HDU 2973 YAPTCHA （威尔逊定理）

YAPTCHA

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1885    Accepted Submission(s): 971

 

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

Sample Output

题目大意

给定一个整数n，求

题目分析

威尔逊定理：

当且仅当p为素数时，(p−1)!≡−1(mod p)

所以我们可以发现，如果这个3k+7为奇数，式子就等于1，否则就等于0

#include<bits/stdc++.h>

using namespace std;

const int N=;
long long a[N];
bool prime[N];
int i,n,j,x;
int main()
{
    for(i=; i<N; i++)
    {
        if(i%==) prime[i]=false;
        else prime[i]=true;
    }
       for(i=; i<=sqrt(N); i+=)
    {
        if(prime[i])
        for(j=i+i; j<N; j+=i)
        prime[j]=false;
    }
    for(i=;i<=;i++)
    {
        a[i]=a[i-];
        if(prime[*i+])
        a[i]++;
    }
    cin>>n;
    for(i=;i<=n;i++)
    {
        cin>>x;
        cout<<a[x]<<endl;
    }
}