CF 102697157 - Computing Sin

Rating: -
Tags: -
Solve time: 1m 1s
Verified: yes

Solution

Problem Understanding

The task is to approximate sin(x) using exactly n terms of its Maclaurin, or Taylor-at-zero, series. The input consists of the number n of terms followed by a real number x, where x is measured in radians. The required output is the value obtained by truncating the sine series after those n terms.

The sine series is

sin(x)=x− 3! x 3 + 5! x 5 − 7! x 7 +⋯

If the requested number of terms is n, the answer is

k=0 ∑ n−1 (−1) k (2k+1)! x 2k+1 .

For example, with n = 5 and x = 1, we calculate

1− 3! 1 + 5! 1 − 7! 1 + 9! 1 ,

which gives the sample output 0.8414710097001764.

The published problem gives a one-second time limit and 256 MB of memory, but does not state explicit numeric bounds for n or x. The useful algorithmic target is consequently linear in the number of requested terms. A quadratic implementation may still work for small inputs, but there is no reason to repeatedly rebuild powers and factorials when consecutive Taylor terms have a simple relationship.

There are several numerical edge cases that a careless implementation can mishandle. With one term, for example,

the correct output is 2.0, because only the first term x is included. A loop that always subtracts the cubic term would accidentally calculate a two-term approximation.

A zero angle is another simple boundary case:

The correct output is 0.0. Every Taylor term contains a positive power of x, so every term is zero. An implementation that initializes the answer incorrectly, such as to 1, would produce the wrong result.

Negative angles test both the alternating signs and the odd powers:

2-1

The correct output is -0.8333333333333334, corresponding to

−1− 3! (−1) 3 =−1+ 6 1 =− 6 5 .

A solution that applies the sign separately while also using x with its original sign can accidentally negate every other term incorrectly.

Finally, computing factorials and powers directly can cause unnecessary intermediate growth. The mathematical expression is harmless for a small angle, but explicitly constructing (2k+1)! and x^(2k+1) at every iteration performs much more work than necessary and can also create unnecessarily large floating-point intermediates.

Approaches

The direct approach follows the formula literally. For every term, calculate the odd exponent 2k+1, calculate x^(2k+1), calculate (2k+1)!, apply the alternating sign, and add the result.

This is correct because it evaluates exactly the mathematical expression defining the requested Taylor approximation. The problem is repeated work. The factorial for the next term contains almost the entire factorial from the previous term, and the next power contains almost the entire previous power.

If the k-th factorial is rebuilt from scratch, its cost is proportional to k. Across n terms, that gives approximately

1+2+⋯+n= 2 n(n+1) ,

which is Θ(n²) arithmetic operations. Recomputing powers from scratch introduces additional work as well. For large n, this quadratic behavior is unnecessary.

The key observation is that two consecutive Taylor terms can be obtained directly from one another. Let

T k =(−1) k (2k+1)! x 2k+1 .

Then

T k+1 =−T k (2k+2)(2k+3) x 2 .

This identity removes the need to calculate either a new factorial or a new high power. Once the first term, x, is known, every subsequent term costs only a constant number of arithmetic operations.

The brute-force solution works because it evaluates every term independently, but it fails to reuse information between consecutive terms. The recurrence observes exactly what changes from one term to the next, reducing the entire computation to Θ(n) time and constant extra space.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²)	O(1)	Too slow for large n
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

Read the number of Taylor terms n and the radian value x. The first term of the series is always x, so initialize both term = x and answer = x.
For every remaining term, update the current term using

T k+1 =−T k (2k+2)(2k+3) x 2 .

If the current term corresponds to index k, its denominator is (2k+1)!. Moving to the next odd power multiplies the numerator by x² and the factorial by (2k+2)(2k+3). The minus sign accounts for the alternating signs of the sine series. 3. Add the newly calculated term to answer. At this point answer contains exactly the first k+2 Taylor terms. 4. After n-1 updates, print answer. No call to Python's built-in sin function is needed because the problem specifically asks for the Taylor approximation rather than the actual sine value.

Why it works

The invariant is that at the beginning of each iteration, term is exactly the latest Taylor-series term

(−1) k (2k+1)! x 2k+1 .

Multiplying it by

− (2k+2)(2k+3) x 2

produces

(−1) k+1 (2k+3)! x 2k+3 ,

which is precisely the next term. Since the algorithm starts with the correct first term and repeatedly generates the correct next term, the accumulated sum is exactly the requested finite Taylor approximation, up to ordinary floating-point rounding.

Python Solution

Pythonimport sysinput = sys.stdin.readline

def solve():    n = int(input())    x = float(input())
    term = x    ans = x    x2 = x * x
    for k in range(n - 1):        term *= -x2 / ((2 * k + 2) * (2 * k + 3))        ans += term
    print(ans)

if __name__ == "__main__":    solve()

The first two assignments initialize the first Taylor term. This also handles n = 1 naturally, because the loop executes zero times and the answer remains x.

x2 stores x² once rather than multiplying x by itself on every iteration. The loop runs exactly n - 1 times because the first term was already included before the loop.

The denominator uses (2*k + 2) * (2*k + 3), not (2*k + 1) * (2*k + 2). The latter would describe the wrong transition. For example, the transition from x to -x³/3! must multiply by -x²/(2*3), giving the required denominator 6.

Python's float is used because the input is a real number and the output is a floating-point approximation. Python integers would avoid integer overflow for factorials, but the optimal solution never constructs a factorial at all.

The recurrence also preserves the sign automatically. There is no separate even/odd branch, so negative x is handled correctly without special cases.

The problem's sample is reproduced by the recurrence:

which produces 0.8414710097001764.

Worked Examples

For the first example, n = 5 and x = 1. The algorithm starts with the first term and generates four additional terms.

k	term before update	New term	answer
0	1	-1/6 = -0.16666666666666666	0.8333333333333334
1	-0.16666666666666666	1/120 = 0.008333333333333333	0.8416666666666667
2	0.008333333333333333	-1/5040 = -0.0001984126984126984	0.841468253968254
3	-0.0001984126984126984	1/362880 = 2.7557319223985893e-06	0.8414710097001764

The final value matches the required sample output. The trace also shows the recurrence's central property: each term is derived from the immediately preceding term, while the sign alternates automatically.

For a second example, consider two terms with x = -1.

2-1

The first term is -1. The recurrence multiplies it by

− 2⋅3 (−1) 2 =− 6 1 ,

so the second term is 1/6.

k	term before update	New term	answer
0	-1.0	0.16666666666666666	-0.8333333333333334

The result is -0.8333333333333334. The negative input demonstrates why the sign should not be handled separately from the recurrence. The odd power already carries the correct sign, while the recurrence's negative ratio supplies the alternating Taylor sign.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each of the `n-1` additional terms requires a constant number of arithmetic operations.
Space	O(1)	Only the current term, accumulated answer, and `x²` are stored.

The algorithm scales linearly with the requested number of terms, which is the appropriate complexity when the input can ask for many Taylor terms. It also avoids storing the series or constructing increasingly large factorials, so memory usage remains constant under the problem's 256 MB limit.

Test Cases

Pythonimport sysimport io

def solve():    input = sys.stdin.readline
    n = int(input())    x = float(input())
    term = x    ans = x    x2 = x * x
    for k in range(n - 1):        term *= -x2 / ((2 * k + 2) * (2 * k + 3))        ans += term
    print(ans)

def run(inp: str) -> str:    old_stdin = sys.stdin    old_stdout = sys.stdout
    try:        sys.stdin = io.StringIO(inp)        sys.stdout = io.StringIO()        solve()        return sys.stdout.getvalue()    finally:        sys.stdin = old_stdin        sys.stdout = old_stdout

# Provided sampleassert run("5\n1\n") == "0.8414710097001764\n", "sample 1"
# Minimum number of termsassert run("1\n2\n") == "2.0\n", "one-term approximation"
# Zero inputassert run("10\n0\n") == "0.0\n", "zero angle"
# Negative input and alternating signsassert run("2\n-1\n") == "-0.8333333333333334\n", "negative angle"
# Three termsassert run("3\n1\n") == "0.8416666666666667\n", "three-term approximation"
# Large number of terms, where later terms become negligibleassert run("10000\n0.1\n") == "0.09983341664682817\n", "large term count"

Test input	Expected output	What it validates
`1\n2\n`	`2.0`	Minimum term count and loop boundary
`10\n0\n`	`0.0`	Zero input and repeated zero terms
`2\n-1\n`	`-0.8333333333333334`	Negative input and alternating signs
`3\n1\n`	`0.8416666666666667`	Correct transition between consecutive terms
`10000\n0.1\n`	`0.09983341664682817`	Large term count and constant-time term generation

Edge Cases

For one requested term,

the initialization gives term = 2 and answer = 2. The loop condition range(n - 1) becomes range(0), so no extra term is added. The output is 2.0, exactly the first Taylor term.

For zero,

the initial term is zero and x² is also zero. Every recurrence update keeps term equal to zero, so the accumulated answer remains zero. The algorithm never needs a special zero case.

For a negative angle,

2-1

the initial term is -1. The recurrence multiplies it by -1/6, producing 1/6, so the answer becomes -5/6. This confirms that the odd powers and alternating Taylor signs are both represented correctly by the recurrence.

For the boundary between terms, consider

The required expression is

1− 3! 1 + 5! 1 .

Starting from 1, the first update divides by 2*3 and negates the term, producing -1/6. The second update divides by 4*5 and negates again, producing 1/120. The final answer is

1− 6 1 + 120 1 =0.8416666666666667.

This specifically catches an off-by-one error where the implementation accidentally generates n+1 terms instead of exactly n.