有几条能被窥孔化简

按模型，[y=x*1, z=x*0, r=a+b]，count_simplified 交回几？

贯穿本节的 IR 模型（判题机没 gcc，这是它的确定模型）：程序是三地址码元组列表，每条 (dst, op, a, b)——op ∈ const/copy/+/-/*，操作数 a/b 是整数（常量）或字符串（变量名）。

窥孔优化：peep1 单条化简（x*1/x+0 → copy x；x*0 → const 0）；peephole 逐条化简；is_nop 判 x=copy x 空操作、strip_nops 删空操作、count_simplified 数能化简的。

def peep1(ins):
    """单条窥孔化简：x*1 或 x+0 -> copy x；x*0 -> const 0；其它原样。"""
    (dst, op, a, b) = ins
    if op == "*" and b == 1:
        return (dst, "copy", a, None)
    if op == "+" and b == 0:
        return (dst, "copy", a, None)
    if op == "*" and b == 0:
        return (dst, "const", 0, None)
    return ins


def peephole(prog):
    """对每条指令做一次窥孔化简。"""
    return [peep1(ins) for ins in prog]


def is_nop(ins):
    """x = copy x 是空操作（什么也没做）。"""
    return ins[1] == "copy" and ins[0] == ins[2]


def strip_nops(prog):
    """把空操作删掉。"""
    return [ins for ins in prog if not is_nop(ins)]


def count_simplified(prog):
    """有几条能被窥孔化简（化简后和原来不一样）。"""
    return sum(1 for ins in prog if peep1(ins) != ins)

print(count_simplified([("y", "*", "x", 1), ("z", "*", "x", 0), ("r", "+", "a", "b")]))