Overview

This problem asks you to evaluate a mathematical expression written in Polish notation (prefix notation). You need to compute the evaluation results of both the original expression and the expression where operators at specified positions are replaced with their paired operators.

Analysis

Evaluating Polish Notation

A Polish notation expression can be evaluated by reading tokens from left to right using a stack. In the usual Reverse Polish Notation (postfix notation), you read tokens, push them onto the stack, and when an operator appears, pop two values and compute. However, in prefix notation, operators come first, so a slight modification is needed.

Key Insights

1. Both expressions can be evaluated simultaneously: The original expression and the flipped expression have exactly the same structure (tree shape), differing only in operators. Therefore, both can be evaluated simultaneously in a single parse.

2. Stack-based evaluation: When reading prefix notation from left to right, upon encountering an operator, push it onto the stack as a “waiting for operands” state. When encountering a number, advance the computation based on the stack’s state.

Stack State Management

Each entry on the stack is one of three types:

• val: An already-evaluated value (a pair of original value and flipped value)
• op1: An operator has been read, but the first operand has not yet been received
• op2: The first operand is determined, waiting for the second operand

As an example, let’s trace the processing of * + 3 5 2:

Token read	Stack transition
*	[op1: *]
+	[op1: *, op1: +]
3	3 is a value → merges with op1: + → [op1: *, op2: +(3)]
5	5 is a value → merges with op2: +(3) → 3+5=8 becomes a value → merges with op1: * → [op2: *(8)]
2	2 is a value → merges with op2: *(8) → 8*2=16 becomes a value → [val: 16]

The final result is \(16\).

Division Caveat

/ is integer division that truncates toward zero. Python’s // is floor division (truncation toward negative infinity), so results differ when the signs are different. For example, \((-7) / 2\) is \(-3\) (toward zero), but Python’s (-7) // 2 is \(-4\) (floor). This must be handled correctly.

Algorithm

1. Read the input and store the positions to be flipped in a set.
2. Process tokens from left to right in order:
   • If it’s an operator: Record both the original operator and the flipped operator, and push it onto the stack as op1.
   • If it’s a number: Create it as a val. If the stack top is op2 (waiting for the second operand), compute and merge. If the merged result can further resolve another op2, repeat. Finally, if the stack top is op1, promote it to op2.
3. The single val remaining on the stack is the answer.

Complexity

• Time complexity: \(O(N)\) — each token is processed with a constant number of stack operations
• Space complexity: \(O(N)\) — \(O(N)\) for the stack and the set of flip positions

Implementation Notes

• Implementing truncated division: Since Python’s // is floor division, results are off when the dividend and divisor have different signs. Use (a ^ b) >= 0 to check if they have the same sign, and for different signs, divide the absolute values and then apply the sign.

• Stack reduction loop: After reading a number, it can be merged with op2 entries on the stack top in a chain, so process this with a while loop. This correctly evaluates nested expressions like + 1 + 2 + 3 4.

• Non-existence of -0: The problem constraints guarantee that -0 is never given, so if a token is the single character -, it can safely be determined to be an operator.

  Source Code

import sys
from typing import List, Tuple

def solve():
    input_data = sys.stdin.buffer.read().decode()
    tokens_input = input_data.split()
    idx = 0
    N = int(tokens_input[idx]); idx += 1
    toks = tokens_input[idx:idx+N]; idx += N
    K = int(tokens_input[idx]); idx += 1
    
    flipped = set()
    if K > 0:
        for i in range(K):
            p = int(tokens_input[idx]); idx += 1
            flipped.add(p - 1)  # 0-indexed
    
    def flip_op(op):
        if op == '+': return '-'
        if op == '-': return '+'
        if op == '*': return '/'
        if op == '/': return '*'
    
    def is_operator(t):
        return t in ('+', '-', '*', '/')
    
    # Build the expression tree iteratively using a stack-based approach
    # We'll evaluate both original and flipped expressions simultaneously
    
    # First, let's parse the Polish notation into a tree structure
    # Then evaluate both versions
    
    # For efficiency, we do iterative evaluation using a stack
    # We process tokens left to right. 
    # Stack contains entries: either a resolved value or a partially applied operator
    
    # Each stack entry: (type, ...)
    # type 'val': ('val', original_value, flipped_value)
    # type 'op1': waiting for first operand - ('op1', orig_op, flip_op)
    # type 'op2': has first operand, waiting for second - ('op2', orig_op, flip_op, orig_val1, flip_val1)
    
    def trunc_div(a, b):
        # Truncated division toward zero
        if b == 0:
            raise ZeroDivisionError
        # Python's // is floor division, we need truncation toward zero
        if (a ^ b) >= 0:
            return a // b
        else:
            return -((-a) // b) if a < 0 else -(a // (-b))
    
    def apply_op(op, a, b):
        if op == '+': return a + b
        if op == '-': return a - b
        if op == '*': return a * b
        if op == '/': return trunc_div(a, b)
    
    stack = []
    
    for i in range(N):
        t = toks[i]
        if is_operator(t):
            orig_op = t
            if i in flipped:
                flip = flip_op(t)
            else:
                flip = t
            stack.append(('op1', orig_op, flip))
        else:
            # It's a number
            val = int(t)
            # Try to reduce
            current = ('val', val, val)
            
            while stack and stack[-1][0] == 'op2':
                entry = stack.pop()
                _, orig_op, flip, ov1, fv1 = entry
                ov2 = current[1]
                fv2 = current[2]
                orig_res = apply_op(orig_op, ov1, ov2)
                flip_res = apply_op(flip, fv1, fv2)
                current = ('val', orig_res, flip_res)
            
            if stack and stack[-1][0] == 'op1':
                entry = stack.pop()
                _, orig_op, flip = entry
                stack.append(('op2', orig_op, flip, current[1], current[2]))
            else:
                stack.append(current)
    
    # The stack should have exactly one 'val' entry
    result = stack[0]
    print(result[1])
    print(result[2])

solve()

---

This editorial was generated by claude4.6opus-thinking.