（素数求解）I - Dirichlet's Theorem on Arithmetic Progressions(1.5.5)

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u

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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 aand 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 <cmath>

using namespace std;

bool ss(int x)

{

    if(x < 2)

        return 0;

	else

	{

        for(int i = 2; i <= sqrt((float)x); i++)

            if(x % i == 0)

                return 0;

        return 1;

    }

}

int main()

{

    int a, d, n;

    while (cin>>a>>d>>n)

	{

        if (a==d&&d==n&&n==0)

		{

            return 0;

        }

        int i, count = 0;

        for (i = a; count < n; i += d) {

            if (ss(i))

			{

                count++;

            }

        }

    cout<<i-d<<endl;

    }

    return 0;

}

巴特西

（素数求解）I - Dirichlet's Theorem on Arithmetic Progressions(1.5.5)

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巴特西

（素数求解）I - Dirichlet&#39;s Theorem on Arithmetic Progressions(1.5.5)

最新文章

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（素数求解）I - Dirichlet's Theorem on Arithmetic Progressions(1.5.5)