【HDU6608 Fansblog】求很大很大的数的阶乘

Fansblog

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Description

Farmer John keeps a website called ‘FansBlog’ .Everyday , there are many people visited this blog.One day, he find the visits has reached P , which is a prime number.He thinks it is a interesting fact.And he remembers that the visits had reached another prime number.He try to find out the largest prime number Q ( Q < P ) ,and get the answer of Q! Module P.But he is too busy to find out the answer. So he ask you for help. ( Q! is the product of all positive integers less than or equal to n: n! = n * (n-1) * (n-2) * (n-3) *… * 3 * 2 * 1 . For example, 4! = 4 * 3 * 2 * 1 = 24 )

Standard Input

First line contains an number T(1<=T<=10) indicating the number of testcases. Then T line follows,each contains a positive prime number P (1e9≤p≤1e14)

Standard Output

For each case output a line with one integer, means the factorial of Q modulo P for one line.

Sample Input

1 1000000007

Sample Output

328400734

emmm。。。。。。。。。。。。

首先，$Q!$ 是肯定不能从1乘过去的，也没有方法可以正着算过去。。。
于是聪明的zzy小公举就想，可不可以倒着推呢。
我们知道，$P!modP=0$，好像可以从这里入手耶。对于P取模的话，其实也只需要在$P!$这里乘好多逆元~
但是$P!modP=0$。。。怎么乘也不行啊。。。
于是小公举就入手了数论四大玄学定理中的威尔逊定理！（但他比赛的时候并没有入手，居然自己瞎推推出来了神奇的$Q!=p-(Q+1{)}^{-1}(Q+2{)}^{-1}...(p-2{)}^{-1}(p-1{)}^{-1}$）
由此可见，瞎推还是很重要的，嗯。
那么传说中的威尔逊定理呢？
他告诉我们：

$(p-1)!\equiv -1(modp)$

于是我们就可以这么写，，然后这么算出来。。

$Q![(Q+1)(Q+2)...(p-1)]=p-1(modp)$

$Q!(Q+1)(Q+2)...(p-2)=(p-1)\ast (p-1{)}^{-1}=1$

$Q!=1\ast (Q+1{)}^{-1}(Q+2{)}^{-1}...(p-2{)}^{-1}$

由于P和Q不会离很远，于是直接这么乘，用大数乘法就好了~