第五次作业 - 卡希勒

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task1：

 1 #include <stdio.h>
 2 #define N 5
 3 
 4 void input(int x[], int n);
 5 void output(int x[], int n);
 6 void find_min_max(int x[], int n, int *pmin, int *pmax);
 7 
 8 int main() {
 9 int a[N];
10 int min, max;
11 
12 printf("录入%d个数据:\n", N);
13 input(a, N);
14 
15 printf("数据是: \n");
16 output(a, N);
17 
18 printf("数据处理...\n");
19 find_min_max(a, N, &min, &max);
20 
21 printf("输出结果:\n");
22 printf("min = %d, max = %d\n", min, max);
23 
24 return 0;
25 }
26 
27 void input(int x[], int n) {
28     int i;
29 
30     for(i = 0; i < n; ++i)
31         scanf("%d", &x[i]);
32 }
33 
34 void output(int x[], int n) {
35     int i;
36     
37     for(i = 0; i < n; ++i)
38         printf("%d ", x[i]);
39     printf("\n");
40 }
41 
42 void find_min_max(int x[], int n, int *pmin, int *pmax) {
43     int i;
44     
45     *pmin = *pmax = x[0];
46     
47     for(i = 0; i < n; ++i)
48         if(x[i] < *pmin)
49             *pmin = x[i];
50         else if(x[i] > *pmax)
51             *pmax = x[i];
52 }

task1_2

 1 #include <stdio.h>
 2 #define N 5
 3 
 4 void input(int x[], int n);
 5 void output(int x[], int n);
 6 int *find_max(int x[], int n);
 7 
 8 int main() {
 9     int a[N];
10     int *pmax;
11 
12     printf("录入%d个数据:\n", N);
13     input(a, N);
14     
15     printf("数据是: \n");
16     output(a, N);
17     
18     printf("数据处理...\n");
19     pmax = find_max(a, N);
20     
21     printf("输出结果:\n");
22     printf("max = %d\n", *pmax);
23 
24     return 0;
25 }
26 
27 void input(int x[], int n) {
28     int i;
29     for(i = 0; i < n; ++i)
30         scanf("%d", &x[i]);
31 }
32 
33 void output(int x[], int n) {
34     int i;
35     for(i = 0; i < n; ++i)
36         printf("%d ", x[i]);
37     printf("\n");
38 }
39 
40 int *find_max(int x[], int n) {
41     int max_index = 0;
42     int i;
43 
44     for(i = 0; i < n; ++i)
45         if(x[i] > x[max_index])
46             max_index = i;
47 
48     return &x[max_index];
49 }

task2_1

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define N 80
 4 
 5 int main() {
 6 char s1[N] = "Learning makes me happy";
 7 char s2[N] = "Learning makes me sleepy";
 8 char tmp[N];
 9 
10 printf("sizeof(s1) vs. strlen(s1): \n");
11 printf("sizeof(s1) = %d\n", sizeof(s1));
12 printf("strlen(s1) = %d\n", strlen(s1));
13 
14 printf("\nbefore swap: \n");
15 printf("s1: %s\n", s1);
16 printf("s2: %s\n", s2);
17 
18 printf("\nswapping...\n");
19 strcpy(tmp, s1);
20 strcpy(s1, s2);
21 strcpy(s2, tmp);
22 
23 printf("\nafter swap: \n");
24 printf("s1: %s\n", s1);
25 printf("s2: %s\n", s2);
26 
27 return 0;
28 }

task2_2

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define N 80
 4 int main() {
 5 char *s1 = "Learning makes me happy";
 6 char *s2 = "Learning makes me sleepy";
 7 char *tmp;
 8 
 9 printf("sizeof(s1) vs. strlen(s1): \n");
10 printf("sizeof(s1) = %d\n", sizeof(s1));
11 
12 printf("strlen(s1) = %d\n", strlen(s1));
13 printf("\nbefore swap: \n");
14 printf("s1: %s\n", s1);
15 printf("s2: %s\n", s2);
16 
17 printf("\nswapping...\n");
18 tmp = s1;
19 s1 = s2;
20 s2 = tmp;
21 
22 printf("\nafter swap: \n");
23 printf("s1: %s\n", s1);
24 printf("s2: %s\n", s2);
25 
26 return 0;
27 }

task3

 1 #include <stdio.h>
 2 int main() {
 3     int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
 4     int i, j;
 5     int *ptr1; // 指针变量，存放int类型数据的地址
 6     int(*ptr2)[4]; // 指针变量，指向包含4个int元素的一维数组
 7 
 8     printf("输出1: 使用数组名、下标直接访问二维数组元素\n");
 9     for (i = 0; i < 2; ++i) {
10         for (j = 0; j < 4; ++j)
11             printf("%d ", x[i][j]);
12         printf("\n");
13     }
14     
15     printf("\n输出2: 使用指针变量ptr1(指向元素)间接访问\n");
16     for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ++ptr1, ++i) {
17         printf("%d ", *ptr1);
18         
19         if ((i + 1) % 4 == 0)
20             printf("\n");
21     }
22     
23     printf("\n输出3: 使用指针变量ptr2(指向一维数组)间接访问\n");
24     for (ptr2 = x; ptr2 < x + 2; ++ptr2) {
25         for (j = 0; j < 4; ++j)
26             printf("%d ", *(*ptr2 + j));
27         printf("\n");
28     }
29     
30     return 0;
31 }

task4

 1 #include <stdio.h>
 2 #define N 80
 3 
 4 void replace(char *str, char old_char, char new_char); // 函数声明
 5 
 6 int main() {
 7     char text[N] = "Programming is difficult or not, it is a question.";
 8 
 9     printf("原始文本: \n");
10     printf("%s\n", text);
11 
12     replace(text, 'i', '*'); // 函数调用 注意字符形参写法，单引号不能少
13     printf("处理后文本: \n");
14     printf("%s\n", text);
15 
16     return 0;
17 }
18 // 函数定义
19 void replace(char *str, char old_char, char new_char) {
20     int i;
21 
22     while(*str) {
23         if(*str == old_char)
24             *str = new_char;
25         str++;
26     }
27 }

task5

 1 #include <stdio.h>
 2 #define N 80
 3 
 4 char *str_trunc(char *str, char x);
 5 
 6 int main() {
 7     char str[N];
 8     char ch;
 9 
10     while(printf("输入字符串: "), gets(str) != NULL) {
11         printf("输入一个字符: ");
12         ch = getchar();
13         
14         printf("截断处理...\n");
15         str_trunc(str, ch); // 函数调用
16         
17         printf("截断处理后的字符串: %s\n\n", str);
18         getchar();
19     }
20     
21     return 0;
22 }
23 
24 char *str_trunc(char *str, char x){
25     int i;
26     
27     for(i=0;str[i]!='\0'; i++){
28         if(str[i]==x){
29             str[i]=0;
30             break;
31         }
32     }
33     return str;
34     
35 }

task6

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define N 5
 4 
 5 int check_id(char *str); // 函数声明
 6 
 7 int main()
 8 {
 9     char *pid[N] = {"31010120000721656X",
10     "3301061996X0203301",
11     "53010220051126571",
12     "510104199211197977",
13     "53010220051126133Y"};
14     int i;
15     
16     for (i = 0; i < N; ++i)
17         if (check_id(pid[i])) // 函数调用
18             printf("%s\tTrue\n", pid[i]);
19         else
20             printf("%s\tFalse\n", pid[i]);
21     
22     return 0;
23 }
24 
25 
26 int check_id(char *str) {
27     char *p = str;
28     int length = 0;
29     
30     while(*p){
31         length++;
32         p++;
33     }
34     if(length != 18)
35         return 0;
36         
37     for(p=str;p < str + 17; p++){
38         if(*p < '0' || *p > '9')
39             return 0;
40     }
41     
42     return (*p == 'X') || (*p >= '0' && *p <= '9');
43 }

task7

 1 #include <stdio.h>
 2 #define N 80
 3 
 4 void encoder(char *str, int n); // 函数声明
 5 void decoder(char *str, int n); // 函数声明
 6 
 7 int main() {
 8     char words[N];
 9     int n;
10     
11 while (1) {
12     printf("输入英文文本: ");
13     if (gets(words) == NULL) {
14         break; 
15     }
16     
17     printf("输入n: ");
18     if (scanf("%d", &n) != 1) {
19         break; 
20     }
21     while (getchar() != '\n');
22     
23     printf("编码后的英文文本: ");
24     encoder(words, n); // 函数调用
25     printf("%s\n", words);
26     
27     printf("对编码后的英文文本解码: ");
28     decoder(words, n); // 函数调用
29     printf("%s\n\n", words);
30     }
31     
32     return 0;
33 }
34 
35 void encoder(char *str, int n) {
36     int i;
37     n%=26;
38     
39     for (i = 0; str[i] != '\0'; i++) {
40         if (str[i] >= 'A' && str[i] <= 'Z') {
41             str[i] = 'A' + (str[i] - 'A' + n + 26) % 26;
42         } 
43         else if (str[i] >= 'a' && str[i] <= 'z') {
44             str[i] = 'a' + (str[i] - 'a' + n + 26) % 26;
45         }
46     }
47     
48 }
49 
50 void decoder(char *str, int n) {
51     int i;
52     n %= 26;
53     
54     for (i = 0; str[i] != '\0'; i++) {
55         if (str[i] >= 'A' && str[i] <= 'Z') {
56             str[i] = 'A' + (str[i] - 'A' - n + 26) % 26;
57         } else if (str[i] >= 'a' && str[i] <= 'z') {
58             str[i] = 'a' + (str[i] - 'a' - n + 26) % 26;
59         }
60     }
61 }

task8

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <string.h>
 4 
 5 int compare(const void *a, const void *b) {
 6     return strcmp(*(const char**)a, *(const char**)b);
 7 }
 8 
 9 int main(int argc, char *argv[]) {
10     int i;
11     
12     if (argc < 2) return 0;
13 
14     char **names = malloc((argc-1) * sizeof(char*));
15     
16     for (i = 1; i < argc; ++i)
17         names[i-1] = argv[i];
18     
19     qsort(names, argc-1, sizeof(char*), compare);
20 
21     for (i = 0; i < argc-1; ++i)
22         printf("hello, %s\n", names[i]);
23 
24     free(names);
25     return 0;
26 }

posted on 2025-05-19 03:00 卡希勒阅读(3) 评论(0) 收藏举报

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