m + 1, 998244353))
|
FUNC_DEF IF VAR NUMBER RETURN BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR VAR VAR VAR VAR VAR VAR VAR RETURN VAR FUNC_DEF IF VAR NUMBER RETURN LIST NUMBER ASSIGN VAR LIST WHILE VAR NUMBER FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP BIN_OP FUNC_CALL VAR VAR VAR BIN_OP NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR VAR BIN_OP NUMBER VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR BIN_OP NUMBER VAR VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
P = 998244353
def max_set_bit(n: int):
k = 0
while n:
n >>= 1
k += 1
return k - 1
def numbers_with_kth_bit_set(M: int, k: int):
M += 1
b = 1 << k
n = M // b
return n // 2 * b + M % b * (n % 2)
def mod_pow(b, e, p):
ans = 1
while e > 0:
if e % 2:
ans = ans * b % p
e //= 2
b = b * b
return ans
def mod_pow(b, e, p):
if e == 0:
return 1
ans = mod_pow(b, e // 2, p)
ans = ans * ans % p
if e % 2:
ans = ans * b % p
return ans
t = int(input())
for _ in range(t):
N, M = list(map(int, input().split()))
answer = 0
base = 1
for k in range(31):
m = numbers_with_kth_bit_set(M, k)
counts = mod_pow(m, N, P)
answer = (answer + base * counts) % P
base = base * 2 % P
k += 1
if m == 0:
break
print(answer)
|
ASSIGN VAR NUMBER FUNC_DEF VAR ASSIGN VAR NUMBER WHILE VAR VAR NUMBER VAR NUMBER RETURN BIN_OP VAR NUMBER FUNC_DEF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR VAR RETURN BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER IF BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR RETURN VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR IF BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
def get_possibilities(M):
result = [0] * M.bit_length()
bits = [b for b in f"{M:b}"]
for i, b in enumerate(bits):
if b == "1":
result[i] += 1
if i == M.bit_length():
break
factor = 1 << M.bit_length() - i - 1
resid = M % factor
result[i] += resid
for j in range(i + 1, len(result)):
result[j] += factor // 2
return result[::-1]
T = int(input())
for _ in range(T):
N, M = map(int, input().split(" "))
possibilities = get_possibilities(M)
result = 0
modulo = 998244353
for i, p in enumerate(possibilities):
result += pow(2, i, modulo) * pow(p, N, modulo)
result %= modulo
print(result)
|
FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR ASSIGN VAR VAR VAR VAR STRING FOR VAR VAR FUNC_CALL VAR VAR IF VAR STRING VAR VAR NUMBER IF VAR FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR NUMBER VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
for i in range(int(input())):
n, m = map(int, input().split())
m1 = m
c = 1
xx = 1
while 1:
if m1 < xx:
c -= 1
break
elif m1 == xx:
break
c += 1
xx *= 2
ans = 0
p = 1
k = {}
h = 2
for y in range(0, c):
if (m + 1) % h == 0:
l = (m + 1) // h
l = l * (h // 2)
k[y] = l
else:
l = (m + 1) // h
l = l * (h // 2)
r = (m + 1) % h
if r <= h // 2:
r = 0
else:
r = r - h // 2
k[y] = l + r
h *= 2
f = {}
n1 = n
z = 998244353
for q in range(c):
n1 = n
t = 1
kk = k[q]
while n1:
if n1 & 1 == 1:
t = t * kk % z
n1 = n1 >> 1
kk = kk * kk % z
ans = ans + t * p
p *= 2
print(ans % z)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE NUMBER IF VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR DICT ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR WHILE VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
def coun(i, j):
return (i // (1 << j + 1) << j) + max(0, i % (1 << j + 1) - (1 << j) + 1)
def powe(b, n):
s = 1
while n:
if n % 2 == 1:
s = s * b % 998244353
b = b * b % 998244353
n = n // 2
return s
t = int(input())
for h in range(t):
n, m = map(int, input().split())
s = 0
for j in range(31):
s = (s + (powe(coun(m, j), n) << j) % 998244353) % 998244353
print(s)
|
FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP NUMBER VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER WHILE VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER 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 FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
MOD = 998244353
def K(m, i):
nai = 2**i
naii = nai * 2
return m // naii * nai + max(0, m % naii - (nai - 1))
for _ in range(int(input())):
n, m = map(int, input().split())
result = 0
for i in range(32):
result += pow(K(m, i), n, MOD) * 2**i
result %= MOD
print(result)
|
ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN BIN_OP BIN_OP BIN_OP VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
for _1 in range(int(input())):
n, m = map(int, input().split())
def ans(a, b):
return (a // (1 << b + 1) << b) + max(0, a % (1 << b + 1) - (1 << b) + 1)
def list1(c, d):
cou = 1
while d:
if d % 2 == 1:
cou = cou * c % 998244353
c = c * c % 998244353
d = d // 2
return cou
v = 0
for l in range(31):
v = (v + (list1(ans(m, l), n) << l) % 998244353) % 998244353
print(v)
|
FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP NUMBER VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER WHILE VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
mod = 998244353
for _ in range(int(input())):
n, m = map(int, input().split())
ans = 0
for bit in range(30):
ct = m // (2 << bit) * (1 << bit)
ct += max(0, m % (2 << bit) - (1 << bit) + 1)
ans += pow(ct, n, mod) << bit
print(ans % mod)
|
ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP NUMBER VAR BIN_OP NUMBER VAR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER VAR BIN_OP NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
def count(i, j):
return (i // (1 << j + 1) << j) + max(0, i % (1 << j + 1) - (1 << j) + 1)
for _ in range(int(input())):
n, m = list(map(int, input().split()))
s = 0
for i in range(0, 32):
s += pow(count(m, i), n, 998244353) << i
print(s % 998244353)
|
FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER
|
Given two positive integers N and M, let S denote the set of all the arrays of size N such that each element of the array lies in the range [1, M]. Since there are M^{N} such arrays, the size of S is M^{N}.
Let X_{i} denote the [bitwise AND] of all elements of the i^{th} array in the set, where 1 ≤ i ≤ M^{N}.
Find the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains two integers N and M, the size of the array, and the maximum limit of elements.
------ Output Format ------
For each test case, print the value \sum_{i = 1}^{M^{N}} X_{i}. Since the answer can be huge, output the answer modulo 998244353.
------ Constraints ------
$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 2\cdot10^{5}$
$1 ≤ M ≤ 10^{9}$
----- Sample Input 1 ------
2
2 2
2 3
----- Sample Output 1 ------
3
12
----- explanation 1 ------
Test case $1$: The set $S$ contains $\{[1,1], [1,2], [2,1], [2,2]\}$. The array $X = [1\& 1, 1\& 2, 2\& 1, 2\& 2] = [1, 0, 0, 2]$. Thus, sum of all elements of $X$ is $1+0+0+2 = 3$.
Test case $2$: The set $S$ contains $\{[1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3]\}$. The array $X = [1\& 1, 1\& 2, 1\& 3, 2\& 1, 2\& 2, 2\& 3, 3\& 1, 3\& 2, 3\& 3] = [1, 0, 1, 0, 2, 2, 1, 2, 3]$. Thus, sum of all elements of $X$ is $12$.
|
t = int(input())
mod = 998244353
for _ in range(t):
n, m = map(int, input().split())
ans = 0
for bit in range(31):
k = m // (2 << bit) * (1 << bit) + max(0, m % (2 << bit) - (1 << bit) + 1)
ct = pow(k, n, mod)
ans = (ans + ct * (1 << bit)) % mod
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER VAR BIN_OP NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER VAR BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR VAR
|
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