循环和模数

Question

我正在学习C,我很难理解循环和模数的使用.我知道循环用于缩短程序,Modulo用于排除剩余部分.我的任务是"编写一个C程序来查找单个正整数的总和".

我花了几个小时试图理解这个问题.我也做过实验.

    int  n,d=0,s=0;
    printf("\nEnter a number\n\n");
    scanf("%d",&n);


while(n>0)
     { 
           d = n%10;
       s = s+d;
       n = n/10;
     }

printf("\n sum of the individual digits = %d",s);

我的问题是:

任何人都可以帮我理解这个程序的流程吗？为什么要使用Modulo？为什么有= n/10

我做过的经历:

当我删除d = n%10; 输出行打印出数字seperatley.因此它不算.

即123 = 6 - >它给了我136

当我删除行n = n/10时它没有显示输出.printf语句有一个参数's'

提前致谢!

Answer 1

以在模数d = n % 10使得d等于的最后一位n在底座10 n = n / 10去除的最后一位数字n.

模数基本上是剩余的,所以我们说n = 123.然后n / 10是12和n % 10是3.

删除每次循环运行之间不会改变的n = n / 10平均值n,因此循环条件n > 0始终为true,因此循环一直持续到手动终止程序为止.

这是一个程序的跟踪n = 123.最初d并且s都是零.

 while (n > 0) {  // n is 123, which is greater than zero, so we enter the loop
     d = n % 10;  // 123 % 10 is 3, so d is now 3
     s = s + d;   // 0 + 3 is 3, so s is now 3
     n = n / 10;  // 123 / 10 is 12, so n is now 12.
 }                // go back to the top of the loop
 while (n > 0) {  // n is 12, which is still greater than zero
     d = n % 10;  // 12 % 10 is 2, so d is now 2
     s = s + d    // 3 + 2 is 5, so s is now 5
     n = n / 10;  // 12 / 10 is 1, so n is now 1
 }                // go back to the top again
 while (n > 0) {  // n is 1, which is still greater than zero
     d = n % 10;  // 1 % 10 is 1, so d is now 1
     s = s + d;   // 5 + 1 is 6, so s is now 6
     n = n / 10;  // 1 / 10 is 0, so n is now 0
 }                // go back to the top
 while (n > 0) {  // n is 0, which is not greater than zero, so we skip
                  // to after the loop body
 printf("\n sum of the individual digits = %d",s);