YAPTCHA HDU - 2973 (威尔逊定理）

最新推荐文章于 2024-08-10 11:30:00 发布

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威尔逊定理

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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

这里写图片描述
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

这个感觉没什么意思，就是关键需要知道一个定理：

(p - 1)! \equiv - 1 (mod p), p 为 素 数

$(p-1)!\equiv-1\pmod p,p为素数$

所以当 $3k+7$ 为素数时：

(3 k + 6)! + 1 = t (3 k + 7)

$(3k+6)!+1=t(3k+7)$

(3 k + 6)! = (t - 1) (3 k + 7) + 3 k + 6

$(3k+6)!=(t-1)(3k+7)+3k+6$
所以当

3k+7 $3k+7$ 为素数时:

[( 3 k + 6 ) ! + 1 3 k + 7 - [( 3 k + 6 ) ! 3 k + 7]] = 1

$[\frac{(3k+6)!+1}{3k+7}-[\frac{(3k+6)!}{3k+7}]]=1$
反之为0，所以只需要统计哪些数是素数即可。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#define INF 0x7fffffff
#define N 3000010
using namespace std;
bool prime[N];
int ans[1000005];
void init()
{
    for(int i=2;i<N;i++)
        if(!prime[i])
            for(int j=i+i;j<N;j+=i)
                prime[j]=true;
    for(int i=1;i<=1000000;i++)
    {
        if(!prime[i*3+7])
            ans[i]++;
        ans[i]+=ans[i-1];
    }   
}
int main()
{
    init();
    int t;
    int n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("%d\n",ans[n]);
    }
    return 0;   
}