2013
11-12

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 the nth prime number in this arithmetic sequence for given positive integers a, d, and n.

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.

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.

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

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

//* @author popop0p0popo
import java.util.*;
import java.io.*;

public class Main{
public static void main(String[] args){
while (true){
int a=scanner.nextInt();
int d=scanner.nextInt();
int n=scanner.nextInt();
if (a+d+n==0){
break;
}
int index=0;
int i=0;
while (index< n){
if (isZh(a+d*i)){
index++;
}
i++;
}
System.out.println(a+d*(i-1));
}
}

public static boolean isZh(int p){
if (p==1){
return false;
}
if (p==2){
return true;
}
if (p%2==0){
return false;
}
for (int i=3;i*i<=p ;i=i+2 ){
if (p%i==0){
return false;
}
}
return true;
}
}

1. 思路二可以用一个长度为k的队列来实现，入队后判断下队尾元素的next指针是否为空，若为空，则出队指针即为所求。

2. 学算法中的数据结构学到一定程度会乐此不疲的，比如其中的2－3树，类似的红黑树，我甚至可以自己写个逻辑文件系统结构来。