CF 102697088 - Radioactivity

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Solve time: 37s
Verified: yes

Solution

Problem Understanding

An isotope contains two kinds of particles relevant to this problem: protons and neutrons. Its mass number counts both of them, so

[ m = p + n, ]

where (p) is the number of protons, (n) is the number of neutrons, and (m) is the mass number.

The input gives (p) and (m), with (p \le m). The required output is the number of neutrons. Rearranging the definition immediately gives

[ n = m - p. ]

The published statement specifies a one-second time limit and 256 MB of memory, but it does not give a restrictive upper bound for the two integers beyond (p \le m). The computation itself is constant time and constant extra space, so the size of the integers is the only practical consideration. Python's integers support arbitrary precision, so there is no fixed-width overflow issue in the implementation.

The main edge case is equality. For example, with input 5 5, the mass number consists of five protons and no neutrons, so the correct output is 0. A careless solution that assumes the answer must be positive because the statement describes the result as a positive integer would fail on this boundary if equality is allowed by the stated condition.

Another boundary case is when there is exactly one neutron. With input 92 93, the correct output is 1. The subtraction must be performed in the correct order, because p - m would produce -1 even though the number of neutrons is positive.

Approaches

The brute-force interpretation would be to somehow search for a possible number of neutrons and check whether the resulting total matches the mass number. For example, we could try (n=0,1,2,\ldots) until (p+n=m). This is correct because the definition of mass number tells us exactly when a candidate neutron count is valid. However, in the worst case it can perform (m-p+1) iterations, which is unnecessary work and becomes arbitrarily large because the statement does not impose a small numerical bound.

The structure of the problem makes that search unnecessary. The equation defining the mass number already contains the desired value directly. Since (m=p+n), subtracting the known number of protons from the known total leaves exactly the number of neutrons. The brute-force works because it searches for a value satisfying the equation, but fails because it ignores that the equation can be solved algebraically in one operation. The observation that the quantities form a simple sum reduces the entire problem to one subtraction.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	(O(m-p))	(O(1))	Unnecessarily slow
Optimal	(O(1))	(O(1))	Accepted

Algorithm Walkthrough

Read the number of protons (p) and the mass number (m). These are the two quantities needed to reconstruct the isotope's neutron count.
Compute (m-p). The mass number contains every proton and every neutron, so removing all (p) protons leaves precisely the neutrons.
Print the result. No additional processing, search, or data structure is required.

The key invariant is the definition (m=p+n). At the moment we calculate (m-p), we are algebraically removing the known proton contribution from the total mass number. Thus the remaining value is exactly (n), so the printed answer cannot represent anything other than the required neutron count.

Python Solution

import sys
input = sys.stdin.readline

p, m = map(int, input().split())
print(m - p)

The first line reads the two integers from the only input line. map(int, ...) converts them to Python integers, which also avoids any concern about fixed-width integer overflow.

The subtraction follows directly from the algorithm's second step. It is deliberately written as m - p, not p - m, because the mass number is the total and the proton count is the part being removed.

There is no loop and no auxiliary data structure. The program performs one input operation, one arithmetic operation, and one output operation.

Worked Examples

For Sample 1, the isotope has 92 protons and a mass number of 238.

(p)	(m)	(m-p)	Output
92	238	146	146

The equation is (238=92+n), so (n=146). The subtraction directly recovers the neutron count.

For Sample 2, the isotope has 36 protons and a mass number of 85.

(p)	(m)	(m-p)	Output
36	85	49	49

Again, (85=36+n), giving (n=49). The same invariant applies without any special handling.

Complexity Analysis

Measure	Complexity	Explanation
Time	(O(1))	One subtraction is performed after reading the two integers.
Space	(O(1))	Only the two input integers and the result are stored.

The solution is far below the one-second and 256 MB limits. Since the algorithm does not depend on the numerical magnitude through iteration, it remains constant time even when the input integers are large. Python's arbitrary-precision integers also remove the overflow concern that would arise in languages using fixed-width integer types.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve():
    input = sys.stdin.readline
    p, m = map(int, input().split())
    print(m - p)

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout
    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    solve()
    out = sys.stdout.getvalue()

    sys.stdin = old_stdin
    sys.stdout = old_stdout
    return out

# provided samples
assert run("92 238\n") == "146\n", "sample 1"
assert run("36 85\n") == "49\n", "sample 2"

# minimum-size style case
assert run("1 1\n") == "0\n", "zero neutrons"

# all protons except one particle is a neutron
assert run("92 93\n") == "1\n", "exactly one neutron"

# large values, checking arithmetic without iteration
assert run("1000000000000000000 1000000000000000000\n") == "0\n", "large equal values"

# large gap between proton count and mass number
assert run("123456789 1000000000\n") == "876543211\n", "large neutron count"

Test input	Expected output	What it validates
`1 1`	`0`	Minimum-size boundary and equality case
`92 93`	`1`	Smallest positive neutron count
`1000000000000000000 1000000000000000000`	`0`	Large integers and equality
`123456789 1000000000`	`876543211`	Large subtraction and correct operand order

The statement does not specify a finite maximum value for (p) and (m), so there is no literal maximum-size test that can be derived from the published constraints. The (10^{18})-scale test is a representative stress case that exercises the same implementation property that would matter for any larger permitted integer.

Edge Cases

When the number of protons equals the mass number, there are no neutrons. For the exact input

5 5

the algorithm reads (p=5) and (m=5), computes (5-5=0), and prints

This catches an incorrect assumption that the answer must be strictly positive. The arithmetic definition permits zero neutrons whenever (p=m).

When there is exactly one neutron, consider

92 93

The algorithm computes (93-92=1) and prints

This catches the common operand-order mistake. The mass number is the total quantity, so it must be the minuend.

For a large input such as

1000000000000000000 1000000000000000000

the same calculation gives zero. Python represents these integers exactly, and the program does not loop based on their magnitude. The result is therefore

For a large neutron count,

123456789 1000000000

the calculation is

[ 1{,}000{,}000{,}000-123{,}456{,}789=876{,}543{,}211, ]

so the output is

876543211

This confirms that the solution handles the entire problem through the defining equation rather than through enumeration.