表达数字100而没有太多嵌套函数？
干得好：
>>> test = lambda f: f(lambda x: x + 1)(0)>>> z = lambda f: lambda x: x>>> test(z)0>>> succ = lambda n: lambda f: lambda x: f(n(f)(x))>>> _1 = succ(z)>>> test(_1)1>>> _2 = succ(_1)>>> test(_2)2>>> plus = lambda m: lambda n: lambda f: lambda x: m(f)(n(f)(x))>>> _3 = plus(_1)(_2)>>> test(_3)3>>> mult = lambda m: lambda n: lambda f: lambda x: m(n(f))(x)>>> _6 = mult(_2)(_3)>>> test(_6)6>>> _5 = plus(_2)(_3)>>> _25 = mult(_5)(_5)>>> _4 = plus(_2)(_2)>>> _100 = mult(_25)(_4)>>> test(_100)100>>>
