什么是结束我的for循环的适当条件?

时间:2016-02-19 06:29:37

标签: c# for-loop

我需要在for循环中添加以下间隔(我尝试使用if语句,老师让我减少编码量。)

Intervals (Pizza Diameters)   
12 - 20 = 8 slices  
21 - 24 = 12 slices   
25 - 30 = 16 slices  
31 - 36 = 24 slices

以下代码为我提供了非常接近我需要的输出,除了它不符合上述条件。例如,如果我输入24,它应该只给出8和12个切片的输出。

int[] pizzaSlices = new int[4] { 8, 12, 16, 24 };

for (int i = pizzaSlices[0]; i < inputDiameter; i++) {
    if (i == pizzaSlices[0] || i == pizzaSlices[1] || i == pizzaSlices[2] ||
        i == pizzaSlices[3]) {

       Console.WriteLine("cut in " + i + " slices results in a slice area of " +
       Math.Round(areaOfThePizza / i, 2) + " per slices");

    }
}

当前输出:

期望的输出:

2 个答案:

答案 0 :(得分:2)

我无法抗拒添加一些LINQ'y解决方案:

[Test]
[TestCase(37, Result = 0)]
[TestCase(36, Result = 24)]
[TestCase(35, Result = 24)]
[TestCase(30, Result = 16)]
[TestCase(29, Result = 16)]
[TestCase(26, Result = 16)]
[TestCase(22, Result = 12)]
[TestCase(12, Result = 8)]
[TestCase(11, Result = 0)]
[TestCase(10, Result = 0)]
public int GetNumberOfSlices(int diameter)
{
    var pizzas = new[] {
        new[] { 11, 0 }, 
        new[] { 20, 8 }, 
        new[] { 24, 12 }, 
        new[] { 30, 16 }, 
        new[] { 36, 24 }
    };
    var pizza = pizzas.FirstOrDefault(p => diameter <= p[0]);

    return pizza == null ? 0 : pizza[1];
}

答案 1 :(得分:1)

我添加了另一个int数组来保存披萨直径。 for循环从0迭代到pizzaSlices的长度,因此03

if循环中的for语句检查用户输入的直径是否在当前迭代的pizzaDims对应范围内。

int[] pizzaSlices = new int[4] { 8, 12, 16, 24 };
int[] pizzaDims = new int[4] { 12, 21, 25, 31 }; // pizza diameters array

if(inputDiameter >= smallest) {
    for (int i = 0; i <= pizzaSlices.Length; i++) {
        if(inputDiameter >= pizzaDims[i] {
            Console.WriteLine("cut in " + pizzaSlices[i] + 
                              " slices results in a slice area of " +
                              Math.Round(areaOfThePizza / pizzaSlices[i], 2) +
                              " per slices");
        } else {
            break; // no need to continue iterating, if one condition is false then
                   // the rest will be as well
        }
    }
}