打表问题O

打表问题O - Dirichlet's Theorem on Arithmetic Progressions

If a and d are relatively prime positive integers, the arithmetic sequence beginning with a and increasing by d, i.e., a, a + d, a + 2d, a + 3d, a + 4d, ..., contains infinitely many prime numbers. This fact is known as Dirichlet's Theorem on Arithmetic Progressions, which had been conjectured by Johann Carl Friedrich Gauss (1777 - 1855) and was proved by Johann Peter Gustav Lejeune Dirichlet (1805 - 1859) in 1837.

For example, the arithmetic sequence beginning with 2 and increasing by 3, i.e.,

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, ... ,

contains infinitely many prime numbers

2, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, ... .

Your mission, should you decide to accept it, is to write a program to find thenth prime number in this arithmetic sequence for given positive integers a, d, and n.

Input

The input is a sequence of datasets. A dataset is a line containing three positive integers a, d, and n separated by a space. a and d are relatively prime. You may assume a <= 9307, d <= 346, and n <= 210.

The end of the input is indicated by a line containing three zeros separated by a space. It is not a dataset.

Output

The output should be composed of as many lines as the number of the input datasets. Each line should contain a single integer and should never contain extra characters.

The output integer corresponding to a dataset a, d, n should be the nth prime number among those contained in the arithmetic sequence beginning with a and increasing by d.

FYI, it is known that the result is always less than 10⁶ (one million) under this input condition.

Sample Input

367 186 151179 10 203271 37 39103 230 127 104 185253 50 851 1 19075 337 210307 24 79331 221 177259 170 40269 58 1020 0 0

Sample Output

92809
6709
12037
103
93523
14503
2
899429
5107
412717
22699
25673

 这道题如果写函数，然后每次判断该数是否是素数，肯定会超时，所以需要打表提高效率。

在题目限定范围内打素数表，基本思想是除了该数本身，它的所有倍数全部设为非素数状态，在进入该数的倍数设置阶段时，首先判断该数是不是素数，以免前面已经将该数设为非素数，然后又进行循环，

具体代码如下：

#include<iostream>

usingnamespace std;

intnum[1000005];

intmain()

{

        inta,b,n;

        for(inti = 0; i < 1000005; i++)

               num[i] = 1;

        num[1] =num[0] = 0;

        for(inti = 2; i < 1000005 / 2; i++)

        {

               if(num[i])

               for(intj = i + i;j < 1000005; j += i)//除了它本身，自身的所有倍数全部设置为0；

                       num[j] = 0;

        }

        while(cin>> a >> b >> n && (a|| b || n))

        {

               intcount = 0;

               for(a;;a += b)

               {

                       if(num[a])

                              count++;

                       if(count== n) { cout << a << endl; break; }

               }

        }

        return0;

}