手算开2、3次方的公式验证

最新推荐文章于 2025-01-06 13:39:51 发布

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最新推荐文章于 2025-01-06 13:39:51 发布

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原文链接：https://my.oschina.net/sqhua/blog/1137844

为什么80%的码农都做不了架构师？>>>

目前来看，以下算法开2次方能在百分之2以内的精度，开3次方能在百分之6以内的精度，而且误差主要出在20以内。200以上则是千分位误差

using System;

using System.Collections.Generic; using System.Linq; using System.Text;

namespace KaiFangTest { class Program { static void Main(string[] args) { for (int i = 4; i <= 100000; i++) { float i2 = get2(i); //i2 = (float)Math.Truncate((double)i210)/10; //i2 = (i2 + i / i2 ) / 2; float wuCha = Math.Abs(((i2 * i2) - i) / i); if (Math.Abs(wuCha) > 0.01) System.Console.WriteLine(i.ToString() + " " + i2.ToString("N2") + " " + (wuCha).ToString("N4")); } System.Console.WriteLine("********"); for (int i = 8; i <= 10000; i++) { float i3 = get3(i); //i3 = (float)Math.Truncate((double)i310)/10; //i3 = (2 * i3 + i / i3 / i3) / 3; float wuCha = Math.Abs((( i3* i3 * i3) - i) / i); if (wuCha > 0.1) System.Console.WriteLine(i.ToString() + " " + i3.ToString("N3") + " " + (wuCha).ToString("N4")); }

        //for (int i = 2; i < 100; i++)
        //    System.Console.WriteLine(i.ToString() + " " + find2(i));
    }

    static float get2(int t)
    {
        float a = find2(t);
        return (a + t / a) / 2;
    }

    static float find2(int t)
    {
        for (int i = 0; i < t; i++)
            if (i * i == t)
                return i;
            else 
                if (i * i > t)
            {
                int small = i - 1;
                int result =i;
                if ( (result * result - t) != Math.Min(result * result - t, t - small * small))
                    result = small;
                float misus = (i * i  - t) / (t - small * small );
                if (misus > 0.5 && misus < 2)
                    return (float)(small + 0.5);
                return result;
            }
        return 0;
    }

    static float get3(float t)
    {
        float a = find3(t);
        return (a*2 + t / a/a) / 3;
    }

    static float find3(float t)
    {
        for (int i = 0; i < t; i++)
            if (i * i*i == t)
                return i;
            else if (i * i*i > t)
            {
                int small = i - 1;
                int result = i;
                if ((result * result * result - t) != Math.Min(result * result * result - t, t - small* small * small))
                    result = small;
                float misus = (i*i*i-t)/(t - small* small * small);
                if (misus > 0.3 && misus < 3)
                    return (float)(small + 0.5);
                return result;
            }
        return 0;
    }
}

} `

转载于:https://my.oschina.net/sqhua/blog/1137844