【ProjectEuler】ProjectEuler_016

// 2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
// 
// What is the sum of the digits of the number 2^1000?
using System;
using System.Collections.Generic;
using System.Text;

namespace projecteuler016
{
    class Program
    {
        /// <summary>
        /// 每个数字在2^n中的个数
        /// </summary>
        static int[] x = new int[10];

        static void Main(string[] args)
        {
            F1();
        }

        private static void F1()
        {
            Console.WriteLine(new System.Diagnostics.StackTrace().GetFrame(0).GetMethod());
            DateTime timeStart = DateTime.Now;

         /*
            for (int i = 0; i < 100; i++)
                     {
                         long p = (long)Math.Pow(2, i);
                         Console.WriteLine(i + ":" + p + ":" + digitTotal(p) + ":" + y(p));
                     }*/
           /*
            for (int i = 1; i < 100; i++)
                       {
                           Console.WriteLine(y(i));
                       }*/
           
        /*
            x = xn(1);
                    for (int i = 0; i < 10; i++)
                    {
                        Console.WriteLine(x[i]);
                    }*/


            string result="1";
            int sum=0;

            for (int i=0;i<1000;i++)
            {
                result=longAdd(result,result,0);
            }

            for (int i = result.Length - 1; i >= 0;i-- )
            {
                sum += result[i] - '0';
            }

            Console.WriteLine("2^1000 = "+result);
            Console.WriteLine("The sum of these digits is " + sum);

        
            Console.WriteLine("Total Milliseconds is " + DateTime.Now.Subtract(timeStart).TotalMilliseconds + "\n\n");
        }

        private static long digitTotal(long value)
        {
            int total = 0;
            while (value > 0)
            {
                total += (int)(value % 10);
                value /= 10;
            }
            return total;
        }

        private static int y(long value)
        {
            /*
              int total = 0;
                        while (value > 0)
                        {
                            if (value % 10 >= 5)
                            {
                                total++;
                            }
                            value /= 10;
                        }
                        return total;*/

            int[] x = xn(value);
            int result = 0;
            for (int i = 5; i < 10; i++)
            {
                result += x[i];
            }
            return result;
        }

        private static int[] xn(long n)
        {
            int[] result = new int[10];
            if (n == 0)
            {
                result[1] = 1;
                return result;
            }

            int[] lastN = xn(n - 1);

            for (int i = 0; i < 10; i++)
            {
                if (i < 5)
                {
                    result[2 * i] = lastN[i] + lastN[2 * i];
                }
                else
                {
                    result[2 * i - 10] = lastN[i] + lastN[2 * i - 10];
                    result[1]++;
                }
            }
            return result;
        }

        
        /// <summary>
        /// 超大数字表达式加法
        /// </summary>
        /// <param name="s1">数1</param>
        /// <param name="s2">数2</param>
        /// <param name="c">后面的数的进位,范围0~2</param>
        /// <returns></returns>
        private static string longAdd(string s1, string s2, int c)
        {
            //s1,s2的末位Index
            int l1 = s1.Length - 1;
            int l2 = s2.Length - 1;
            int x;

            //如果2个数都不为空
            if (l1 >= 0 && l2 >= 0)
            {
                x = s1[l1] - '0' + s2[l2] - '0' + c;
                return longAdd(s1.Substring(0, l1), s2.Substring(0, l2), x / 10) + (x % 10).ToString();
            }


            //下面的情况,s1和s2中至少有一个为空,我们要做的是找到那个不为空的进行下一步操作
            //把不为空的值放到s1里
            if (l2 >= 0)
            {
                s1 = s2;
                l1 = l2;
            }
            else if (l1 == -1)   //表示全为空,判断最后的进位,如果没进位,返回空白,结束递归
            {
                if (c!=0)
                {
                    return c.ToString();
                }
                return "";
            }

            x = s1[l1] - '0' + c;

            //如果s1只有1位
            if (l1 == 0)
            {
                return x.ToString();
            }

            //如果s1有不止一位
            return longAdd(s1.Substring(0, l1), "", x / 10) + (x % 10).ToString();
        }
    }
    
}


/*
Void F1()
2^1000 = 10715086071862673209484250490600018105614048117055336074437503883703510
51124936122493198378815695858127594672917553146825187145285692314043598457757469
85748039345677748242309854210746050623711418779541821530464749835819412673987675
59165543946077062914571196477686542167660429831652624386837205668069376
The sum of these digits is 1366
Total Milliseconds is 166.021

By GodMoon
*/

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