Poj 3006 Dirichlet's Theorem on Arithmetic Progressions

1.Link:

http://poj.org/problem?id=3006

2.Content:

Dirichlet's Theorem on Arithmetic Progressions

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15795		Accepted: 7932

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

92809

6709

12037

103

93523

14503

2

899429

5107

412717

22699

25673

Source

Japan 2006 Domestic

3.Method:

篩素數法，以空間換時間

4.Code:

#include<iostream>

#include<cstdio>

#include<cmath>



#define MAX_NUM 1000000



using namespace std;



int main()

{

    //freopen("D://input.txt","r",stdin);

    

    int i,j;

    

    bool arr_prime[MAX_NUM + 1];

    

    for(i = 3; i <= MAX_NUM; i += 2) arr_prime[i] = true;

    for(i = 4; i <= MAX_NUM; i += 2) arr_prime[i] = false;

    arr_prime[2] = true;

    

    arr_prime[1] = false; //!!

    

    int sqrt_mn = sqrt(MAX_NUM);

    for(i = 3; i <= sqrt_mn; i += 2)

    {

        if(arr_prime[i])

        {

            for(j = i + i; j <= MAX_NUM; j += i) arr_prime[j] = false;

        }

    }

    

    int a,d,n;

    cin >> a >> d >> n;

    

    while(a != 0 || d != 0 || n != 0)

    {

        while(n)

        {

            if(arr_prime[a]) --n;

            a += d;

        }

        cout << a - d << endl;

        

        cin >> a >> d >> n;

    } 

    

    return 0;

} 

 

5:Reference: