Fansblog

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 779 Accepted Submission(s): 287

Problem Description

Farmer John keeps a website called ‘FansBlog’ .Everyday , there are many people visited this blog.One day, he find the visits has reached PPP , 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)Q ( Q < P )Q(Q<P) ,and get the answer of Q!Q!Q! Module PPP.But he is too busy to find out the answer. So he ask you for help. ( Q!Q!Q! is the product of all positive integers less than or equal to n:n!=n∗(n−1)∗(n−2)∗(n−3)∗…∗3∗2∗1n: n! = n * (n-1) * (n-2) * (n-3) *… * 3 * 2 * 1n:n!=n∗(n−1)∗(n−2)∗(n−3)∗…∗3∗2∗1. For example, 4!=4∗3∗2∗1=244! = 4 * 3 * 2 * 1 = 244!=4∗3∗2∗1=24)

Input

First line contains an number #T(1≤T≤10)T(1\leq T\leq 10)T(1≤T≤10) indicating the number of testcases.
Then TTT line follows, each contains a positive prime number P(109≤p≤1014)P (10^9\leq p\leq10^{14})P(109≤p≤1014)

Output

For each testcase, output an integer representing the factorial of QQQ modulo PPP.

Sample Input

1
1000000007

Sample Output

328400734

Source

2019 Multi-University Training Contest 3

题意

给你一个109−101410^9-10^{14}109−1014内的质数ppp，求小于ppp的最大质数的阶乘取模ppp

题解

威尔逊定理+Miller_RabinMiller\_RabinMiller_Rabin素数测试
威尔逊定理就是对于任意的正质数kkk，有
((k−1)!)%k=k−1((k-1)!)\%k=k-1 ((k−1)!)%k=k−1
然后对于本题先用Miller_RabinMiller\_RabinMiller_Rabin找到小于ppp的最大质数qqq，然后用威尔逊定理推一下式子：
((p−1)!)%p=q!×(q+1)×(q+2)×...×(p−1)%p=p−1\begin{aligned}((p-1)!)\%p &= q!\times(q+1)\times(q+2)\times...\times(p-1)\%p \\ & =p-1 \\ \end{aligned} ((p−1)!)%p=q!×(q+1)×(q+2)×...×(p−1)%p=p−1
q!≡1(q+1)×(q+2)×...×(p−2)(modp)\begin{aligned}q! &\equiv \frac{1}{(q+1)\times(q+2)\times...\times(p-2)}(mod\ p) \end{aligned} q!≡(q+1)×(q+2)×...×(p−2)1(mod p)

代码

#include <cstdio>
#include <cstdlib>
#include <map>
using namespace std;
long long gcd(long long a,long long b) {if (b == 0) return a;return gcd(b,a%b);
}
long long mul(long long a,long long b,long long mod){long long ret=0;while(b) {if(b & 1) ret=(ret+a)%mod;a=(a+a)%mod;b >>= 1;}return ret;
}
long long pow(long long a,long long b,long long mod) {long long ret = 1;while(b) {if(b & 1) ret = mul(ret,a,mod);a = mul(a,a,mod);b >>= 1;}return ret;
}
bool check(long long a,long long n){long long x = n - 1;int t = 0;while((x & 1) == 0) {x >>= 1;t ++;}x = pow(a,x,n);long long y;for(int i=1;i<=t;i++) {y = mul(x,x,n);if(y == 1 && x != 1 && x != n - 1) return true;x = y;}if(y != 1) return true;return false;
}
bool Miller_Rabin(long long n) {if(n == 2) return true;if(n == 1 || !(n & 1)) return false;const int arr[12] = {2,3,5,7,11,13,17,19,23,29,31,37};for(int i = 0; i < 12; i++) {if (arr[i] >= n) break;if(check(arr[i], n)) return false;}return true;
}int main() {int t;long long n;scanf("%d",&t);while(t--) {scanf("%lld",&n);long long p=n-1;while(!Miller_Rabin(p)) p--;long long ans=1;for(long long i=p+1;i+1<n;i++) ans=mul(ans,pow(i,n-2,n),n);printf("%lld\n",ans);}
}

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「hdu6608」Fansblog【Miller_Rabin+威尔逊定理】