数字的总和

本文介绍了数字的总和的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

int num，sum = 0 ，r; Console.WriteLine（ 输入数字：）; num = int .Parse（Console.ReadLine（））; while （num！= 0 ） {r = num ％ 10 ; num = num / 10 ; sum = sum + r; }

让我们拿num = 123 r = 123％10 = 12.3，然后如何计算其他两行，我没有得到，你能解释我吗，我是programmimg的新手 我尝试了什么： i尝试了相同的代码并输出正确，但我无法理解公式如何工作

解决方案

实际％ = mod 123％10的结果为3，因为mod等于剩余的。 num 是一个int。所以不能保存任何小数位 int / int的结果总是一个int（四舍五入的四舍五入刚刚丢失）

所以123/10 = 12

所以这样做是取最后一位数并添加直到没有剩余数字为止：

sum = 0; num = 123;

pass1：

r = 123％10 = 3 num = 123/10 = 12 sum = 3

pass2： r = 12％10 = 2 num = 123/10 = 1 sum = 5

pass3：

r = 1％10 = 1 num = 123/10 = 0 sum = 6

pass4： 终止为num = 0 我希望有帮助 Andy ^ _ ^

不，模数运算符不会像那样工作。 让我们拿num = 123 r = 123％10 = 12.3，

不对：

r = 123％10 = 3

模数运算符'％'返回除法的余数，执行123/10 == 12余数3： 如果是模数10，它返回最低有效数字，因为输入数字是十进制数。

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int num, sum = 0, r; Console.WriteLine("Enter a Number : "); num = int.Parse(Console.ReadLine()); while (num != 0) { r = num % 10; num = num / 10; sum = sum + r; }

let we take num = 123 r=123%10=12.3, then how is it calculate other two line, i didn't get, can you explain me,, i am new in programmimg What I have tried: i tried the same code and output got correct, but i am unable to understand the how the formula works

解决方案

Actually % = mod the result of 123 % 10 is 3 as mod equals the remainder only. num is an int. and so can't hold any decimal places the result of int / int is always an int (rounded down as the decimal place is just lost)

so 123 / 10 = 12

so what this does is take the last digit and add it to the total until there are no digits left:

sum = 0; num = 123;

pass1:

r = 123%10 = 3 num = 123/10 = 12 sum = 3

pass2:

r = 12%10 = 2 num = 123/10 = 1 sum = 5

pass3:

r = 1%10 = 1 num = 123/10 = 0 sum = 6

pass4: terminated as num = 0 I hope that helps Andy ^_^

No, the modulus operator doesn;t work like that.

let we take num = 123 r=123%10=12.3,

Is not right:

r = 123 % 10 = 3

The Modulus operator '%' returns the remainder of the division, do 123 / 10 == 12 remainder 3: In the case of modulus 10, it returns the least significant digit, since the input number is in base ten.

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