随笔分类 - 语言入门
摘要：1.说明下列每对scanf格式串是否等价？如果不等价，请指出它们的差异。 (c) "%f"与"%f "。 在 `scanf` 函数中，`"%f"` 和 `"%f "` 这两种格式的区别在于后面的空格。 1. `scanf("%f", &variable);` 这种情况下，`scanf` 会读取并解析
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摘要：(define (list-change total denoms) (define (cons-all num ls) (let ((num-val num)) (cond ((null? ls) nil) (else (cons (cons num (car ls)) (cons-all num
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摘要：def mutate_reverse(link): """Mutates the Link so that its elements are reversed. >>> link = Link(1) >>> mutate_reverse(link) >>> link Link(1) >>> link
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摘要：(define-macro (switch expr cases) (cons 'cond (map (lambda (case) (cons (eq? (eval expr) (car case)) (cdr case))) cases)) ) 这段代码是一个用于 Scheme 语言的宏定义，可以
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摘要：(define-macro (def func args body) `(define ,(cons func args) ,body)) 分析： 定义一个万能的函数定义，那就要模拟函数定义的样子。ok，函数定义是什么样子的呢？ eg: (define (filter-lst fn lst) (if
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摘要：#lang sicp (define (unique s) (if (null? s) nil (cons (car s) (unique (filter (lambfa (x) (not (eq? x (car s)))) (cdr s))) ) ) ) 这是一个Scheme函数，名为unique
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摘要：Write sub-all, which takes a list s, a list of old words, and a list of new words; the last two lists must be the same length. It returns a list with
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摘要：def reduce_armor(self, amount): """Reduce armor by AMOUNT, and remove the FireAnt from its place if it has no armor remaining. Make sure to damage eac
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摘要：题目： def is_bst(t): """Returns True if the Tree t has the structure of a valid BST. >>> t1 = Tree(6, [Tree(2, [Tree(1), Tree(4)]), Tree(7, [Tree(7), Tr
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摘要：重点：： 观察到：在带入函数之后，没有任何返回值，是直接对参数本身进行作用。遇到这种情况就必须在基线条件下多加一个 【return】语句 why? 其实就是人为两种情况，第一种就是t是节点，此时只需要将节点平方就可以了。第二种情况就是有分支，首先呢就将root平方，再将分支进行处理。 1 def l
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摘要：#include<iostream> using namespace std; int f(int** r, int ** s){ int temp= **r; int temp2=**s; int * z=*r; *r=*s; *s=z; printf("**r= %d\n",**r); prin
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摘要：题目：about generator Write the generator function make_generators_generator, which takes a zero-argument generator function g and returns a generator th
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摘要：def close(n, smallest=10, d=10): """ A sequence is near increasing if each element but the last two is smaller than all elements following its subsequ
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摘要：Problem 2 题目描述： 代码： 1 def inc_subseqs(s): 2 """Assuming that S is a list, return a nested list of all subsequences 3 of S (a list of lists) for which
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摘要：Problem 6 题目描述： 注意事项： 1.不能使用for while循环 2.达到limit时，必须立即返回，不能继续递归 代码： 1.我的代码，很繁琐 1 def shifty_shifts(start, goal, limit): 2 """A diff function for auto
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摘要：Q6 题目描述： Write a function has_path that takes in a tree t and a string phrase. It returns True if there is a path that starts from the root where the
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摘要：1. >>> odds = [1, 3, 5, 7, 9] >>> [x+1 for x in odds] [2, 4, 6, 8, 10] 2. >>> [x for x in odds if 25 % x == 0] [1, 5] 3 >>> digits = [1, 8, 2, 8] >>>[
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摘要：Q1 题目描述： 代码实现： 1 def coords(fn, seq, lower, upper): 2 """ 3 >>> seq = [-4, -2, 0, 1, 3] 4 >>> fn = lambda x: x**2 5 >>> coords(fn, seq, 1, 9) 6 [[-2,
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摘要：一：例题 例题1： 题目描述： #递归算法def g(n): """Return the value of G(n), computed recursively. >>> g(1) 1 >>> g(2) 2 >>> g(3) 3 >>> g(4) 10 >>> g(5) 22 >>> from co
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摘要：题目描述： 代码： 1 def num_eights(x): 2 """Returns the number of times 8 appears as a digit of x. 3 4 >>> num_eights(3) 5 0 6 >>> num_eights(8) 7 1 8 >>> num
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