1 /*只需要在原程序中的/前后各加入一个空格即可 */
 2 /* Adds two fractions */
 3 
 4 #include <stdio.h>
 5 
 6 int main(void)
 7 {
 8     int num1, denom1, num2, denom2, result_num, result_denom;
 9 
10     printf("Enter first fraction: ");
11     scanf("%d / %d", &num1, &denom1);     /*至于要在原程序中的/前后各加入一个空格即可 */ 
12 
13     printf("Enter second fraction: ");
14     scanf("%d / %d", &num2, &denom2);
15 
16     result_num = num1 * denom2 + num2 * denom1;
17     result_denom = denom1 * denom2;
18     printf("The sum is %d/%d\n", result_num, result_denom);
19 
20     return 0;
21 }

 1 /*月日年转成年月日*/
 2 #include <stdio.h>
 3 
 4 int main(void)
 5 {
 6     int day = 0, month = 0, year = 0;
 7 
 8     printf("Enter a date (mm/dd/yyyy): ");
 9     scanf("%d/%d/%d", &month, &day, &year);
10 
11     printf("You entered the date %.2d%.2d%d", year, month, day);   /*月和日用%.2d显示，保证月份和日期会以两位数形式显示 */
12 
13     return 0;
14 }

 1 /*输入信息并显示，产品编号和日期左对齐，价格右对齐*/
 2 
 3 #include <stdio.h>
 4 
 5 int main(void)
 6 {
 7     /* 声明item number变量item_num，unit price变量price，日期变量day、month和year */
 8     int item_num = 0, day = 0, month = 0, year = 0;
 9     /* unit price包含小数点，所以其变量应为float型 */
10     float price = 1.0f;
11 
12     /* 用户输入数据，程序录入对应变量中 */
13     printf("Enter item number: ");
14     scanf("%d", &item_num);
15     printf("Enter unit price: ");
16     scanf("%f", &price);
17     printf("Enter purchase date (mm/dd/yyyy): ");
18     scanf("%d/%d/%d", &month, &day, &year);
19 
20     printf("Item\t\tUnit\t\tPurchase\n");   /* 第一行显示，中间通过两个制表符\t分隔 */
21     printf("\t\tprice\t\tDate\n");          /* 第二行显示，中间通过两个制表符\t分隔 */
22     /* 编号和日期是左对齐，所以要在前面加上负号。价格采用%7.2f显示，是因为最大价格是9999.99共七位，在小于七位时用空格占位，.2则是保证小数点后只显示2位数字。 */
23     /* 日期中的.2是为了让最少以两位数形式显示，年份为四位数，所以采用.4保证最少显示四位数 */
24     printf("%-d\t\t$%7.2f\t%-.2d/%-.2d/%-.4d\n", item_num, price, month, day, year);
25 
26     return 0;
27 }

 1 /*ISBN分解*/
 2 
 3 #include <stdio.h>
 4 
 5 int main(void)
 6 {
 7     /* 第一组数字的变量定义为gs1，第二组数字的变量定义为group_id，第三组数字的变量定义为pub_code，第四组数字的变量定义为item_num，校验位定义为check_d */
 8     int gs1 = 0, group_id = 0, pub_code = 0, item_num = 0, check_d = 0;
 9 
10     printf("Enter ISBN: ");
11     /* 每组数字看成一个整体，通过-来识别每组数字，并将其存入对应的变量中，这样每组数字个数即使发生变化也能正确的显示出来 */
12     scanf("%d-%d-%d-%d-%d", &gs1, &group_id, &pub_code, &item_num, &check_d);
13 
14     /* 直接将每组变量分别显示出来即可 */
15     printf("GS1 prefix: %d\nGroup identifier: %d\nPublish code: %d\nItem number: %d\nCheck digit: %d\n", gs1, group_id, pub_code, item_num, check_d);
16 
17     return 0;
18 }

 1 /*输入电话号码(xxx) xxx-xxxx并转换格式显示xxx.xxx.xxxx */
 2 
 3 #include <stdio.h>
 4 
 5 int main(void)
 6 {
 7     /* 将电话号码拆分成3组，分别定义成num1、num2和num3 */
 8     int num1 = 0, num2 = 0, num3 = 0;
 9 
10     printf("Enter phone number [(xxx) xxx-xxxx]: ");
11     /* 根据提示的格式分别将3组数字存入变量中 */
12     scanf("(%d) %d-%d", &num1, &num2, &num3);
13 
14     printf("You entered: %d.%d.%d\n", num1, num2, num3);
15 
16     return 0;
17 }

 1 /*输入1到16所有整数，以4*4显示，并计算每行、每列、和对角线的和 */
 2 
 3 #include <stdio.h>
 4 
 5 int main(void)
 6 {
 7     /* 定义十六个数字的变量，从num1到num16 */
 8     int num1 = 0, num2 = 0, num3 = 0, num4 = 0, num5 = 0, num6 = 0, num7 = 0, num8 = 0;
 9     /* 拆成两行定义是因为写在一行里实在是太长了，不方便阅读 */
10     int num9 = 0, num10 = 0, num11 = 0, num12 = 0, num13 = 0, num14 = 0, num15 = 0, num16 = 0;
11     /* 定义4×4矩阵的每行和的变量，从row1到row4 */
12     int row1 = 0, row2 = 0, row3 = 0, row4 = 0;
13     /* 定义4×4矩阵的每列和的变量，从col1到col4 */
14     int col1 = 0, col2 = 0, col3 = 0, col4 = 0;
15     /* 定义两个对角线和的变量dia1和dia2 */
16     int dia1 = 0, dia2 = 0;
17 
18     printf("Enter the numbers from 1 to 16 in any order: ");
19     /* 将十六个数字分别存入到16个变量里 */
20     scanf("%d%d%d%d%d%d%d%d", &num1, &num2, &num3, &num4, &num5, &num6, &num7, &num8);
21     /* 同样因为太长了，所以拆成两行写 */
22     scanf("%d%d%d%d%d%d%d%d", &num9, &num10, &num11, &num12, &num13, &num14, &num15, &num16);
23 
24     /* 显示出矩阵，根据题目示例，每个数字应该保证两个字符宽，右对齐。通过制表符来使每行和每列对齐 */
25     printf("%2d\t%2d\t%2d\t%2d\t\n%2d\t%2d\t%2d\t%2d\t\n", num1, num2, num3, num4, num5, num6, num7, num8);
26     printf("%2d\t%2d\t%2d\t%2d\t\n%2d\t%2d\t%2d\t%2d\t\n\n\n", num9, num10, num11, num12, num13, num14, num15, num16);
27 
28     /* 计算每行的和 */
29     row1 = num1 + num2 + num3 + num4;
30     row2 = num5 + num6 + num7 + num8;
31     row3 = num9 + num10 + num11 + num12;
32     row4 = num13 + num14 + num15 + num16;
33 
34     /* 计算每列的和 */
35     col1 = num1 + num5 + num9 + num13;
36     col2 = num2 + num6 + num10 + num14;
37     col3 = num3 + num7 + num11 + num15;
38     col4 = num4 + num8 + num12 + num16;
39 
40     /* 计算两个对角线的和 */
41     dia1 = num1 + num6 + num11 + num16;
42     dia2 = num4 + num7 + num10 + num13;
43 
44     /* 显示结果 */
45     printf("Row sums: %d  %d  %d  %d  \nColumn sums: %d  %d  %d  %d  \n", row1, row2, row3, row4, col1, col2, col3, col4);
46     printf("Diagonal sums: %d  %d  \n", dia1, dia2);
47 
48     return 0;
49 }

 1 /*同时输入两个分数并计算*/
 2 
 3 #include <stdio.h>
 4 
 5 int main(void)
 6 {
 7     /* 定义两个分数的变量，分子分别是x1和x2，分母是y1和y2 */
 8     int x1 = 0, y1 = 0, x2 = 0, y2 = 0;
 9     /* 计算出的分子用result_num变量表示，分母用result_denom变量表示 */
10     int result_num = 0, result_denom = 0;
11 
12     printf("Enter two fractions separated by a plus sign: ");
13     /* 读取两个分数 */
14     scanf("%d/%d+%d/%d", &x1, &y1, &x2, &y2);
15 
16     /* 计算两个分数的和 */
17     result_num = x1 * y2 + x2 * y1;
18     result_denom = y1 * y2;
19     printf("The sum is: %d/%d\n", result_num, result_denom);
20 
21     return 0;
22 }