求一算法

Question

如题：
求一算法来代码如下计算方法switch (i) {
                            case 21:
                                num = 1;
                                break;
                            case 20:
                                num = 2;
                                break;
                            case 19:
                                num = 4;
                                break;
                            case 18:
                                num = 8;
                                break;
                            case 17:
                                num = 16;
                                break;
                            case 16:
                                num = 32;
                                break;
                            case 15:
                                num = 64;
                                break;
                            case 14:
                                num = 128;
                                break;
                            case 13:
                                num = 256;
                                break;
                            case 12:
                                num = 512;
                                break;
                            case 11:
                                num = 1024;
                                break;
                            case 10:
                                num = 2048;
                                break;
                            case 9:
                                num = 4096;
                                break;
                            case 8:
                                num = 8192;
                                break;
                            case 7:
                                num = 16384;
                                break;
                            case 6:
                                num = 32768;
                                break;
                            case 5:
                                num = 65536;
                                break;
                            case 4:
                                num = 131072;
                                break;
                            case 3:
                                num = 262144;
                                break;
                            case 2:
                                num = 524288;
                                break;
                            case 1:
                                num = 1048576;
                                break;
                            default:
                                num = 1;
                                break;                        }

Answer 1

i=21-->2的0次方
i=20-->2的1次方
..............
i=1-->2的20次方else 2的0次方

Answer 2

num = Convert.ToInt32(Math.Pow((double)(i - 21), (double)2));

Answer 3

其实这个规律就是2的倍数。我的I是一个变量来的。如果i=23的话。那就是23时，num=1,22时，num=2,21时 NUM=4.........

Answer 4

double d = (i > 0 && i < 22) ? Math.Pow(21 - i, 2) : 1;

Answer 5

            int i = 2;
            double result = 0;
            if (i >= 1 && i <= 21)
            {
                result = Math.Pow(2, 21 - i);
            }
            else
            {
                result = 1;
            }

Answer 6

虽然写出来了，但是说实话，你这么判断没有什么意义，因为你是在程序中对i进行判断，因此程序只会走case i这一步，按你所说，那么，返回值只会是1.

Answer 7

   &  用这个符号计算
    

Answer 8

temp = 1 << ((21 - i) > 0 ? (21 - i) : 0);
可以再优化优化，在上班没有办法多想，见谅。

Answer 9

for(int i=21;i>=0;i--)
{
   console.writeline("此时i的值为:{0},num为：{1}",i,2^(21-i));
}

Answer 10

这样应该容易明白点
static void Main(string[] args)
        {
            for (int i = 21; i > 0; i--)
            {
                Console.WriteLine(Test(i));
            }
            Console.Read();
        }        static int Test(int i)
        {
            int temp = 21 - i;
            int result = 1;
            while (temp > 0)
            {
                result = result * 2;
                temp--;
            }
            return result;
        }

Answer 11

if(i<=20&&i>0)
{
   num=2;    
   for(int j=1;j<(21-i);j++)
   {
       num=num*2;
   }
}
else
    num=1;
这个算法可以，试过的。