1^=1 2^=1+3 3^=1+3+5 编程判断一个数是否是平方数

Question

using System;namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args) { Console.WriteLine(foo(1)); }        static bool foo(int input)
        {
            if (input < 0)
            {
                return false;
            }            var i = (int)Math.Sqrt(input);            return i * i == input;
        }
    }
}

Answer 1

using System;
using System.Collections.Generic;namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args) { Console.WriteLine(foo(1)); }        static bool foo(uint input)
        {
            foreach (var i in getList(input))
            {
                if (i > input)
                {
                    return false;
                }                if (i == input)
                {
                    return true;
                }
            }
            return false;
        }        static IEnumerable<uint> getList(uint input)
        {
            uint sum = 1;
            uint i = 1;            while (true)
            {
                yield return sum;
                i += 2;
                sum += i;
            }
        }
    }
}

Answer 2

(（1+（2n-1））/2)^2=1+3+5+............+2n-1
反过来写吧
1+3+5+............+ (2(n -1) -1)  + 2n-1
= 1+3+5+............+ (2n -3)  + (2n-1)
头尾两数为一组相加都等于2n，且不难看出数列共有n个数，共能组成多少队？
= 2n * (n/2)
= n^;
证毕

Answer 3

正着写也一样啊
(（1+（2n-1））/2)^2
不就是n*n吗