題目描述

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.

輸入

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

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

輸出

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.

樣例輸入

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

樣例輸出

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

演算法實現：

#include <iostream>
#include <cstring> 
#define MAX 1000000
bool prime[MAX+1];//定義一個素數表 ，用來記錄從0到1000000的數是否是素數 
using namespace std;
int main()
{
	 int i,j;
	 memset(prime,true,sizeof(prime));//將prime陣列全部初始化為true,假定全為素數，找出合數定義為false則素數表建成 
	 prime[0]=prime[1]=false; prime[2]=true;
	 
	 for(i=2;i*i<MAX;i++)/*從2開始遍歷，i*i<=MAX等價於sqrt(MAX)
	 但用前者更不容易出錯*/ 
	 {
	 	if(prime[i])//如果沒有被標記的話，就把它的倍數（一直找到MAX為止）標記為false 
	 	{
	 		for(j=i+i;j<MAX;j=j+i)
	 		prime[j]=false; 
		 }
	 }
	 
	long x,y,n,count=0;//count用來計數 統計遍歷了多少個素數 
	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;
}