Codeforces 1780 D: Bit Guessing Game

In this problem, we are given the number of unit bits in a hidden positive integer, and we must guess the integer within 30 moves. Each move, we can subtract an integer x from the hidden integer, and the number of unit bits in the new integer is given to us as a hint. The constraint is that 1≤n≤10^9.

One approach to solve this problem is to start by guessing the maximum value of n, which is 2^{cnt - 1}. We can do this by subtracting 2^{cnt - 1} from the hidden integer and check if the number of unit bits decreases by cnt. If it does, then our guess is correct, and we can print the result and move on to the next test case. If not, we can check if the number of unit bits decreases by cnt-1, cnt-2, and so on, until we find the correct value of n or we run out of moves.

Here is a sample implementation in Python:

t = int(input())
for _ in range(t):
    cnt = int(input())
    n = 0
    for i in range(30):
        n += 2**i
        if (bin(n).count("1") == cnt):
            print("!", n)
            break
        print("-", 2**i)
        cnt = int(input())

Explanation:

• First, we take the input for the number of test cases, t, and start a for loop that runs t times.
• For each test case, we take the input for the initial number of unit bits in the binary notation of the hidden integer, cnt.
• We initialize a variable n to keep track of the number that we are guessing.
• We start another for loop that runs 30 times. This is because we are allowed to perform at most 30 operations to guess the number.
• Inside the for loop, we add 2 to the power of the loop variable i to n. This is because we are only allowed to subtract integers from n in this problem, and adding 2 to the power of i is equivalent to subtracting 2 to the power of i from n.
• We check if the number of ones in the binary representation of n is equal to cnt. If it is, we print ! and the number n, and break out of the loop.
• If the number of ones in the binary representation of n is not equal to cnt, we print - and the number 2 to the power of i, as that is the number we are subtracting from n.
• We take the input for the updated number of unit bits in the binary notation of the updated value of n, and update the value of cnt for the next iteration of the for loop.

This approach works for the given constraints of the problem, since the maximum value of n is 2^{30}, which is less than 2^{31}, so we can represent it in an int data type.

Also, this approach is optimal because we are using the information given to us (number of unit bits) to narrow down the possibilities, and we are always subtracting the maximum possible value of n, which decreases the number of unit bits by cnt, cnt-1, cnt-2, and so on, which guarantees that we will find the correct value of n within 30 moves.