如何在 C# 中计算整数的除法和模数?

发布于 2024-10-26 01:01:24 字数 26 浏览 4 评论 0原文

如何在 C# 中计算整数的除法和模数?

How can I calculate division and modulo for integer numbers in C#?

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徒留西风 2024-11-02 01:01:24

以下是 MSDN 文档 的答案。

当您将两个整数相除时,结果始终是一个整数。例如,7 / 3 的结果是 2。要确定 7 / 3 的余数,请使用余数运算符 (%)。

int a = 5;
int b = 3;

int div = a / b; //quotient is 1
int mod = a % b; //remainder is 2

Here's an answer from the MSDN documentation.

When you divide two integers, the result is always an integer. For example, the result of 7 / 3 is 2. To determine the remainder of 7 / 3, use the remainder operator (%).

int a = 5;
int b = 3;

int div = a / b; //quotient is 1
int mod = a % b; //remainder is 2
坏尐絯℡ 2024-11-02 01:01:24

还有 Math.DivRem

quotient = Math.DivRem(dividend, divisor, out remainder);

There is also Math.DivRem

quotient = Math.DivRem(dividend, divisor, out remainder);
初见你 2024-11-02 01:01:24

有趣的事实!

“取模”运算定义为:

a % n ==> a - (a/n) * n

参考:模数运算

所以你可以推出自己的,虽然它会比内置的 % 运算符慢得多:

public static int Mod(int a, int n)
{
    return a - (int)((double)a / n) * n;
}

编辑:哇,原来在这里说错了,感谢 @joren 抓住我

现在在这里我依赖于这样一个事实,即 C# 中的 除法 + 强制转换为 int 是等价于Math.Floor(即,它删除了分数),但“true”实现将类似于:

public static int Mod(int a, int n)
{
    return a - (int)Math.Floor((double)a / n) * n;
}

事实上,您可以通过以下方式看到%和“true modulus”之间的差异以下:

var modTest =
    from a in Enumerable.Range(-3, 6)
    from b in Enumerable.Range(-3, 6)
    where b != 0
    let op = (a % b)
    let mod = Mod(a,b)
    let areSame = op == mod
    select new 
    { 
        A = a,
        B = b,
        Operator = op, 
        Mod = mod, 
        Same = areSame
    };
Console.WriteLine("A      B     A%B   Mod(A,B)   Equal?");
Console.WriteLine("-----------------------------------");
foreach (var result in modTest)
{
    Console.WriteLine(
        "{0,-3} | {1,-3} | {2,-5} | {3,-10} | {4,-6}", 
        result.A,
        result.B,
        result.Operator, 
        result.Mod, 
        result.Same);
}

结果:

A      B     A%B   Mod(A,B)   Equal?
-----------------------------------
-3  | -3  | 0     | 0          | True  
-3  | -2  | -1    | -1         | True  
-3  | -1  | 0     | 0          | True  
-3  | 1   | 0     | 0          | True  
-3  | 2   | -1    | 1          | False 
-2  | -3  | -2    | -2         | True  
-2  | -2  | 0     | 0          | True  
-2  | -1  | 0     | 0          | True  
-2  | 1   | 0     | 0          | True  
-2  | 2   | 0     | 0          | True  
-1  | -3  | -1    | -1         | True  
-1  | -2  | -1    | -1         | True  
-1  | -1  | 0     | 0          | True  
-1  | 1   | 0     | 0          | True  
-1  | 2   | -1    | 1          | False 
0   | -3  | 0     | 0          | True  
0   | -2  | 0     | 0          | True  
0   | -1  | 0     | 0          | True  
0   | 1   | 0     | 0          | True  
0   | 2   | 0     | 0          | True  
1   | -3  | 1     | -2         | False 
1   | -2  | 1     | -1         | False 
1   | -1  | 0     | 0          | True  
1   | 1   | 0     | 0          | True  
1   | 2   | 1     | 1          | True  
2   | -3  | 2     | -1         | False 
2   | -2  | 0     | 0          | True  
2   | -1  | 0     | 0          | True  
2   | 1   | 0     | 0          | True  
2   | 2   | 0     | 0          | True  

Fun fact!

The 'modulus' operation is defined as:

a % n ==> a - (a/n) * n

Ref:Modular Arithmetic

So you could roll your own, although it will be FAR slower than the built in % operator:

public static int Mod(int a, int n)
{
    return a - (int)((double)a / n) * n;
}

Edit: wow, misspoke rather badly here originally, thanks @joren for catching me

Now here I'm relying on the fact that division + cast-to-int in C# is equivalent to Math.Floor (i.e., it drops the fraction), but a "true" implementation would instead be something like:

public static int Mod(int a, int n)
{
    return a - (int)Math.Floor((double)a / n) * n;
}

In fact, you can see the differences between % and "true modulus" with the following:

var modTest =
    from a in Enumerable.Range(-3, 6)
    from b in Enumerable.Range(-3, 6)
    where b != 0
    let op = (a % b)
    let mod = Mod(a,b)
    let areSame = op == mod
    select new 
    { 
        A = a,
        B = b,
        Operator = op, 
        Mod = mod, 
        Same = areSame
    };
Console.WriteLine("A      B     A%B   Mod(A,B)   Equal?");
Console.WriteLine("-----------------------------------");
foreach (var result in modTest)
{
    Console.WriteLine(
        "{0,-3} | {1,-3} | {2,-5} | {3,-10} | {4,-6}", 
        result.A,
        result.B,
        result.Operator, 
        result.Mod, 
        result.Same);
}

Results:

A      B     A%B   Mod(A,B)   Equal?
-----------------------------------
-3  | -3  | 0     | 0          | True  
-3  | -2  | -1    | -1         | True  
-3  | -1  | 0     | 0          | True  
-3  | 1   | 0     | 0          | True  
-3  | 2   | -1    | 1          | False 
-2  | -3  | -2    | -2         | True  
-2  | -2  | 0     | 0          | True  
-2  | -1  | 0     | 0          | True  
-2  | 1   | 0     | 0          | True  
-2  | 2   | 0     | 0          | True  
-1  | -3  | -1    | -1         | True  
-1  | -2  | -1    | -1         | True  
-1  | -1  | 0     | 0          | True  
-1  | 1   | 0     | 0          | True  
-1  | 2   | -1    | 1          | False 
0   | -3  | 0     | 0          | True  
0   | -2  | 0     | 0          | True  
0   | -1  | 0     | 0          | True  
0   | 1   | 0     | 0          | True  
0   | 2   | 0     | 0          | True  
1   | -3  | 1     | -2         | False 
1   | -2  | 1     | -1         | False 
1   | -1  | 0     | 0          | True  
1   | 1   | 0     | 0          | True  
1   | 2   | 1     | 1          | True  
2   | -3  | 2     | -1         | False 
2   | -2  | 0     | 0          | True  
2   | -1  | 0     | 0          | True  
2   | 1   | 0     | 0          | True  
2   | 2   | 0     | 0          | True  
乜一 2024-11-02 01:01:24

使用/运算符执行除法:

result = a / b;

使用%运算符执行除法:

result = a % b;

Division is performed using the / operator:

result = a / b;

Modulo division is done using the % operator:

result = a % b;
坠似风落 2024-11-02 01:01:24

余数:a%b
示例: 5 % 3 = 2

除法:

两个变量都是整数的情况: 5/3 = 1

小数和所需整数的情况: Math.Floor(5/3)

或 5/3 - (5%3)/3 如果您不想使用数学课

Remainder: a % b
example: 5 % 3 = 2

Divide:

case both variables are integer: 5/3 = 1

case decimal and neeeded integer: Math.Floor(5/3)

or 5/3 - (5%3)/3 if you dont want to use the Math Class

怪我鬧 2024-11-02 01:01:24

从用户处读取两个整数。然后计算/显示余数和商,

// When the larger integer is divided by the smaller integer
Console.WriteLine("Enter integer 1 please :");
double a5 = double.Parse(Console.ReadLine());
Console.WriteLine("Enter integer 2 please :");
double b5 = double.Parse(Console.ReadLine());

double div = a5 / b5;
Console.WriteLine(div);

double mod = a5 % b5;
Console.WriteLine(mod);

Console.ReadLine();

Read two integers from the user. Then compute/display the remainder and quotient,

// When the larger integer is divided by the smaller integer
Console.WriteLine("Enter integer 1 please :");
double a5 = double.Parse(Console.ReadLine());
Console.WriteLine("Enter integer 2 please :");
double b5 = double.Parse(Console.ReadLine());

double div = a5 / b5;
Console.WriteLine(div);

double mod = a5 % b5;
Console.WriteLine(mod);

Console.ReadLine();
~没有更多了~
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