Dirichlet's Theorem on Arithmetic Progressions 素数筛选法-优快云博客

这篇博客介绍了Dirichlet定理，即对于互质的正整数a和d，以a为首项，d为公差的算术序列包含无限多个素数。文章提出了编写程序找出该序列中第n个素数的任务，给出了输入输出格式，并提到了在给定条件下结果始终小于10^6。

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题目描述

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.

输入

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.

输出

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.

样例输入

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

样例输出

注意：一定要使用素数筛选法建立从0到10^6数是否是素数（bool )型的一个数组表；否则使用普通方法会运行时间过长

算法实现：

#include <iostream>
#include <cstring> 
#define MAX 1000000
bool prime[MAX+1];//定义一个素数表 ，用来记录从0到1000000的数是否是素数 
using namespace std;
int main()
{
	 int i,j;
	 memset(prime,true,sizeof(prime));//将prime数组全部初始化为true,假定全为素数，找出合数定义为false则素数表建成 
	 prime[0]=prime[1]=false; prime[2]=true;
	 
	 for(i=2;i*i<MAX;i++)/*从2开始遍历，i*i<=MAX等价于sqrt(MAX)
	 但用前者更不容易出错*/ 
	 {
	 	if(prime[i])//如果没有被标记的话，就把它的倍数（一直找到MAX为止）标记为false 
	 	{
	 		for(j=i+i;j<MAX;j=j+i)
	 		prime[j]=false; 
		 }
	 }
	 
	long x,y,n,count=0;//count用来计数 统计遍历了多少个素数 
	while(cin>>x>>y>>n,x!=0,y!=0,n!=0)
	{
		long long i;
		int count=0;
        for(i=x;;i=i+y)
    	{
    		if(prime[i]) 
        	count++;
        	if(count==n) break;
     	}
     	cout<<i<<endl;
     }
	return 0;
}