HDU6608 Fansblog【Miller_Rabin素性測試算法+威爾遜定理】

Fansblog

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2801 Accepted Submission(s): 1158

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

Source
2019 Multi-University Training Contest 3

問題鏈接：HDU6608 Fansblog
問題簡述：給定一個10⁹−10¹⁴內的質數p,求小於p的最大質數的階乘取模p。
問題分析：
    這個問題的解決需要用到Miller_Rabin素性測試算法和威爾遜定理，這裏不做解釋。
程序說明：
    快速模冪計算過程中需要做乘法，有可能會產生溢出，所以需要用一個函數來實現。還有一種方法是使用128位整數計算。
參考鏈接：（略）
題記：（略）

AC的C++語言程序如下：

/* HDU6608 Fansblog */

#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ULL;

const int TIME = 5;

ULL gcd(ULL a, ULL b) {
  return b == 0 ? a : gcd(b, a % b);
}

// 快速乘積，計算a*b
ULL mulmod(ULL a, ULL b, ULL m)
{
    ULL ret = 0;
    while(b) {
        if(b & 1) {ret += a; ret %= m;}
        a = (a << 1) % m;
        b >>= 1;
    }
    return ret;
}

// 快速模冪計算
ULL powmod(ULL a, ULL b, ULL mod)
{
    ULL ret = 1;
    while(b) {
        if(b & 1) ret = mulmod(ret, a, mod);
        a = mulmod(a, a, mod);
        b >>= 1;
    }
    return ret;
}

ULL random(ULL n)
{
    return (ULL)((double) rand() / RAND_MAX * n + 0.5);
}

bool check(ULL a, ULL n)
{
    ULL m = n - 1;
    int t = 0;
    while((m & 1) == 0) {t++; m >>= 1;}
    ULL x = powmod(a, m, n);
    if(x == 1 || x == n - 1)
        return false;
    while(t--) {
        x = mulmod(x, x, n);
        if(x == n - 1)
            return false;
    }
    return true;
}

bool Miller_Rabin(ULL n)
{
    if(n < 2) return false;
    if(n == 2) return true;
    if((n & 1) == 0) return false;
    ULL x = n - 1;
    ULL t = 0;
    while((x&1) == 0) {
        x >>= 1;
        t++;
    }
    for(int i = 0; i <= TIME; i++)
    {
        ULL a = random(n - 2) + 1;
        if(check(a, n)) return false;
    }
    return true;
}

int main()
{
    int t;
    ULL n;
    scanf("%d", &t);
    while(t--) {
        scanf("%llu", &n);
        ULL p = n - 1;
        while(!Miller_Rabin(p)) p--;

        ULL ans = 1;
        for(ULL i = p + 1; i < n - 1; i++)
            ans = mulmod(ans, powmod(i, n - 2, n), n);

        printf("%llu\n", ans);
    }

    return 0;
}

AC的C++語言程序如下：

/* HDU6608 Fansblog */

#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ULL;

ULL gcd(ULL a,ULL b) {
  return b == 0 ? a : gcd(b, a % b);
}

// 快速乘積，計算a*b
ULL mulmod(ULL a, ULL b, ULL m)
{
    ULL ret = 0;
    while(b) {
        if(b & 1) {ret += a; ret %= m;}
        a = (a << 1) % m;
        b >>= 1;
    }
    return ret;
}

// 快速模冪計算
ULL powmod(ULL a, ULL b, ULL mod)
{
    ULL ret = 1;
    while(b) {
        if(b & 1) ret = mulmod(ret, a, mod);
        a = mulmod(a, a, mod);
        b >>= 1;
    }
    return ret;
}

bool check(ULL a, ULL n)
{
    ULL m = n - 1;
    int t = 0;
    while((m & 1) == 0) {t++; m >>= 1;}
    ULL x = powmod(a, m, n);
    if(x == 1 || x == n - 1)
        return false;
    while(t--) {
        x = mulmod(x, x, n);
        if(x == n - 1)
            return false;
    }
    return true;
}

bool Miller_Rabin(ULL n)
{
    if(n == 2) return true;
    if(n == 1 || !(n & 1)) return false;
    const ULL prime[12] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
    for(int i = 0; i < 12; i++) {
        if (prime[i] >= n) break;
        if(check(prime[i], n)) return false;
    }
    return true;
}

int main()
{
    int t;
    ULL n;
    scanf("%d", &t);
    while(t--) {
        scanf("%llu", &n);
        ULL p = n - 1;
        while(!Miller_Rabin(p)) p--;

        ULL ans = 1;
        for(ULL i = p + 1; i < n - 1; i++)
            ans = mulmod(ans, powmod(i, n - 2, n), n);

        printf("%llu\n", ans);
    }

    return 0;
}