CF 102697102 - Pocket Calculator

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Solve time: 4m 46s
Verified: yes

Solution

Problem Understanding

The input is one line containing a sequence such as 3 + 4 * 3 - 6 / 3 * 3. Every number occupies one token and consists of exactly one digit, with exactly one space between consecutive tokens. The first and last tokens are numbers.

We need to print the integer represented by the expression after applying the required precedence rules. The statement guarantees that the final answer fits in a signed 32-bit integer and is itself an integer.

There is no explicit upper bound on the number of tokens in the statement. That makes an algorithm whose running time grows quadratically with the expression length unnecessarily risky. With a one-second limit, we should aim for a single pass through the expression, which is linear in its length. The fact that every number has one digit makes tokenization particularly simple, but it does not change the need to handle an arbitrarily long sequence efficiently.

There are two edge cases that are easy to mishandle.

Consider:

3 - 6 * 3

The correct output is:

-15

A left-to-right calculator that simply performs every operation as it encounters it would calculate (3 - 6) * 3 = -9, which violates multiplication precedence.

A subtler case is:

5 / 2 * 2

The correct result is:

Division and multiplication have equal precedence, so they must be evaluated from left to right: (5 / 2) * 2. The intermediate value is 5/2, even though the final answer is an integer. An implementation using integer division would turn 5 / 2 into 2 and incorrectly produce 4. Keeping the current multiplicative value as an exact fraction avoids this problem.

Another useful boundary case is an expression containing only one number:

Its answer is simply 7. Code that assumes there is always an operator can accidentally access a nonexistent token or initialize the result incorrectly.

Approaches

A direct approach is to repeatedly locate the next multiplication or division, perform it, replace the three involved tokens with their result, and then do the same for addition and subtraction. This is correct because it explicitly follows the precedence rules. If an expression contains k operators, however, each search through the remaining expression can inspect O(k) tokens, and there can be O(k) such searches. In the worst case this performs roughly k + (k - 1) + ... + 1 = k(k + 1)/2 token inspections, which is O(k²).

The brute-force method works because the precedence structure is simple, but it repeatedly reprocesses parts of the expression that have already been understood. The key observation is that when we encounter + or -, every preceding * and / operation has already been resolved. We can keep the unresolved multiplication/division chain in one variable and commit that entire chain to the final sum whenever an addition or subtraction is reached.

There is one additional detail because division can create a fractional intermediate result. We should represent the current multiplication/division chain as an exact rational number rather than using integer division or floating point. Python's Fraction provides exactly that representation. Since the final result is guaranteed to be an integer, we can print its numerator after confirming the denominator is one.

The resulting algorithm scans the expression once, performs each operation immediately, and never needs to rescan or rearrange tokens.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(k²)	O(k)	Too slow for long expressions
Optimal	O(k)	O(k)	Accepted

Here k is the number of operators, equivalently Θ(k) tokens in the expression.

Algorithm Walkthrough

Split the input line into tokens. Numbers appear at even positions and operators at odd positions, because the input contains exactly one space between every token.
Initialize total = 0 and term to the first number. The variable term represents the complete multiplication/division chain that has been read but has not yet been added to the final answer.
Process each remaining operator together with the number immediately after it. If the operator is *, multiply term by that number. If it is /, divide term by that number using an exact rational representation.
When the operator is + or -, add the current term to total, with the appropriate sign, and start a new term from the next number. We can commit term at this point because every multiplication or division belonging to it has already been processed.
After the last token has been processed, add the final term to total. There is no following + or - operator to trigger this operation, so it must be done explicitly.
Print the resulting integer. The problem guarantees that the final mathematical result is an integer, so the exact fraction has denominator one.

The invariant is that before processing an addition or subtraction, term is exactly the value of the current maximal multiplication/division segment, while total is exactly the value of all completed addition/subtraction segments. Since multiplication and division are resolved immediately and addition/subtraction only combines completed terms, the algorithm maintains the required precedence at every point.

Python Solution

Pythonimport sysinput = sys.stdin.readline
from fractions import Fraction

def solve():    tokens = input().split()
    total = Fraction(0)    term = Fraction(int(tokens[0]))
    i = 1    while i < len(tokens):        op = tokens[i]        value = int(tokens[i + 1])
        if op == '*':            term *= value        elif op == '/':            term /= value        elif op == '+':            total += term            term = Fraction(value)        else:  # op == '-'            total += term            term = Fraction(-value)
        i += 2
    total += term
    print(total.numerator // total.denominator)

if __name__ == "__main__":

The first number initializes term rather than total because multiplication and division have to be completed before that value participates in addition or subtraction.

For * and /, the current term is modified immediately. This naturally gives left-to-right evaluation for operations of equal precedence. For example, 8 / 4 * 3 becomes 2 * 3, then 6, rather than interpreting it as 8 / (4 * 3).

For +, the completed term is added to total, and the next number starts a fresh term. For -, the same idea is used, except the new term starts negative. This avoids needing a separate sign variable.

Fraction is the subtle part of the implementation. Python's // would perform integer floor division, which is not ordinary arithmetic division, and / on integers produces a floating-point value. An expression such as 5 / 2 * 2 demonstrates why neither is appropriate. Fraction stores 5/2 exactly, then multiplication by 2 returns exactly 5.

The input guarantees a valid expression, so division by zero does not occur. Python integers also have arbitrary precision, so there is no overflow issue even though the official final result is guaranteed to fit in 32 bits.

Worked Examples

For Sample 1:

7 + 5

The expression has only one addition, so the first term can be committed immediately.

Operator	Value	term before	total after	term after
start	7	7	0	7
`+`	5	7	7	5
finish		5	12	5

The final addition of term gives 7 + 5 = 12, matching the sample output.

For Sample 3:

3 + 4 * 3 - 6 / 3 * 3

The multiplicative chains are 4 * 3 and 6 / 3 * 3. The second chain also demonstrates left-to-right handling of division and multiplication.

Operator	Value	term before	total after	term after
start	3	3	0	3
`+`	4	3	3	4
`*`	3	4	3	12
`-`	6	12	15	-6
`/`	3	-6	15	-2
`*`	3	-2	15	-6
finish		-6	9	-6

The final result is 9. The trace shows why addition happens only after the preceding multiplication chain is complete, and why 6 / 3 * 3 is evaluated from left to right.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(k)	Every token is processed exactly once, with constant-many arithmetic operations per token.
Space	O(k)	The tokenized input is stored, while the arithmetic state itself uses O(1) additional variables.

Here k denotes the number of operators or, equivalently up to a constant factor, the expression length. Since the statement gives no explicit maximum expression length, the linear scan is the appropriate choice for the one-second limit.

Test Cases

Pythonimport sysimport iofrom fractions import Fraction

def solve():    tokens = input().split()
    total = Fraction(0)    term = Fraction(int(tokens[0]))
    i = 1    while i < len(tokens):        op = tokens[i]        value = int(tokens[i + 1])
        if op == '*':            term *= value        elif op == '/':            term /= value        elif op == '+':            total += term            term = Fraction(value)        else:            total += term            term = Fraction(-value)
        i += 2
    total += term    print(total.numerator // total.denominator)

def run(inp: str) -> str:    global input    old_stdin = sys.stdin

Test input	Expected output	What it validates
`0`	`0`	Minimum-size expression containing a single number
`2 * 2 * 2 * 2`	`16`	Repeated equal values and left-to-right multiplication
`5 / 2 * 2`	`5`	Fractional intermediate result and exact division
`1 - 5 / 2 * 2 + 4`	`0`	Negative terms combined with multiplication and division
`1 + 1 + ... + 1` with 10,001 numbers	`10001`	Large input and linear-time processing

Edge Cases

A single-number expression such as

starts with term = 7, the loop has no operators to process, and the final total += term produces 7. The algorithm never assumes that an operator exists.

For precedence, consider

3 - 6 * 3

The algorithm starts with term = 3. The subtraction commits 3 to total and starts term = -6. The multiplication then changes that term to -18. At the end, total + term is 3 - 18 = -15. The multiplication never gets evaluated after the subtraction, so the required precedence is preserved.

For equal-precedence operations, consider

5 / 2 * 2

The initial term is 5. Division changes it to the exact fraction 5/2, and multiplication changes it to 5. The final output is 5. An integer-division implementation would instead obtain 2 after the division and incorrectly output 4.

Finally, consider a negative multiplication/division chain such as

1 - 5 / 2 * 2 + 4

After the subtraction, term is -5. Division produces -5/2, multiplication produces -5, and the next addition commits that term, giving 1 - 5 = -4. Starting the next term at 4 gives the final answer 0. The exact rational representation handles the negative intermediate value without relying on language-specific integer division rules.