51nod 1556 计算（默慈金数）

来源：互联网发布：android 仿淘宝搜索编辑：程序博客网时间：2024/06/05 19:19

这题要用到默慈金数，又涨知识了。
默慈金数：http://blog.csdn.net/acdreamers/article/details/41213667
题解：http://blog.csdn.net/qingshui23/article/details/52068031
在这里默慈金数是用来排除错误走法的。当时我没想明白为啥错误的走法是M[n-2],就在51nod问了一下，回答如下：
这里写图片描述

using System;using System.IO;namespace timeless{    class Program    {        private static Int64 MAXN = (Int64)1e6+10;        private static Int64 mod = (Int64)1e9 + 7;        private static Int64 extend_gcd(Int64 a, Int64 b, ref Int64 x, ref Int64 y)        {            if (a == 0 && b == 0) return -1;            if (b == 0)            {                x = 1;                y = 0;                return a;            }            Int64 d = extend_gcd(b,a%b,ref y,ref x);            y = y - a / b * x;            return d;        }        private static Int64 mod_reverse(Int64 a)        {            Int64 x = 0,y = 0;            Int64 d = extend_gcd(a,mod,ref x, ref y);            if(d == 1) return (x%mod+mod)%mod;            else return -1;        }        static void Main(string[] args)        {            StreamReader sr = new StreamReader(Console.OpenStandardInput());            StreamWriter sw = new StreamWriter(Console.OpenStandardOutput());            Int64 n = Convert.ToInt64(sr.ReadLine());            Int64[] ans = new Int64[MAXN];            Int64[] M = new Int64[MAXN];            Int64 x;            M[1] = 1;            M[2] = 2;            for (Int64 i = 3; i <= n; ++i)            {                x = mod_reverse(i+2);                M[i] = (((2 * i + 1) * M[i - 1]) % mod + ((3 * i - 3) * M[i - 2]) % mod) * x;                M[i] = (M[i] % mod + mod) % mod;            }            ans[1] = 1;            ans[2] = 2;            for (Int64 i = 3; i <= n; ++i)            {                ans[i] = (3 * ans[i - 1] - M[i - 2]);                ans[i] = (ans[i] % mod + mod) % mod;            }            sw.WriteLine(ans[n]);            sw.Flush();            sr.Close();            sw.Close();        }    }}

阅读全文

0 0