CF 102409A - Easy Math
For each test case, we have three positive integers (A), (B), and (C). We first add (A) and (B), then divide that sum by (C). The required output is the decimal representation of the result with exactly 50 digits after the decimal point.
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Solve time: 2m 13s
Verified: yes
Solution
Problem Understanding
For each test case, we have three positive integers (A), (B), and (C). We first add (A) and (B), then divide that sum by (C). The required output is the decimal representation of the result with exactly 50 digits after the decimal point. If the exact decimal expansion continues beyond those 50 digits, everything after the 50th digit is discarded rather than rounded.
The numbers can contain up to 50 decimal digits, so ordinary machine-sized integer types are not sufficient in languages with fixed-width integers. The input can contain up to (10^4) test cases, which also rules out algorithms whose running time depends on the numerical value of (A), (B), or (C). In Python, arbitrary-precision integers make the size of the numbers manageable, but we still want only a constant amount of work per test case. Since the output itself always contains 50 fractional digits, an algorithm doing (O(50)) long-division steps per case is a natural target.
There are several cases where an apparently reasonable implementation can fail. The first is an exact integer result. For example, with input
1
1 3 2
the answer is
2.00000000000000000000000000000000000000000000000000
A program that simply prints the integer quotient forgets that the output must always contain exactly 50 digits after the decimal point.
The second issue is truncation instead of rounding. For example,
1
1 1 6
represents (1/3), so the required answer begins with 50 copies of the digit 3:
0.33333333333333333333333333333333333333333333333333
A floating-point conversion followed by formatting can introduce rounding and, more fundamentally, floating-point numbers cannot represent arbitrary 50-digit integer inputs or 50 decimal places exactly.
The third issue is that the fractional part may contain leading zeroes. For
1
1 1 100
the value is (0.02), so the output must be
0.02000000000000000000000000000000000000000000000000
Those zeroes are part of the 50-digit fractional field and cannot be removed.
Finally, the 50th digit itself must be retained. With
1
1 1 99999999999999999999999999999999999999999999999999
the numerator is 2 and the denominator has exactly 50 nines. The first nonzero fractional digit appears at position 50, so the fractional part ends in 2. An implementation that generates only 49 digits has an off-by-one error.
Approaches
A direct brute-force interpretation of decimal division would repeatedly subtract the denominator to determine each quotient digit. For the integer part, finding one quotient digit could require up to (C) subtractions. The same problem occurs for every fractional digit. With 50 fractional positions, this can require (O(50C)) subtractions for one test case. Since (C) can be almost (10^{50}), the worst case is on the order of (5 \times 10^{51}) subtractions for one case, and with (10^4) cases the theoretical total reaches about (5 \times 10^{55}). The approach is mathematically valid, but the numerical size of the input makes it completely impractical.
The useful observation is that decimal division does not require searching for each digit by subtraction. Ordinary long division already tells us exactly what to do. First compute the integer quotient using integer division. The remainder after that division contains all the information needed for the fractional part.
Suppose the current remainder is (r). To obtain the next decimal digit, multiply (r) by 10 and divide by (C). The quotient is exactly the next digit, and the new remainder is what is left over. Repeating this process 50 times produces precisely the first 50 digits after the decimal point.
The reason this works is the same reason manual long division works. If the current remainder represents the unresolved part of the number, multiplying it by 10 shifts that unresolved part one decimal position to the left. Dividing by (C) extracts the next digit without ever using floating-point arithmetic.
The two approaches can be compared as follows.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | (O(50C)) | (O(1)) | Too slow |
| Long Division | (O(50)) big-integer operations per test case | (O(1)) apart from output | Accepted |
Algorithm Walkthrough
- Read (A), (B), and (C), and compute (N=A+B). Python's integer type handles the resulting value exactly, even though it can contain up to 51 digits.
- Compute the integer part with
N // C. This gives everything before the decimal point. - Compute the initial remainder with
N % C. The division identity [ N = (\lfloor N/C\rfloor)C + (N\bmod C) ] tells us that this remainder is exactly the part still needed to construct the fractional digits. - Repeat the following operation exactly 50 times. Multiply the current remainder by 10 and divide by (C). The quotient is the next decimal digit. Store that digit, then replace the remainder by the division remainder.
If the remainder is zero, every later digit is also zero. We can still perform all 50 iterations, which keeps the implementation simple and guarantees the required output length. 5. Convert the 50 generated digits into a string and concatenate the integer part, a decimal point, and the fractional string. Since exactly 50 digits were generated, the output always has the required format.
Why it works
After the integer division, we have (N=qC+r), where (q) is the integer part and (0\le r<C). The fractional part is (r/C). At any fractional step, suppose the current remainder is (r). Then
\frac{\lfloor 10r/C\rfloor}{10} + \frac{(10r\bmod C)}{10C}. ]
The first term gives the next decimal digit, while the second term has exactly the same form as the fractional problem we started with, just with a new remainder. Thus every iteration preserves the same invariant: the generated digits are exactly the leading decimal digits of the original fraction, and the stored remainder represents everything not generated yet. After 50 iterations, the stored digits are exactly the first 50 fractional digits, which is precisely the requested truncated result.
Python Solution
import sys
input = sys.stdin.readline
def solve():
t = int(input())
out = []
for _ in range(t):
a, b, c = map(int, input().split())
n = a + b
integer_part = n // c
remainder = n % c
digits = []
for _ in range(50):
remainder *= 10
digit = remainder // c
remainder %= c
digits.append(str(digit))
out.append(f"{integer_part}." + "".join(digits))
sys.stdout.write("\n".join(out))
if __name__ == "__main__":
solve()
The first two arithmetic operations implement the integer part and initial remainder from steps 2 and 3. They must be performed before generating fractional digits because the remainder after dividing (A+B) by (C) is the exact starting state for long division.
Inside the 50-iteration loop, remainder *= 10 performs the decimal shift. Integer division by c extracts the next digit, and % c keeps the new remainder for the next iteration. The order matters: the digit must be obtained from the multiplied remainder before replacing that remainder with % c.
No floating-point operation appears anywhere in the solution. This avoids both precision loss and accidental rounding. Python also has no fixed 32-bit or 64-bit overflow issue here, so values close to (10^{50}) are handled exactly.
The loop runs 50 times even when the remainder becomes zero. That is intentional. Stopping early would require separately padding the result with zeroes, while fixed iteration count gives the required 50-digit fractional part directly.
The final sys.stdout.write collects all answers and writes them together. With up to (10^4) test cases, this is preferable to repeatedly printing each result independently.
Worked Examples
Consider the first sample case, (A=1), (B=3), (C=2). The sum is 4, so the integer quotient is 2 and the initial remainder is zero.
| Step | Current remainder | After multiplying by 10 | Digit | New remainder |
|---|---|---|---|---|
| Integer division | 4 | 2 | 0 | |
| 1 | 0 | 0 | 0 | 0 |
| 2 | 0 | 0 | 0 | 0 |
| 3 | 0 | 0 | 0 | 0 |
| ... | 0 | 0 | 0 | 0 |
| 50 | 0 | 0 | 0 | 0 |
The fractional part therefore consists entirely of zeroes, producing 2. followed by exactly 50 zeroes. This confirms the exact-division edge case.
Now consider the third sample case, where (A=99), (B=89), and (C=17). The sum is 188. Integer division gives (188//17=11), with remainder (1). The first few long-division iterations are:
| Step | Current remainder | After multiplying by 10 | Digit | New remainder |
|---|---|---|---|---|
| Integer division | 188 | 11 | 1 | |
| 1 | 1 | 10 | 0 | 10 |
| 2 | 10 | 100 | 5 | 15 |
| 3 | 15 | 150 | 8 | 14 |
| 4 | 14 | 140 | 8 | 4 |
| 5 | 4 | 40 | 2 | 6 |
| 6 | 6 | 60 | 3 | 9 |
| 7 | 9 | 90 | 5 | 5 |
| 8 | 5 | 50 | 2 | 16 |
| 9 | 16 | 160 | 9 | 7 |
The generated digits begin with 058823529, giving the sample's 11.058823529.... Continuing the same recurrence for 50 iterations produces the exact required truncated result.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | (O(T \cdot 50)) big-integer arithmetic operations | Every test case performs one integer division and exactly 50 fractional long-division steps. |
| Space | (O(50)) per test case | At most 50 fractional digit characters are stored before producing the answer. |
There are at most (10^4) test cases, so the algorithm performs only about 500,000 fractional iterations. The operands contain at most roughly 51 decimal digits during the calculation, which is tiny for Python's arbitrary-precision integer arithmetic. The resulting work fits comfortably within the 5-second limit, and memory usage is far below 256 MB.
Test Cases
# The solution is organized so that solve_case can be tested directly.
import sys
import io
def solve_case(a, b, c):
n = a + b
integer_part = n // c
remainder = n % c
digits = []
for _ in range(50):
remainder *= 10
digit = remainder // c
remainder %= c
digits.append(str(digit))
return f"{integer_part}." + "".join(digits)
def run(inp: str) -> str:
old_stdin = sys.stdin
try:
sys.stdin = io.StringIO(inp)
t = int(input())
answers = []
for _ in range(t):
a, b, c = map(int, input().split())
answers.append(solve_case(a, b, c))
return "\n".join(answers)
finally:
sys.stdin = old_stdin
sample = (
"3\n"
"1 3 2\n"
"10 25 5\n"
"99 89 17\n"
)
sample_expected = (
"2.00000000000000000000000000000000000000000000000000\n"
"7.00000000000000000000000000000000000000000000000000\n"
"11.05882352941176470588235294117647058823529411764705"
)
assert run(sample) == sample_expected, "sample 1"
assert run("1\n1 1 1\n") == (
"2.00000000000000000000000000000000000000000000000000"
), "all values equal"
assert run("1\n1 1 100\n") == (
"0.02000000000000000000000000000000000000000000000000"
), "leading fractional zero"
assert run("1\n1 1 99999999999999999999999999999999999999999999999999\n") == (
"0.00000000000000000000000000000000000000000000000002"
), "50th fractional digit"
max_value = "9" * 49
assert run(f"1\n{max_value} {max_value} {max_value}\n") == (
"2.00000000000000000000000000000000000000000000000000"
), "maximum-size integers"
assert run("1\n1 1 6\n") == (
"0.33333333333333333333333333333333333333333333333333"
), "truncation rather than rounding"
| Test input | Expected output | What it validates |
|---|---|---|
1 1 1 |
2. followed by 50 zeroes |
Exact integer result and fixed fractional length |
1 1 100 |
0.02 followed by 48 zeroes |
Leading zeroes in the fractional part |
1 1 with a denominator of 50 nines |
0. followed by 49 zeroes and 2 |
Correct handling of the 50th digit |
| Two copies of a 49-digit maximum value divided by the same value | 2. followed by 50 zeroes |
Maximum-size integer arithmetic |
1 1 6 |
0. followed by 50 threes |
Truncation and recurring decimals |
Edge Cases
For an exact division such as
1
1 3 2
we get (N=4), integer_part = 2, and remainder = 0. Every fractional iteration multiplies zero by 10 and extracts digit zero. The result is exactly 2.00000000000000000000000000000000000000000000000000. No special case is required because the remainder invariant naturally handles it.
For a fraction with leading fractional zeroes,
1
1 1 100
we have (N=2), so the integer part is zero and the initial remainder is 2. The first iteration computes (20//100=0), producing the first fractional digit 0. The second computes (200//100=2), producing 2. The remainder then becomes zero, so the remaining 48 digits are zero. The final answer is 0.02000000000000000000000000000000000000000000000000.
For the boundary involving the 50th digit,
1
1 1 99999999999999999999999999999999999999999999999999
the numerator is 2 and the denominator is (10^{50}-1). For the first 49 iterations, multiplying the remainder by 10 still gives a value smaller than the denominator, so every extracted digit is zero. On iteration 50, the remainder has become (2\cdot10^{49}), and multiplying it by 10 gives (2\cdot10^{50}). Dividing by (10^{50}-1) produces digit 2, with a new remainder of 2. The algorithm performs exactly 50 iterations, so that final 2 is retained and everything after it is correctly truncated.
For a recurring decimal such as
1
1 1 6
the initial remainder is 2. Every iteration multiplies it by 10, extracts (20//6=3), and leaves remainder 2 again. The state returns to exactly the same remainder, so every one of the 50 generated digits is 3. The result is 0.33333333333333333333333333333333333333333333333333, with no rounding taking place.
For maximum-size values, Python's arbitrary-precision integers allow the same operations without overflow. If both (A) and (B) are 49-digit numbers, their sum can have 50 digits, and the quotient and remainder operations still work exactly. The algorithm depends on the number of digits, not on repeatedly iterating up to the numerical value of (C), which is what makes it suitable for the given bounds.