HDU 2973 YAPTCHA

YAPTCHA

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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 <= 10⁶). Each query consist of one natural number n (1 <= n <= 10⁶).

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

题意：

实际上就是求题目的公式。

思路：

首先就是看题目，题目的意思是求和，所以能够有威尔逊定理得到( p - 1 )! ≡ 1﹡( p -1 ) ≡ -1 ( mod p ),我们可以设p是3k+7，所以(3k+6)!+1就可以把3k+7整除了，所以只有再3k+7是素数的时候，出现1，其他情况都是0。

#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1e+6 + 10;
int sum[maxn] = {0};
bool judge(int n) {
    if (n == 1 || n == 0) return false;
    for (int i = 2; i <= sqrt(n); i++) {
        if (n % i == 0) return false;
    }
    return true;
}
int main() {
    for (int i = 1; i <= maxn; i++) {
        sum[i] = sum[i - 1] + judge(3 * i + 7);
    }
    int n, x;
    scanf("%d", &n);
    while (n--) {
        scanf("%d", &x);
        printf("%d\n", sum[x]);
    }
    return 0;
}