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()

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 1at 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)

To solve this problem, we can use the fact that any number in base n can be written as a sum of powers of n. So, we can convert k to binary and use the bits of the binary representation of k to determine which powers of n to use. For example, if k = 13 and n = 3, we can convert k to binary to get 1101. Then, we can use the 1's in the binary representation to determine which powers of 3 to use. In this case, we have 1101, which means we can use the powers of 3 corresponding to 3, 2, and 0 (since these are the positions of the 1's in the binary representation). So, the k-th special number for n = 3 is 3^3 + 3^2 + 3^0 = 37.

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)

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()

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 corresponds to a power of two, which becomes a summand in the answer. If the number of summands required is greater than k , then it is impossible to obtain the sum using only powers of two. If we have fewer summands than k, we can repeatedly choose a summand from the set and split it into two summands of equal value until we have exactly k summands. This can be done by maintaining a data structure (e.g., stack, array, queue, multiset) and choosing an arbitrary summand to split. This process terminates when we have exactly k summands, since we will always be able to choose a summand to split until all summands are equal to 1.

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()

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()