Python中int((n+1)/2)与(n+1)//2在大数下为何结果不同？

Great question! The key difference here boils down to floating-point precision limits in Python, which becomes a critical issue when working with extremely large integers like those up to 10^18. Let's break this down:

The Problem with int((n+1)/2) #

• In Python 3, the / operator performs floating-point division, which returns a float (a double-precision 64-bit floating-point number).
• Double-precision floats only have 53 bits of mantissa (the part that stores the actual digits of the number). This means they can only exactly represent integers up to 2^53 (~9 × 10^15). Any integer larger than this will lose precision when converted to a float—some bits get truncated, turning the exact integer into an approximate value.
• When you handle a number like n = 10^18 (way larger than 2^53), n+1 gets converted to a float and loses precision. Dividing by 2 and wrapping the result with int() then gives you a rounded, incorrect value instead of the exact integer quotient.

Why (n+1)//2 Works Perfectly #

• The // operator is Python's integer division operator, which operates entirely within Python's arbitrary-precision integer system.
• Python natively supports integers of any size (the only limit is your system's memory). This means // will always compute the exact quotient of (n+1) divided by 2, no matter how large n is—no precision loss, no rounding errors.

Concrete Example #

Let's use a number just above the float precision limit to see the difference:

• Let n = 2**53 + 1 (an odd number, so your code would process it)
• n+1 = 2**53 + 2, but this value can't be represented exactly as a float—it gets rounded to 2**53
• int((n+1)/2) becomes int(2**53 / 2) = 2**52
• (n+1)//2 becomes (2**53 + 2) // 2 = 2**52 + 1

As you can see, only the // version gives the correct exact result. This is why your first submission failed for large test cases, while the second one was accepted.

内容的提问来源于stack exchange，提问作者gaurav