A) XOR MIXUP
Any element of the array works. Why?
Given an array a, b, c, d, we need to find a number in the array such that the xor of the rest of the elements equals this number. The array's overall xor is zero. So, any element from the array can be selected as the answer since the xor of a number with itself is zero. Therefore, a, b, c, or d can be chosen as the answer.
def solve():
length = int(input())
nums = list(map(int, input().split()))
return nums[0]
def main():
num_tests = int(input())
for _ in range(num_tests):
print(solve())
main()
B) Raising Bacteria
To determine the minimum number of bacteria needed to achieve exactly x bacteria in the box at some point, we can represent x in binary form. We will then iterate through each bit in the binary representation of x, starting from the least significant bit to the most significant bit.
For each bit that is equal to 1, we will add one bacterium into the box on the corresponding day. Specifically, if the bit is at position i, we will add the bacterium on the morning of the (n+1-i)th day, where n is the total number of bits in the binary representation of x.
After iterating through all the bits, the box will contain exactly x bacteria at the noon of the nth day. Therefore, the minimum number of bacteria needed is equal to the number of 1s in the binary representation of x.
x = int(input())
ans = 0
for i in range(32):
if x & (1 << i):
ans += 1
print(ans)
C) Special Numbers
The problem of finding the minimum number of bacteria needed to achieve exactly x bacteria in the box at some point is equivalent to finding the i-th number in base that consists only of 0s and 1s.
To solve this problem, we can represent x in binary form, and then rewrite each power of 2 as a power of 1.414 (which is approximately the square root of 2). This allows us to represent any positive integer using only 0s and 1s in base 1.414.
Therefore, the -th number in base 1.414 that consists only of 0s and 1s will correspond to the minimum number of bacteria needed to achieve exactly x bacteria in the box at some point, where is the number of 1s in the binary representation of x.
MOD = 1000000007
noOfTests = int(input())
for _ in range(noOfTests):
n, k = map(int, input().split())
ans = 0
mult = 1
for i in range(32):
if k & (1 << i):
ans = (ans + mult) % MOD
mult *= n
mult %= MOD
print(ans)
D) Rock and Lever
In order for the given condition "a & b >= a ^ b" to hold true, it is necessary for the most significant bits of both a and b to be set. This is because if we have both the i-th bits of a and b as 1, then the result of their "AND" logical operation would also be 1 and the result of their "XOR" logical operation would be 0. Additionally, since the most significant bit holds the highest value in the binary representation, if the most significant bit of a & b is greater than the most significant bit of ai^bi, then a & b would be greater than a ^ b.
In order to solve this problem, we need to iterate through the array and for each number, check if there is another element in the array that has the same most significant bit. If we are able to find such a pair, then we can conclude that the condition of "a & b >= a ^b" holds true for those two numbers.
from collections import defaultdict
def most_significant_bit(num):
msb = 0
while num != 0:
num >>= 1
msb += 1
return msb
def solve():
msb_dict = defaultdict(int)
count = 0
length = int(input())
nums = list(map(int, input().split()))
for num in nums:
msb = most_significant_bit(num)
count += msb_dict[msb]
msb_dict[msb] += 1
return count
def main():
num_tests = int(input())
for _ in range(num_tests):
print(solve())
main()
E) Powers of Two
To calculate the minimum number of powers of two needed to obtain a sum of , we can use the binary representation of n. Each bit in the binary representation of corresponds to a power of two, which becomes a summand in the answer.
If the number of summands required is greater than , then it is impossible to obtain a sum of using only powers of two.
If we have fewer summands than, we can repeatedly choose a summand from the set of summands greater than and split it into two summands of equal value until we have exactly summands. This can be done by maintaining a data structure (e.g., stack, array, queue, multiset) to store the summands greater than and choosing an arbitrary summand to split. This process terminates when we have exactly summands, since we will always be able to choose a summand to split until all summands are equal to n.
def solve():
n, k = map(int, input().split())
if k > n:
print("NO")
return
ans = []
for i in range(32):
if n & (1 << i):
ans.append(1 << i)
if len(ans) > k:
print("NO")
return
print("YES")
i = 0
while len(ans) < k:
while ans[i] == 1:
i += 1
ans[i] //= 2
ans.append(ans[i])
print(*sorted(ans))
solve()
F) Little Gril and Maximum XOR
consider the binary representation of start and end. Let p = be the position of the leftmost set bit in start, and Let q be the position of the leftmost set bit in end.
Then, all pairs of integers a and b between start and end, inclusive, have the same leftmost p bits and the same leftmost q bits. The remaining bits in a and b can take any value between 0 and 2^k — 1, inclusive, where k is the position of the leftmost set bit in the XOR of start and end. Therefore, the maximum value of a xor b for all pairs of integers a and b between start and end, inclusive, is 2^k — 1.
from collections import defaultdict
def most_significant_bit(num):
msb = 0
while num != 0:
num >>= 1
msb += 1
return msb
def main():
start, end = map(int, input().split())
xor = start ^ end
msb = most_significant_bit(xor)
max_xor = (1 << msb) - 1
print(max_xor)
main()