威尔逊定理数论

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本文探讨了威尔逊定理的应用，特别是在解决一个特定的数学难题中，即计算给定自然数n时Sn的值，Sn涉及到对3k+7形式的数进行判断是否为素数。通过实现一个辅助程序，我们能够高效地计算出任意给定n的Sn值，关键在于利用威尔逊定理快速判断素数。

威尔逊定理

n为素数时

$(n - 1)! \equiv -1 (mod$ $n)$

$(n - 2) ! \equiv 1 (mod$ $n)$

应用

$\sum _{k=1}^{n} \left \lfloor \frac{(3 \cdot k + 6)! + 1}{(3 \cdot k+7)} +\left \lfloor \frac{(3 \cdot k + 6)!}{(3\cdot k + 7)} \right \rfloor \right \rfloor$

YAPTCHA

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1858 Accepted Submission(s): 960

Problem Description

The math department has been having problems lately. Due to immense amount of unsolicited automated programs which were crawling across their pages, they decided to put Yet-Another-Public-Turing-Test-to-Tell-Computers-and-Humans-Apart on their webpages. In short, to get access to their scientific papers, one have to prove yourself eligible and worthy, i.e. solve a mathematic riddle.

However, the test turned out difficult for some math PhD students and even for some professors. Therefore, the math department wants to write a helper program which solves this task (it is not irrational, as they are going to make money on selling the program).

The task that is presented to anyone visiting the start page of the math department is as follows: given a natural n, compute

where [x] denotes the largest integer not greater than x.

Input

The first line contains the number of queries t (t <= 10^6). Each query consist of one natural number n (1 <= n <= 10^6).

Output

For each n given in the input output the value of Sn.

Sample Input

 13 1 2 3 4 5 6 7 8 9 10 100 1000 10000

Sample Output

 0 1 1 2 2 2 2 3 3 4 28 207 1609

Source

Central European Programming Contest 2008

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lcy

题解：3k + 7 为质数时，又威尔逊定理左式等于0，右式等于-1 整式等于1

3k+7为合数时，（3k+6)!是3k +7的倍数所以 $\left \lfloor \frac{(3 \cdot k + 6)!}{(3\cdot k + 7)} \right \rfloor =\frac{(3 \cdot k + 6)!}{(3\cdot k + 7)}$ ，原式等于 $\left \lfloor \frac{1}{3k+7} \right \rfloor = 0$

所以这题就是求给定集合的素数个数

/*
 *威尔逊定理
 */
#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e6 +10;
const int maxm = 1e6 + 10;
int v[maxn];
vector <int> pri;
int cnt, n;
int ans[maxn];
void prime(){
    for (int i =2 ; i < maxn; i++){
        if (!v[i]) pri.push_back(i);
        for (int j = 0; j < pri.size() && pri[j] * i < maxn; j++){
            v[i * pri[j]] = 1;
            if (i % pri[j] == 0) break;
        }
    }
}

int main()
{
    prime();
    cnt = pri.size();
    int T;cin >> T;
    for (int i = 1; i < maxm; i++){
        int k = 3 * i + 7;
        if (!v[k]) ans[i] = ans[i - 1] + 1;
        else ans[i] = ans[i -1];
    }
    while (T--){
        scanf("%d", &n);
        cout << ans[n] << endl;
    }
    return 0;
}