HDU 2973 / UVa 1434 / CERC 2008 YAPTCHA (威尔逊定理及其逆定理)

http://acm.hdu.edu.cn/showproblem.php?pid=2973

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=516&page=show_problem&problem=4180

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

思路：

当n是合数时，((n-1)!+1)/n不为整数，所以在题目的和式中该部分为0；n是质数时就为1了。

所以筛一遍素数，在筛的过程中，找出形如3k+7的素数即可。

额外说一句，当a,b互素时，形如an+b的素数有无穷个。（wiki:Dirichlet's theorem on arithmetic progressions）

完整代码：

/*796ms,8060KB*/

#include<cstdio>
#include<cmath>
const int mx = 3 * 1000000 + 7 + 1;
const int sqrt_mx = (int)sqrt((double)mx);
const int mx_n = 1000000 + 1;

bool a[mx_n], vis[mx];
int s[mx_n];

void init()
{
	int i, j;
	for (i = 2; i < mx; ++i)
	{
		if (!vis[i])
		{
			if ((i - 7) % 3 == 0) a[(i - 7) / 3] = true;
			if (i <= sqrt_mx)
				for (j = i * i; j < mx ; j += i) vis[j] = true;
		}
	}
}

int main()
{
	init();
	for (int i = 2; i < mx_n; ++i)
		s[i] = s[i - 1] + a[i];
	int t, n;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		printf("%d\n", s[n]);
	}
	return 0;
}