HDU多校训练赛第三场

最新推荐文章于 2022-08-12 13:20:43 发布

原创最新推荐文章于 2022-08-12 13:20:43 发布 · 177 阅读

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Fansblog

题目描述

*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 )

输入

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)

输出

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

思路

这个题目目前我知道有两种做法。做法一暴力求逆元，做法二用米勒罗宾素数判断。但我现在只会做法一，做法二等有空就去学习。做法二的时间特别短，做法一就有点麻烦了。
首先通过暴力打表或者是威尔逊定理，我们都可以发现当P和Q同为孪生素数的时候，Q！%P=1；
且P-Q>=2。所以我们可以将Q！=（P-2）！/（Q+1）(Q+2)…(P-2);
根据（P-2）！%P==1；那么我们只要算（1/（Q+1）(Q+2)…(P-2) ）%P；
那么就是算分母的逆元。但是由于数据范围很大，longlong直接相乘的话会爆，就得把乘法变为加法。
接下来附上AC代码

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
const int N=1e6+5;
typedef long long ll;
ll mul(ll a,ll b,ll mod)
{
    ll sum=0;c
    a%=mod;b%=mod;
    while(b)
    {
        if(b&1) sum=(sum+a)%mod;
        b>>=1;
        a=(a*2)%mod;
    }
    return sum;
}
ll qpow(ll a,ll b,ll mod)
{
    ll sum=1;
    while(b)
    {
        if(b&1) sum=mul(sum,a,mod);
        b>>=1;
        a=mul(a,a,mod);
    }
    return sum;
}
ll niyuan(ll a,ll b,ll mod)
{
    ll sum=1;
    for(register ll i=b+1;i<=a-2;i++)
    {
        sum=mul(sum,i,mod);
    }
    sum=qpow(sum,a-2,mod);
    return sum;
}
bool judge(ll a)
{
    for(register ll i=2;i<=10000000;i++)
    {
        if(a%i==0) return false;
    }
    return true;
}
int main()
{
    int x;
    scanf("%d",&x);
    register ll p,q;
    while(x--)
    {
        scanf("%lld",&p);
        for(q=p-1;;q--)
        {
            if(judge(q)) break;
        }
        //cout<<q<<endl;
        if(q==p-2) printf("1\n");
        else{
            ll ans=niyuan(p,q,p);
            printf("%lld\n",ans);
        }
    }
    return 0;
}