I - Dirichlet's Theorem on Arithmetic Progressions(1.5.5)-优快云博客

本文介绍如何利用狄利克雷定理找到特定等差数列中的第n个质数，并提供了一段C++代码实现。该算法适用于首项与公差互质的情况，通过判断数列中每个数是否为质数来确定第n个质数。

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Description

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 the nth 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

367 186 151
179 10 203
271 37 39
103 230 1
27 104 185
253 50 85
1 1 1
9075 337 210
307 24 79
331 221 177
259 170 40
269 58 102
0 0 0

Sample Output

#include <iostream>
#include <string>
using namespace std;
bool is_prime(int n)
{
    int i;
    if(n<=1)    return false;
    else if(n==2)   return true;
    else if(n%2==0) return false;
    else
    {
        for(i=3;i*i<=n;i+=2)
            if(n%i==0)  return false;
        return true;
    }
}
int main()
{
    long a,d,n,cnt,s;
    while(cin>>a>>d>>n&&(a||d||n))
    {
        for(s=a,cnt=0;cnt<n;s+=d)  //cnt用来计数素数的个数
        {
            if(is_prime(s))
                cnt++;
        }
        cout<<s-d<<endl;  // 最后一次执行完毕后多加了一个d，所以要减去
    }
    return 0;
}