Solution

The precipitation on day \(i\) (\(1\)-indexed) is the \(i\)-th term of an arithmetic sequence with first term \(A\) and common difference \(B\):

\[A + (i-1)B\]

So it suffices to output this value for \(i = 1, 2, \ldots, N\) in order.

In the implementation, we initialize a variable x to \(A\) and add \(B\) to it after printing each day. This generates the same sequence without evaluating the formula each time.

The maximum value is \(A + (N-1)B \le 10^9 + 99 \cdot 10^9 = 10^{11}\), which does not fit in a 32-bit integer, so in C++ we use a 64-bit integer type (long long). Python’s int has arbitrary precision, so it works as is.

Proof of correctness

Loop invariant: at the start of the \(i\)-th iteration, x \(= A + (i-1)B\) holds (for \(i = 1, \ldots, N\)).

• Initialization: x \(= A\), and for \(i = 1\) we have \(A + 0 \cdot B = A\), so it holds.
• Maintenance: the \(i\)-th iteration prints x \(= A + (i-1)B\) and then adds \(B\), so at the start of the next iteration x \(= A + iB\).
• Termination: the loop runs exactly \(N\) times.

Hence the value printed on line \(i\) is \(A + (i-1)B\), which matches the problem’s definition of the day-\(i\) precipitation (\(A\) on day 1, increasing by \(B\) each day). Therefore the program is correct.

Complexity

• Time: \(O(N)\) (one output per day)
• Space: \(O(1)\) auxiliary (no input array is stored; only the running variable x is updated)

There is no tie-breaking involved (the output is uniquely determined).