pset2 substitution.c
1.extension：To Do Tasks
- 推荐一个vscode里面一个很好用的插件！！！写出解决的步骤，不但理清楚思路。还可以提高效率！
- 特别是针对一些文本比较长的pset，要求多且零碎，好像在做阅读理解>< 而且反复切换页面（浏览器到vscode）就有点太累了，而且截图固定的话，整个页面也看着不舒服，太乱了，所以这个插件可以节省时间。
- 把pset页面的的长文本要求先复制到创建好的new.todo 里面，快速进行删减，直到看到关键信息，然后分段小标题head1…,每个小标题下是需要一些小任务：就初步成形~
- 在初步成形的小任务后面可以写出解决的步骤（或者你不确定也没有关系），有一些必要的步骤可以先开始试试看，一步一步的，每次做完@done都有一个时间，你会有慢慢的有成就感，所以说，把可以解决的先解决，最后自然就出来了。
- 其实写的时间不长，但是中间小错误需要调试的点很多，有时候都不知道在哪，所以说不要太盯着结果了，耐心的去找过程，这个时候找到错误的速度就会出奇的快！Unbelievable!好像有点理解过程比结果更重要了><
使用方法：首先的话，是先创建new.todo文件，然后就是一些快捷键，如下图，还是比较好上手哒！
pset2例子：
2 code
#include <cs50.h> //for get_sting
#include <ctype.h> //for tolower
#include <stdio.h> //for printf
#include <stdlib.h> //for malloc
#include <string.h> //for strlen
int main(int argc, string argv[])
{// 1 command-line argumentif (argc != 2){printf("./substitution key\n");return 1;}/*2 check if valid keyget the user input:argv[1]* contain 26 digits of characters purely :isapha()* no matter is uppercase or lowercase* A -Z:65-90 a-z:97-122* a letter in the key can only show once:frequency(every letter) == 1* if not ,printf message and return 1*/size_t length = strlen(argv[1]);if (length != 26){printf("Key need to contain 26 letters!\n");return 1;}int *frequency = (int *) calloc(length, sizeof(int));if (frequency == NULL) {printf("failed to allocate memory");return 1;}for (int i = 0; i < length; i++){if (!isalpha(argv[1][i])){printf("Key must be purely alphabetically!\n");return 1;}argv[1][i] = tolower(argv[1][i]);char ch = argv[1][i];frequency[(int) ch - 97]++;// printf("count: %d\n", frequency[(int)ch - 97]);if (frequency[(int) ch - 97] != 1){printf("The letters in key can not repeated!\n");return 1;}}// store the keystring key = malloc(length + 1);if (key == NULL){printf("failed to allocate memory\n");return 1;}for (int k = 0; k < length; k++){key[k] = argv[1][k];}// 3 get the plaintextstring s = get_string("Plaintext: ");// printf("plaintext: %s\n", s);/*4 output text* calculate the length of s* allocate new memory* convert new[]:This 26-character key means thatA (the first letter of the alphabet)should be converted into N (the first character of the key)...key:NQXPOMAFTRHLZGECYJIUWSKDVB key[position] = 'H'-> positionalp:ABCDEFGHIJKLMNOPQRSTUVWXYZexample:HELLO --> FOLLE new[i] = key[int(s[i]) - 97]tolower-alpha:A B C D E F G H I J K L M N O P Q R S T U V W X Y Z-97index:0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25*/size_t s_length = strlen(s);string new = malloc(strlen(s) + 1);if (new == NULL){printf("failed to allocate memory");return 1;}printf("ciphertext:");for (int j = 0; j < s_length; j++){// keep the formatif (isupper(s[j])){new[j] = toupper(key[s[j] - 65]);}else if (islower(s[j])){new[j] = tolower(key[s[j] - 97]);}else{new[j] = s[j];}printf("%c", new[j]);}printf("\n");free(frequency);free(key);free(new);return 0;
}3 check:
:) substitution.c exists
:) substitution.c compiles
:) encrypts "A" as "Z" using ZYXWVUTSRQPONMLKJIHGFEDCBA as key
:) encrypts "a" as "z" using ZYXWVUTSRQPONMLKJIHGFEDCBA as key
:) encrypts "ABC" as "NJQ" using NJQSUYBRXMOPFTHZVAWCGILKED as key
:) encrypts "XyZ" as "KeD" using NJQSUYBRXMOPFTHZVAWCGILKED as key
:) encrypts "This is CS50" as "Cbah ah KH50" using YUKFRNLBAVMWZTEOGXHCIPJSQD as key
:) encrypts "This is CS50" as "Cbah ah KH50" using yukfrnlbavmwzteogxhcipjsqd as key
:) encrypts "This is CS50" as "Cbah ah KH50" using YUKFRNLBAVMWZteogxhcipjsqd as key
:) encrypts all alphabetic characters using DWUSXNPQKEGCZFJBTLYROHIAVM as key
:) does not encrypt non-alphabetical characters using DWUSXNPQKEGCZFJBTLYROHIAVM as key
:) handles lack of key
:) handles too many arguments
:) handles invalid key length
:) handles invalid characters in key
:) handles duplicate characters in uppercase key
:) handles duplicate characters in lowercase key
:) handles multiple duplicate characters in key