C#版 - HDUoj 5391 - Zball in Tina Town(素数) - 题解

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HDUoj 5391 - Zball in Tina Town

在线提交:

http://acm.hdu.edu.cn/showproblem.php?pid=5391

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 2790 Accepted Submission(s): 1309

题目大意：

Tina Town 是一个善良友好的地方, 这里的每一个人都互相关心。

Tina有一个球，它的名字叫zball。zball很神奇，它每天会变大一些。在第一天，它和原始大小一样。 在第二天，它的大小将成为第一天的2倍。 在第n天，它的大小将为第(n-1)天大小的n倍。Tina想知道，zball在第n-1天时的大小对n取模是多少呢？

思路：

陶哲轩在他的书Solving mathematical problems 中提到威尔逊定理(n−1)!+1 (mod n)≡0⇔n" role="presentation" style="position: relative;">(n−1)!+1 (mod n)≡0⇔n(n−1)!+1 (mod n)≡0⇔n is prime.

首先，来回忆一下阶乘的定义:

m!=∏k=1mk=1×2×3×⋯×m." role="presentation" style="position: relative;">m!=∏mk=1k=1×2×3×⋯×m.m!=∏k=1mk=1×2×3×⋯×m.

可得出结论: 存在a, b ∈" role="presentation" style="position: relative;">∈∈ [1, m] 使得a⋅b" role="presentation" style="position: relative;">a⋅ba⋅b能整除m!

假定 m=n−1" role="presentation" style="position: relative;">m=n−1m=n−1,

原问题可分类如下:

1. n是质数，则由威尔逊定理知:
   
   (n−1)! (mod n)=−1(mod n)=n−1" role="presentation" style="position: relative;">(n−1)! (mod n)=−1(mod n)=n−1(n−1)! (mod n)=−1(mod n)=n−1.

2. n是合数(composite)，且n不能表示为质数的平方，则∃a,b" role="presentation" style="position: relative;">∃a,b∃a,b使得n=a⋅b | m!" role="presentation" style="position: relative;">n=a⋅b | m!n=a⋅b | m!，即
   
   a⋅b=n | (n−1)!" role="presentation" style="position: relative;">a⋅b=n | (n−1)!a⋅b=n | (n−1)!

3. n是合数，且n可表示成质数p的平方，而且 p > 2+1" role="presentation" style="position: relative;">2–√+12+1, 即 p≥3" role="presentation" style="position: relative;">p≥3p≥3
   
   此时的目标是寻找a, b使得 a⋅b | p2" role="presentation" style="position: relative;">a⋅b | p2a⋅b | p2，不妨假设 a = k1⋅p" role="presentation" style="position: relative;">k1⋅pk1⋅p, b = k2⋅p" role="presentation" style="position: relative;">k2⋅pk2⋅p.
   
   n=p2" role="presentation" style="position: relative;">n=p2n=p2, 则n的约数有1,p,p2" role="presentation" style="position: relative;">1,p,p21,p,p2.
   
   下面用反证法来证明为何a、b均与p线性相关，如果a(≥1" role="presentation" style="position: relative;">≥1≥1)与p线性无关，则b=k⋅p2≥p2(k≥1)" role="presentation" style="position: relative;">b=k⋅p2≥p2(k≥1)b=k⋅p2≥p2(k≥1)，而b≤m=n−1=p2−1" role="presentation" style="position: relative;">b≤m=n−1=p2−1b≤m=n−1=p2−1，矛盾。同理假设b与p线性无关也会出现同样的矛盾，因此a、b均与p线性相关。

   1≤a=k1⋅p&lt;b=k2⋅p≤p2−1" role="presentation" style="position: relative;">1≤a=k1⋅p<b=k2⋅p≤p2−11≤a=k1⋅p<b=k2⋅p≤p2−1
   
   让a尽量小, 则k1=1" role="presentation" style="position: relative;">k1=1k1=1, 令t=k2" role="presentation" style="position: relative;">t=k2t=k2, 于是b可表示为t⋅p" role="presentation" style="position: relative;">t⋅pt⋅p.
   
   ∴t⋅p≤p2−1" role="presentation" style="position: relative;">∴t⋅p≤p2−1∴t⋅p≤p2−1
   
   解上述不等式可得 p≥t+t2+42=f(t) (t≥2)" role="presentation" style="position: relative;">p≥t+t2+4√2=f(t) (t≥2)p≥t+t2+42=f(t) (t≥2), 容易分析得f(t)是递增函数。
   
   当t=2" role="presentation" style="position: relative;">t=2t=2时, p≥2+1" role="presentation" style="position: relative;">p≥2–√+1p≥2+1，即p∈[2+1,+∞)" role="presentation" style="position: relative;">p∈[2–√+1,+∞)p∈[2+1,+∞), 于是b=2⋅p" role="presentation" style="position: relative;">b=2⋅pb=2⋅p.
   
   而当t&gt;2" role="presentation" style="position: relative;">t>2t>2时，p的解集是上述区间的子集, 因此p的解集可取两者中范围较大者, 即[2+1,+∞)" role="presentation" style="position: relative;">[2–√+1,+∞)[2+1,+∞)。
   
   由于p≥2+1" role="presentation" style="position: relative;">p≥2–√+1p≥2+1, 而 p∈Z" role="presentation" style="position: relative;">p∈Zp∈Z, 因而p≥3" role="presentation" style="position: relative;">p≥3p≥3.
   
   因此
   
   a⋅b=2⋅p2=2⋅n|(n−1)!" role="presentation" style="position: relative;">a⋅b=2⋅p2=2⋅n|(n−1)!a⋅b=2⋅p2=2⋅n|(n−1)!
   
   故 (n-1)! (mod n) = 0

4. n是合数，且n可表示成质数p的平方，而且 p < 2+1" role="presentation" style="position: relative;">2–√+12+1, 即 p = 2, n=4.
   
   此时，无法找到满足条件的a和b，4∤3!=6" role="presentation" style="position: relative;">4∤3!=64∤3!=6.

已AC代码：

using System;

namespace hdoj_zball
{
    public class Solution
    {
        public int Mod(int n)
        {
            if (IsPrime(n))
                return n - 1;
            if (n == 4)
                return 2;           

            return 0;
        }

        public bool IsPrime(int m)
        {
            if (m < 2)
                return false;
            for (int i = 2; i * i <= m; i++)
            {
                if (m % i == 0)
                    return false;
            }
            return true;
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());
            while (n-- > 0)
            {
                string[] s = Console.ReadLine().Split();
                int num = int.Parse(s[0]);
                Solution sol = new Solution();
                Console.WriteLine(sol.Mod(num));
            }
        }
    }
}

Rank (C#):

Exe.Time	Exe.Memory
1185ms	26908K

Reference:

elementary number theory - If n≠4" role="presentation" style="position: relative;">n≠4n≠4 is composite, then n" role="presentation" style="position: relative;">nn divides (n−1)!" role="presentation" style="position: relative;">(n−1)!(n−1)!. - Mathematics Stack Exchange

https://math.stackexchange.com/questions/164852/if-n-ne-4-is-composite-then-n-divides-n-1