2019 Multi-University Training Contest 3 1006 Fansblog

                                                           Fansblog

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 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 )

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)

Output

For each testcase, output an integer representing the factorial of Q modulo P.

Sample Input

1 1000000007

Sample Output

328400734

 

 通过oeis找的 a(n) = 1 / (prime[n]-prime[n-1]-1)! % prime[n]

直接暴力实现就好了

 

#include<bits/stdc++.h>

using namespace std;


typedef long long LL;

LL qmul(LL x, LL y, LL mod) { // 乘法防止溢出， 如果p * p不爆LL的话可以直接乘； O(1)乘法或者转化成二进制加法
    return (x * y - (long long)(x / (long double)mod * y + 1e-3) *mod + mod) % mod;
}
LL qpow(LL a, LL n, LL mod) {
    LL ret = 1;
    while(n) {
        if(n & 1) ret = qmul(ret, a, mod);
        a = qmul(a, a, mod);
        n >>= 1;
    }
    return ret;
}
bool Miller_Rabin(LL p) {
    if(p < 2) return 0;
    if(p != 2 && p % 2 == 0) return 0;
    LL s = p - 1;
    while(! (s & 1)) s >>= 1;
    for(int i = 0; i < 5; ++i) {
        if(p == prime[i]) return 1;
        LL t = s, m = qpow(prime[i], s, p);
        while(t != p - 1 && m != 1 && m != p - 1) {
            m = qmul(m, m, p);
            t <<= 1;
        }
        if(m != p - 1 && !(t & 1)) return 0;
    }
    return 1;
}
LL q;
LL Mode(__int128 a, __int128 b, __int128 mode)
{
    __int128 sum = 1;
    while (b) {
        if (b & 1) {
            sum = (sum * a) % mode;
            b--;
        }
        b /= 2;
        a = a * a % mode;
    }
    return sum;
}

LL inv(LL n){
    return Mode(n,q-2,q);
}
LL jc (LL num,LL mod)
{
    LL sum=1;
    for(LL i=2;i<=num;i++)
        sum=(sum*i)%mod;
    return sum;
}
int main()
{
    int n;
    scanf("%d",&n);
    while(n--){
        scanf("%I128d",&q);
        LL p=q-1;
        while(!Miller_Rabin(p))
        {
            p-=1;
        }
        printf("%lld\n",inv(jc(q-p-1,q))%q);
    }
}