AI010. 稳定 Softmax 与交叉熵

# 每行减 max 再 exp 求 softmax;交叉熵不对概率取对数,用 log-sum-exp:−ln p = ln Σexp(z−max) − (z_label−max)。
# 自然对数;概率四位、平均交叉熵六位,注意负零钳制。
import math
import sys


def fmt(value: float, digits: int, eps: float) -> str:
    if abs(value) < eps:
        value = 0.0
    return f"{value:.{digits}f}"


def solve() -> None:
    data = sys.stdin.buffer.read().split()
    if not data:
        return
    it = iter(data)
    r = int(next(it))
    c = int(next(it))
    rows = [[float(next(it)) for _ in range(c)] for _ in range(r)]
    labels = [int(next(it)) for _ in range(r)]

    out = []
    ce_sum = 0.0
    for row, label in zip(rows, labels):
        mx = row[0]
        for v in row:
            if v > mx:
                mx = v
        denom = 0.0
        exps = []
        for v in row:
            e = math.exp(v - mx)
            denom += e
            exps.append(e)
        out.append(" ".join(fmt(e / denom, 4, 0.00005) for e in exps))
        ce_sum += -((row[label] - mx) - math.log(denom))
    out.append(fmt(ce_sum / r, 6, 0.0000005))
    print("\n".join(out))


if __name__ == "__main__":
    solve()