CF 102697151 - Triangle Trigonometry

CF 102697151 - Triangle Trigonometry

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Solve time: 1m 31s
Verified: yes

Solution

Problem Understanding

Despite the name, this problem does not require constructing or analyzing a triangle. The task is simply to evaluate one trigonometric function for a supplied floating-point argument. The first input line identifies the function, using one of sin, cos, tan, cosec, sec, or cotan, and the second line gives the value at which that function should be evaluated. The required output is the resulting floating-point value. The archived statement confirms that the input consists of exactly one operation keyword followed by one floating-point value, and that the output is one floating-point result.

The six operations correspond to the usual definitions

[ \sin(x)=\sin(x), ]

[ \cos(x)=\cos(x), ]

[ \tan(x)=\frac{\sin(x)}{\cos(x)}, ]

[ \cosec(x)=\frac{1}{\sin(x)}, ]

[ \sec(x)=\frac{1}{\cos(x)}, ]

and

[ \cotan(x)=\frac{\cos(x)}{\sin(x)}. ]

The argument is interpreted in radians, because Python's math.sin, math.cos, and math.tan functions use radians. The sample input sin followed by 2 produces approximately 0.9092974268256817, which is exactly the value of (\sin(2)) in radians.

There are no meaningful large input constraints here. Only one operation is evaluated, so even an input with a very large number of digits would not create a combinatorial problem, assuming the value can be represented by the floating-point parser. The one-second time limit is therefore far more than enough for a constant number of arithmetic operations. There is no reason to consider algorithms such as iteration over ranges, dynamic programming, or numerical search.

The main edge cases concern reciprocal trigonometric functions. For example, with

cosec
0

the mathematical value is undefined because (\sin(0)=0). A careless implementation that blindly computes 1 / math.sin(x) will raise a division-by-zero error rather than produce a valid numerical answer. The statement does not specify a special output for undefined values, so such inputs should be regarded as outside the intended valid domain.

A similar issue occurs with

sec
1.5707963267948966

because the argument is approximately (\pi/2), where (\cos(x)=0). The mathematical secant is undefined there. Again, there is no special representation for infinity or undefined values in the statement, so a correct implementation follows the stated mathematical domain rather than inventing an output convention.

A different subtle case is a value very close to a singularity. For example,

tan
1.5707963267948965

is close to (\pi/2), but is not exactly the same floating-point value. The result is a very large finite floating-point number. Comparing the input to a hard-coded value such as pi / 2 and treating nearby values as undefined would be incorrect.

Approaches

The direct approach is already optimal for this problem. A brute-force interpretation would be to try to search for some alternative representation of the requested trigonometric value, perhaps by repeatedly applying identities or approximating the function numerically. Such an approach is unnecessary because the language's standard math library already evaluates the six required functions directly.

The exact operation count of the direct method is constant. We perform one string comparison or dictionary lookup, one call to a standard trigonometric function for sin, cos, or tan, or one additional division for a reciprocal function. Thus the running time is (O(1)) and the auxiliary space is (O(1)).

The usual brute-force-versus-optimized distinction does not really arise here. There is no input size that causes the direct computation to become too slow. If we tried to approximate a trigonometric function using a Taylor series, for example, the number of terms would depend on the desired precision and would introduce unnecessary work. The key observation is simply that the requested operations are exactly the operations provided by the standard floating-point math library.

The only design choice is how to map the keyword to its mathematical operation. A chain of if statements works, but a dictionary makes the relationship between each keyword and its computation explicit and keeps the implementation compact.

Approach	Time Complexity	Space Complexity	Verdict
Numerical approximation or repeated identities	Depends on approximation strategy	Depends on strategy	Unnecessary
Direct standard-library evaluation	(O(1))	(O(1))	Accepted

Algorithm Walkthrough

1. Read the operation keyword from the first line and remove surrounding whitespace. The keyword completely determines which mathematical function must be evaluated.
2. Read the floating-point argument from the second line and convert it with float. Python's trigonometric functions expect a floating-point value measured in radians.
3. Select the requested operation. For sin, cos, and tan, call the corresponding function from math. For cosec, divide one by sin(x), for sec, divide one by cos(x), and for cotan, divide cos(x) by sin(x). These reciprocal functions are not provided directly under the required names, so they must be constructed from sine and cosine.
4. Print the resulting floating-point value. Python's default floating-point formatting provides enough digits for the judge's floating-point comparison.

The invariant is that after the operation has been selected, the stored result is exactly the mathematical function requested by the input keyword, evaluated using the same floating-point representation of the supplied argument. Every possible valid keyword maps to one of the six definitions, so there is no other case that needs to be handled.

Python Solution

import sys
import math

input = sys.stdin.readline

def solve():
    operation = input().strip()
    x = float(input())

    if operation == "sin":
        result = math.sin(x)
    elif operation == "cos":
        result = math.cos(x)
    elif operation == "tan":
        result = math.tan(x)
    elif operation == "cosec":
        result = 1.0 / math.sin(x)
    elif operation == "sec":
        result = 1.0 / math.cos(x)
    else:  # cotan
        result = math.cos(x) / math.sin(x)

    print(result)

if __name__ == "__main__":
    solve()

The first line is read as a string because it is an operation name rather than a numeric value. Calling .strip() removes the newline that readline() leaves at the end.

The second line is converted with float, which is necessary because the input can contain decimal values. The trigonometric functions in math operate on radians, matching the mathematical interpretation used by the problem.

The reciprocal operations deliberately use the definitions rather than trying to find alternate library functions. cosec is 1 / sin(x), sec is 1 / cos(x), and cotan is cos(x) / sin(x). The order of operations matters for cotan, since reversing the numerator and denominator would compute tangent instead.

There is no need for integer arithmetic, modular arithmetic, arrays, loops, or special numerical iteration. Python integers also never overflow, but that is irrelevant here because all computations are performed as floating-point values.

The final else handles cotan, since the statement restricts the keyword to the six specified names. If invalid keywords were possible, an explicit final branch would be preferable, but the input format guarantees the operation is one of those six.

Worked Examples

Sample 1

The supplied sample is

sin
2

The execution is:

Step	Operation	x	Result
1	Read keyword	sin	Not computed yet
2	Read argument	sin, 2.0	Not computed yet
3	Apply sine	sin, 2.0	0.9092974268256817
4	Print	sin, 2.0	0.9092974268256817

The result matches the expected sample output. This demonstrates the basic path where the requested operation is directly available in the standard library.

Sample 2

Consider the valid input

sec
0

The execution is:

Step	Operation	x	Intermediate value	Result
1	Read keyword	sec
2	Read argument	sec	0.0
3	Compute cosine	sec	(\cos(0)=1)
4	Take reciprocal	sec	(1/1)	1.0
5	Print	sec		1.0

This trace demonstrates why reciprocal functions must be implemented from their definitions. sec(0) is simply the reciprocal of cos(0).

Complexity Analysis

Measure	Complexity	Explanation
Time	(O(1))	Exactly one trigonometric computation and at most a constant number of arithmetic operations are performed.
Space	(O(1))	Only the operation name, input value, and result are stored.

The problem contains only one operation and one argument, so the constant-time solution is comfortably within the one-second time limit and uses negligible memory compared with the 256 MB limit specified by the archive.

Test Cases

The official statement provides one sample, so the test suite below includes that sample together with cases for every operation and several boundary-oriented inputs.

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

def solve():
    operation = input().strip()
    x = float(input())

    if operation == "sin":
        result = math.sin(x)
    elif operation == "cos":
        result = math.cos(x)
    elif operation == "tan":
        result = math.tan(x)
    elif operation == "cosec":
        result = 1.0 / math.sin(x)
    elif operation == "sec":
        result = 1.0 / math.cos(x)
    else:
        result = math.cos(x) / math.sin(x)

    print(result)

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

    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    try:
        output = io.StringIO()
        old_stdout = sys.stdout
        sys.stdout = output
        try:
            solve()
        finally:
            sys.stdout = old_stdout
        return output.getvalue()
    finally:
        sys.stdin = old_stdin
        input = old_input

# Provided sample
assert run("sin\n2\n") == "0.9092974268256817\n", "sample 1"

# All-zero-safe direct functions
assert run("sin\n0\n") == "0.0\n", "sin at zero"
assert run("cos\n0\n") == "1.0\n", "cos at zero"

# Boundary between positive and negative sine values
assert run("sin\n3.141592653589793\n") == "1.2246467991473532e-16\n", "sin(pi)"

# Reciprocal functions
assert run("cosec\n1.5707963267948966\n") == "1.0\n", "cosec(pi/2)"
assert run("sec\n0\n") == "1.0\n", "sec(0)"

# Cotangent at pi/4
assert run("cotan\n0.7853981633974483\n") == "1.0000000000000002\n", "cotan(pi/4)"

# Tangent at pi/4
assert run("tan\n0.7853981633974483\n") == "0.9999999999999999\n", "tan(pi/4)"

Test input	Expected output	What it validates
sin / 2	0.9092974268256817	Official sample and direct sine evaluation
sin / 0	0.0	Minimum simple argument and zero handling
cos / 0	1.0	Cosine boundary value
sin / pi	1.2246467991473532e-16	Floating-point behavior near a mathematical zero
cosec / pi/2	1.0	Reciprocal of sine
sec / 0	1.0	Reciprocal of cosine
cotan / pi/4	1.0000000000000002	Reciprocal trigonometric function and floating-point precision
tan / pi/4	0.9999999999999999	Direct tangent computation and floating-point representation

There is no meaningful maximum-size numeric test case specified by the problem, because the statement does not give a bound on the number of digits in the floating-point argument. The computational work remains constant for every normally representable floating-point input.

Edge Cases

For sin(0), the algorithm directly calls math.sin(0.0) and obtains 0.0. For example,

sin
0

produces

0.0

No division is involved, so zero is completely safe for the sine operation.

For cosec(0), the situation is different:

cosec
0

The algorithm first evaluates (\sin(0)=0), then attempts (1/0). The mathematical expression is undefined, and Python raises a division-by-zero exception. The problem statement does not define an output for this case, so there is no correct special value that the solution should invent.

The same reasoning applies to secant at a zero of cosine. With

sec
1.5707963267948966

the argument is the floating-point representation of (\pi/2), and math.cos returns a value extremely close to zero rather than necessarily returning exactly zero. Consequently, Python may produce a very large finite result instead of raising an exception. This is a normal consequence of floating-point arithmetic, not an error in the algorithm.

Finally, values close to singularities must not be treated as singular merely because they are numerically close. For example,

tan
1.5707963267948965

is a valid floating-point input distinct from an exact mathematical (\pi/2) representation. The implementation delegates the calculation to math.tan rather than introducing an arbitrary epsilon check. That preserves the semantics of the supplied floating-point value and avoids incorrectly rejecting legitimate inputs.