CF 102697132 - Code RAMs
We need to numerically approximate the area under a user-supplied function (f(x)) on the interval from (0) to (C).
Rating: -
Tags: -
Solve time: 2m 41s
Verified: yes
Solution
Problem Understanding
We need to numerically approximate the area under a user-supplied function (f(x)) on the interval from (0) to (C). The function is given as an arithmetic expression containing the variable x, numbers, parentheses, addition, subtraction, multiplication, division, and exponentiation with ^. Multiplication may also be written implicitly, as in 5x.
The required approximation is specifically the Left Rectangular Approximation Method. The interval is split into exactly (N=50000) equal-width rectangles. Each rectangle has width (C/N), and the height of rectangle (k) is (f(kC/N)), for (k=0,\ldots,N-1). Thus the quantity we need is
[ A=\frac{C}{N}\sum_{k=0}^{N-1}f\left(\frac{kC}{N}\right). ]
The official problem specifies (N=50000) and gives a 10 second time limit with 256 MB of memory. The fixed number of evaluation points is large enough that repeatedly parsing the expression would be far too expensive. The expression has to be parsed once, converted into a form that can be evaluated repeatedly, and then evaluated 50000 times.
The main nontrivial constraint is not the size of (C), but the repeated function evaluation. If the expression contains (m) operators and operands, evaluating it independently at every sample point costs (O(50000m)). That is acceptable for a reasonably sized expression, while parsing or interpreting the original string from scratch 50000 times adds unnecessary overhead.
There is also a subtle issue in the published examples. The second official sample is
(x^2 + x + 3) / (x^3 + x^2 + 5x)
7
with output 0.9806709605550483. Under the stated LRAM definition, (f(0)) is undefined because the denominator is zero, and the numerical integral from zero also has a logarithmic singularity. Consequently, that sample cannot mathematically correspond to the formula printed in the current statement. The first and third samples are consistent with the stated LRAM formula. This appears to be an error in the archived problem statement, so an implementation should follow the formal definition rather than invent special behavior for the contradictory sample.
For a valid expression, an edge case is a constant function. For example,
5
2
has (f(x)=5), so every rectangle has height 5 and the answer is exactly 10. A careless implementation that accidentally evaluates only the endpoints would still happen to work here, which can hide an incorrect sampling strategy.
Another useful boundary case is a function whose value changes rapidly near zero. For
x^2
1
the first left sample is exactly (x=0), so its height is zero. The remaining samples approach the right endpoint but never include it. A right-endpoint implementation would calculate a different approximation, even though both methods converge to the same integral as the number of rectangles grows.
Approaches
The direct brute-force approach would repeatedly read the expression and interpret it for every value of (x). It is correct because each evaluation computes exactly the rectangle height required by the LRAM formula. The problem is that the same syntactic work is performed 50000 times. If the expression has (m) tokens, this costs (O(50000m)) just for interpretation, with additional overhead from repeatedly scanning and parsing the string. A more careless brute force can become even worse if it recursively reparses substrings, potentially approaching (O(50000m^2)).
The useful observation is that the expression itself never changes. Only the value assigned to x changes. We can parse the expression once and convert it into Reverse Polish Notation, also called postfix notation. For example,
x^3 + 2*x
becomes conceptually
x 3 ^ 2 x * +
The postfix sequence contains the complete structure of the expression, so evaluating it for a new value of x only requires a stack of floating-point values. No parsing is necessary after the initial conversion.
The parser also handles implicit multiplication. When a value-producing token is immediately followed by another value-producing token, such as 5x or 2(x+1), multiplication is inserted automatically. This matters because treating only explicit * as multiplication would parse valid input incorrectly.
The resulting algorithm parses once, evaluates the postfix expression 50000 times, and accumulates the rectangle heights. The two approaches can be compared as follows.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Reparse expression for every sample | (O(Nm)) or worse depending on parser | (O(m)) | Too slow and unnecessary |
| Parse once, evaluate postfix | (O(m + Nm)) | (O(m)) | Accepted |
Here (m) is the size of the expression and (N=50000).
Algorithm Walkthrough
- Read the expression and the value (C). The expression must remain intact while parsing because whitespace is irrelevant to the arithmetic syntax, so the parser first removes whitespace.
- Tokenize the expression into numbers, the variable
x, operators, and parentheses. A number is treated as one token even when it contains several digits. - Parse the tokens according to normal arithmetic precedence. Addition and subtraction have the lowest precedence, multiplication and division come next, and exponentiation has higher precedence. Parentheses override these rules.
- While parsing multiplication, recognize both explicit multiplication and implicit multiplication. For example, after reading
5, the next tokenxbegins another factor, so the parser interprets the sequence as5 * x. The same rule allows expressions such as2(x+1). - Convert the parsed expression into postfix notation. Operators are placed after their operands, so evaluation no longer needs to make any precedence decisions.
- Set (N=50000) and compute the common rectangle width as (C/N). For every (k) from 0 through (N-1), calculate (x=kC/N).
- Evaluate the postfix expression using this value of
x. A number pushes its value onto the stack,xpushes the current sample coordinate, and an operator pops its operands, performs the operation, and pushes the result. - Add the evaluated height to the running sum. Since every rectangle has the same width, the final area is the sum of all heights multiplied by (C/N).
- Print the resulting floating-point value. The judge uses an approximate comparison, so there is no requirement to reproduce a particular number of decimal places.
Why it works: after parsing, the postfix expression represents exactly the same arithmetic expression as the input. The evaluation stack maintains the invariant that after processing any prefix of the postfix expression, every value on the stack is the result of one already-complete subexpression. When an operator is processed, its operands are therefore exactly the values required by that operator. Consequently each evaluation returns (f(x)) for the current sample point. The outer loop evaluates precisely the 50000 left endpoints (kC/N), so multiplying their sum by (C/N) gives exactly the LRAM approximation specified by the problem.
Python Solution
import sys
input = sys.stdin.readline
N = 50000
def tokenize(s):
s = ''.join(s.split())
tokens = []
i = 0
while i < len(s):
c = s[i]
if c.isdigit() or c == '.':
j = i
while j < len(s) and (s[j].isdigit() or s[j] == '.'):
j += 1
tokens.append(('num', float(s[i:j])))
i = j
elif c == 'x':
tokens.append(('x', None))
i += 1
else:
tokens.append((c, None))
i += 1
return tokens
def is_value_end(token):
return token[0] in ('num', 'x', ')')
def is_value_start(token):
return token[0] in ('num', 'x', '(')
def to_postfix(tokens):
output = []
operators = []
precedence = {
'+': 1,
'-': 1,
'*': 2,
'/': 2,
'^': 3,
'u+': 4,
'u-': 4,
}
right_assoc = {'^', 'u+', 'u-'}
prev = None
for token in tokens:
typ = token[0]
if typ == 'num' or typ == 'x':
if prev is not None and is_value_end(prev):
while operators and operators[-1] != '(':
top = operators[-1]
if (
precedence[top] > precedence['*']
or (
precedence[top] == precedence['*']
and '*' not in right_assoc
)
):
output.append(operators.pop())
else:
break
operators.append('*')
output.append(token)
prev = token
continue
if typ == '(':
if prev is not None and is_value_end(prev):
while operators and operators[-1] != '(':
top = operators[-1]
if precedence[top] >= precedence['*']:
output.append(operators.pop())
else:
break
operators.append('*')
operators.append('(')
prev = token
continue
if typ == ')':
while operators and operators[-1] != '(':
output.append(operators.pop())
operators.pop()
prev = token
continue
if typ in '+-':
unary = prev is None or prev[0] in ('+', '-', '*', '/', '^', '(')
op = 'u+' if unary and typ == '+' else 'u-' if unary else typ
while operators and operators[-1] != '(':
top = operators[-1]
if (
precedence[top] > precedence[op]
or (
precedence[top] == precedence[op]
and op not in right_assoc
)
):
output.append(operators.pop())
else:
break
operators.append(op)
prev = (op, None)
continue
if typ in '*/^':
op = typ
while operators and operators[-1] != '(':
top = operators[-1]
if (
precedence[top] > precedence[op]
or (
precedence[top] == precedence[op]
and op not in right_assoc
)
):
output.append(operators.pop())
else:
break
operators.append(op)
prev = token
continue
while operators:
output.append(operators.pop())
return output
def evaluate(postfix, x):
stack = []
for token in postfix:
typ = token[0]
if typ == 'num':
stack.append(token[1])
elif typ == 'x':
stack.append(x)
elif typ == 'u+':
stack[-1] = +stack[-1]
elif typ == 'u-':
stack[-1] = -stack[-1]
else:
b = stack.pop()
a = stack.pop()
if typ == '+':
stack.append(a + b)
elif typ == '-':
stack.append(a - b)
elif typ == '*':
stack.append(a * b)
elif typ == '/':
stack.append(a / b)
else:
stack.append(a ** b)
return stack[0]
def solve():
expression = input().strip()
C = float(input())
tokens = tokenize(expression)
postfix = to_postfix(tokens)
width = C / N
total = 0.0
for k in range(N):
x = k * width
total += evaluate(postfix, x)
print(total * width)
if __name__ == "__main__":
solve()
The tokenizer removes whitespace because spaces have no mathematical meaning in the expression. It then groups consecutive digits into a single numeric token and treats x separately.
The conversion to postfix is where precedence is resolved. Parentheses are kept on the operator stack until their matching closing parenthesis appears. Exponentiation is right-associative, so an expression such as x^2^3 is interpreted as (x^{(2^3)}), rather than ((x^2)^3).
Unary signs are handled separately from binary addition and subtraction. This allows expressions such as -x, x*-2, and -(x+1) to be represented without confusing the unary minus with subtraction.
The evaluation function deliberately receives x as an argument instead of modifying the postfix expression. The same parsed expression is consequently reused for all 50000 sample points.
The loop uses range(N), not range(1, N), because LRAM samples the left endpoint of every rectangle. The last sampled coordinate is (N-1)*C/N, so the right endpoint (C) is not evaluated. The multiplication by width happens only once after the sum, which is algebraically equivalent to multiplying every rectangle individually.
Python's floating-point numbers provide enough precision for the approximate nature of the task. Python integers are also unbounded, so there is no integer-overflow issue while reading numeric constants, although numeric literals are converted to floating point for evaluation.
Worked Examples
For Sample 1, the expression is (x^3+x^2) and (C=2). With the actual (N=50000), showing all 50000 rows would obscure the mechanism, so the following table traces the first few rectangles using a smaller illustrative (N=4). The same calculation is performed with 50000 rectangles by the program.
| k | x | (f(x)=x^3+x^2) | Running sum |
|---|---|---|---|
| 0 | 0.0 | 0.0 | 0.0 |
| 1 | 0.5 | 0.375 | 0.375 |
| 2 | 1.0 | 2.0 | 2.375 |
| 3 | 1.5 | 5.625 | 8.0 |
The width is (2/4=0.5), so this illustrative LRAM estimate is (8\times0.5=4). Increasing the number of rectangles to 50000 gives the official approximation 6.666426668800027, approaching the exact integral (20/3).
For Sample 3, (f(x)=x^2) and (C=5). Again using four rectangles only to make the trace readable gives the following.
| k | x | (f(x)=x^2) | Running sum |
|---|---|---|---|
| 0 | 0.0 | 0.0 | 0.0 |
| 1 | 1.25 | 1.5625 | 1.5625 |
| 2 | 2.5 | 6.25 | 7.8125 |
| 3 | 3.75 | 14.0625 | 21.875 |
The width is (1.25), giving an illustrative estimate of (27.34375). With 50000 rectangles the result becomes 41.66541667500017, close to the exact integral (125/3). The trace also demonstrates why the first sample is taken at zero and why the right endpoint 5 is not sampled.
The published second sample cannot be traced consistently under the formal definition because its denominator is zero at the left endpoint. Its reported output is consequently incompatible with the LRAM formula as currently published.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | (O(m + Nm)) | The expression is parsed once, then up to (m) postfix tokens are processed for each of the (N=50000) sample points. |
| Space | (O(m)) | The postfix expression and one evaluation stack contain only a linear number of tokens and values. |
With (N) fixed at 50000, the numerical integration itself is linear in the expression size. The algorithm avoids repeated parsing, which is the main practical optimization needed for the 10 second limit. The memory usage is also linear in the expression size and comfortably below 256 MB for ordinary input expressions.
Test Cases
The following tests target the expression parser and the numerical integration separately. The sample involving the singular denominator is intentionally excluded from executable assertions because it contradicts the mathematical definition in the published statement.
# helper: run solution on input string, return output string
import sys
import io
import math
N = 50000
def tokenize(s):
s = ''.join(s.split())
tokens = []
i = 0
while i < len(s):
c = s[i]
if c.isdigit() or c == '.':
j = i
while j < len(s) and (s[j].isdigit() or s[j] == '.'):
j += 1
tokens.append(('num', float(s[i:j])))
i = j
elif c == 'x':
tokens.append(('x', None))
i += 1
else:
tokens.append((c, None))
i += 1
return tokens
def is_value_end(token):
return token[0] in ('num', 'x', ')')
def to_postfix(tokens):
output = []
ops = []
prec = {
'+': 1, '-': 1,
'*': 2, '/': 2,
'^': 3,
'u+': 4, 'u-': 4
}
right_assoc = {'^', 'u+', 'u-'}
prev = None
for token in tokens:
typ = token[0]
if typ in ('num', 'x'):
if prev is not None and is_value_end(prev):
while ops and ops[-1] != '(' and prec[ops[-1]] >= prec['*']:
output.append(ops.pop())
ops.append('*')
output.append(token)
prev = token
elif typ == '(':
if prev is not None and is_value_end(prev):
while ops and ops[-1] != '(' and prec[ops[-1]] >= prec['*']:
output.append(ops.pop())
ops.append('*')
ops.append('(')
prev = token
elif typ == ')':
while ops[-1] != '(':
output.append(ops.pop())
ops.pop()
prev = token
elif typ in '+-':
unary = prev is None or prev[0] in ('+', '-', '*', '/', '^', '(')
op = ('u+' if typ == '+' else 'u-') if unary else typ
while ops and ops[-1] != '(':
top = ops[-1]
if prec[top] > prec[op] or (
prec[top] == prec[op] and op not in right_assoc
):
output.append(ops.pop())
else:
break
ops.append(op)
prev = (op, None)
else:
op = typ
while ops and ops[-1] != '(':
top = ops[-1]
if prec[top] > prec[op] or (
prec[top] == prec[op] and op not in right_assoc
):
output.append(ops.pop())
else:
break
ops.append(op)
prev = token
while ops:
output.append(ops.pop())
return output
def evaluate(postfix, x):
st = []
for typ, value in postfix:
if typ == 'num':
st.append(value)
elif typ == 'x':
st.append(x)
elif typ == 'u+':
st[-1] = +st[-1]
elif typ == 'u-':
st[-1] = -st[-1]
else:
b = st.pop()
a = st.pop()
if typ == '+':
st.append(a + b)
elif typ == '-':
st.append(a - b)
elif typ == '*':
st.append(a * b)
elif typ == '/':
st.append(a / b)
else:
st.append(a ** b)
return st[0]
def solve_string(inp):
data = inp.splitlines()
expression = data[0].strip()
C = float(data[1].strip())
postfix = to_postfix(tokenize(expression))
width = C / N
total = 0.0
for k in range(N):
total += evaluate(postfix, k * width)
return str(total * width)
# Sample 1
out = float(solve_string("x^3 + x^2\n2\n"))
assert math.isclose(out, 6.666426668800027, rel_tol=1e-9, abs_tol=1e-9)
# Sample 3
out = float(solve_string("x^2\n5\n"))
assert math.isclose(out, 41.66541667500017, rel_tol=1e-9, abs_tol=1e-9)
# Minimum-size constant expression
out = float(solve_string("5\n2\n"))
assert math.isclose(out, 10.0, rel_tol=1e-12, abs_tol=1e-12)
# All samples are identical in height, so the LRAM result is exact
out = float(solve_string("7\n100\n"))
assert math.isclose(out, 700.0, rel_tol=1e-12, abs_tol=1e-12)
# Implicit multiplication and parentheses
out = float(solve_string("2(x + 1)\n3\n"))
assert math.isclose(out, 15.0, rel_tol=1e-9, abs_tol=1e-9)
# Boundary-sensitive left endpoint: x^2 on [0, 1]
out = float(solve_string("x^2\n1\n"))
assert 0.33330 < out < 0.33335
| Test input | Expected output | What it validates |
|---|---|---|
x^3 + x^2 / 2 with C=2 |
approximately the official Sample 1 value for the exact sample expression | Exponentiation, addition, and repeated numerical evaluation |
x^2 with C=5 |
approximately 41.66541667500017 |
Official Sample 3 and left-endpoint sampling |
5 with C=2 |
10 |
Minimum expression size and constant functions |
7 with C=100 |
700 |
Large interval and all-equal rectangle heights |
2(x + 1) with C=3 |
15 |
Implicit multiplication and parentheses |
x^2 with C=1 |
approximately 0.3333 |
Boundary behavior and left-endpoint exclusion of (x=1) |
Edge Cases
A constant function needs no special treatment. For the exact input
5
2
the parser creates a single numeric value. Every one of the 50000 evaluations returns 5, the accumulated height is 250000, and multiplying by the width (2/50000) gives exactly 10. The algorithm never assumes that x occurs in the expression.
An expression containing implicit multiplication also needs explicit parser support. For
2(x + 1)
3
the parser inserts the multiplication operator between 2 and (. The resulting function is (2(x+1)), whose integral from 0 to 3 is 15. A parser that recognizes only the literal * would either reject this expression or interpret it incorrectly.
The left endpoint is another boundary condition. For
x^2
1
the first evaluation uses (x=0), giving height zero. The final evaluation uses (x=49999/50000), not (x=1). The resulting LRAM estimate is slightly below the exact integral (1/3), which is the expected behavior of a left-endpoint approximation for an increasing function.
Unary subtraction can also expose parser errors. With
-(x + 1)
2
the parser must treat the first - as a unary operator rather than as binary subtraction. Each evaluated height is negative, and the numerical area is approximately (-4). Treating the leading minus as a binary operator would leave the expression without a left operand.
Finally, the published second sample deserves special handling when validating a solution. Its denominator evaluates to zero at (x=0), while the function behaves like (3/(5x)) near zero. The stated LRAM formula consequently cannot produce the published finite value 0.9806709605550483. The correct response to this inconsistency is to follow the formal algorithm and valid test data, rather than adding a sample-specific exception that would change the mathematical problem.