CF 104180A - Weather Forecast

We are given a fixed-length sequence of 28 real numbers, each representing the probability of rain on a particular day in February. Each value lies between 0 and 1. A day is considered “rainy” only if its probability meets or exceeds 0.8.

codeforcescompetitive-programming

CF 104180A - Weather Forecast

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Solve time: 56s
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Solution

Problem Understanding

We are given a fixed-length sequence of 28 real numbers, each representing the probability of rain on a particular day in February. Each value lies between 0 and 1. A day is considered “rainy” only if its probability meets or exceeds 0.8. The task is simply to count how many of these 28 values satisfy that threshold condition.

The structure of the input removes almost all algorithmic freedom. There is no dynamic behavior, no dependencies between days, and no aggregation beyond counting. The output is a single integer, the number of qualifying values.

The constraints are extremely small in practice since the input size is fixed at 28. Even if this were generalized, the operation we need per element is a constant-time comparison, so even for very large inputs this would be linear time and trivially fast.

The only subtle issues come from parsing floating-point input correctly. The values are decimal representations, so implementations that incorrectly tokenize input or rely on integer parsing will fail. Another edge case is boundary handling at exactly 0.8, which must be included, not excluded.

A naive mistake would be to treat the threshold as strictly greater than 0.8 rather than greater-or-equal. For example, a value like 0.8 itself should be counted. Another potential issue is splitting input incorrectly if multiple spaces or formatting variations appear, but standard token-based parsing avoids this.

Approaches

The brute-force approach is also the optimal one in this problem. We iterate through all 28 values, check each one against the threshold 0.8, and maintain a running counter of how many satisfy the condition. Each check is constant time, so the total work is proportional to the number of values.

There is no structure to exploit beyond this direct scan. The key observation is that the problem is fundamentally a filtering operation over a fixed-size list. Since no preprocessing or transformation reduces future work, any algorithm must inspect each element at least once, making a single pass optimal.

The brute-force view becomes optimal because the input size is constant and small, and because the decision rule is independent per element.

Approach Time Complexity Space Complexity Verdict
Brute Force Scan O(28) O(1) Accepted
Single Pass Counter O(28) O(1) Accepted

Algorithm Walkthrough

  1. Read the entire line of input and split it into tokens representing the 28 floating-point values. This ensures we correctly isolate each probability regardless of spacing.
  2. Initialize a counter to zero. This counter will track how many values meet the rainy-day condition.
  3. Iterate through each parsed value. For each value, convert it to a floating-point number.
  4. Compare the value against the threshold 0.8. If the value is greater than or equal to 0.8, increment the counter. The equality condition matters because 0.8 itself is defined as rainy.
  5. After processing all 28 values, output the final counter.

The correctness relies on the fact that every day is evaluated independently. No intermediate computation affects future decisions.

Why it works

Each element is classified solely based on a fixed predicate: whether it is at least 0.8. The algorithm evaluates this predicate exactly once per input value and accumulates the number of true results. Since addition over counts is associative and independent across elements, the final sum exactly matches the number of qualifying days.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    data = input().split()
    count = 0
    for x in data:
        if float(x) >= 0.8:
            count += 1
    print(count)

if __name__ == "__main__":
    solve()

The solution reads all tokens in one pass, avoiding any manual parsing logic. Each token is independently converted to a float, and the comparison is applied immediately.

A common pitfall would be using strict inequality > instead of >=, which would incorrectly exclude exact threshold values. Another subtle issue is forgetting to convert strings to floats, which would lead to lexicographic comparisons rather than numeric ones.

Worked Examples

Example 1

Input:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.25 0.22 0.64 0.43 0 0.99 0.87 0.28 0.83 0.77 0.92 0.45 0.60 0.887 0.935 0.353 0.182

We scan each value and mark those ≥ 0.8.

Day Value ≥ 0.8? Count
1 0 No 0
9 0.8 Yes 1
10 0.9 Yes 2
11 1 Yes 3
17 0.99 Yes 4
18 0.87 Yes 5
20 0.83 Yes 6
22 0.92 Yes 7
25 0.887 Yes 8
26 0.935 Yes 9

Final output is 9.

This confirms that exact threshold values like 0.8 are included and that scattered qualifying values are correctly accumulated.

Example 2

Input:

0.8 0.8 0.8 0.79 0.81 0.7 0.9 0.6 0.85 0.2 0.8 0.1 0.8 0.8 0.3 0.4 0.8 0.8 0.8 0.05 0.95 0.8 0.79 0.8 0.8 0.8 0.8 0.8
Day Value ≥ 0.8? Count
1 0.8 Yes 1
2 0.8 Yes 2
4 0.79 No 2
5 0.81 Yes 3
7 0.9 Yes 4
9 0.85 Yes 5
11 0.8 Yes 6
13 0.8 Yes 7
21 0.95 Yes 8
22 0.8 Yes 9

Final output is 9.

This example stresses repeated threshold comparisons and confirms that multiple identical boundary values are handled consistently.

Complexity Analysis

Measure Complexity Explanation
Time O(28) Each of the 28 values is processed exactly once with constant-time operations
Space O(1) Only a small counter and temporary variables are used

The computation is constant-time in practice since the input size is fixed. Even if generalized, a single linear scan over floating-point values easily fits within typical constraints.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys as _sys
    from math import isclose

    data = _sys.stdin.readline().split()
    count = 0
    for x in data:
        if float(x) >= 0.8:
            count += 1
    return str(count)

# provided sample
assert run("0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.25 0.22 0.64 0.43 0 0.99 0.87 0.28 0.83 0.77 0.92 0.45 0.60 0.887 0.935 0.353 0.182\n") == "9"

# all below threshold
assert run("0 " * 28) == "0"

# all above threshold
assert run("0.8 " * 28) == "28"

# alternating boundary values
assert run("0.8 0.79 " * 14) == "14"

# mixed precision case
assert run("0.799999 0.8000001 " * 14) == "14"
Test input Expected output What it validates
all zeros 0 no false positives
all 0.8 28 inclusive boundary correctness
alternating 0.8/0.79 14 repeated boundary handling
floating precision edge case 14 robustness of comparison

Edge Cases

One important edge case is when values are exactly 0.8. For input like:

0.8 0.8 0.8 ... (28 times)

the algorithm evaluates each value, finds all satisfy >= 0.8, and increments the counter each time. The final output is 28, confirming correct inclusion of boundary values.

Another edge case involves values extremely close to 0.8 due to floating-point representation, such as 0.799999 or 0.8000001. The comparison is still performed directly on float conversions, and the algorithm correctly classifies only those that meet or exceed the threshold after parsing.

A third case is when no values qualify, such as all zeros. The counter remains at zero throughout iteration, and the output is correctly 0, showing that the algorithm handles empty-accumulation scenarios without special casing.