O - Dirichlet's Theorem on Arithmetic Progressions (POJ 3006)

最新推荐文章于 2021-12-21 09:13:22 发布

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最新推荐文章于 2021-12-21 09:13:22 发布

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文章标签： POJ 3006

本文链接：https://blog.youkuaiyun.com/weixin_44011353/article/details/96583012

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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

题解

题意：

给定a，d，n，在a, a + d, a + 2d, a + 3d, a + 4d……中找出第n小的素数

思路：

朴素算法，从a, a + d, a + 2d, a + 3d, a + 4d……中枚举判素数，至于时间嘛，代码中的算法，能过POJ的这个题
关于快速判素数的算法可以参考这篇文章，还有这篇。

Code

#include<iostream>
#include<math.h>
using namespace std;

int prime_num(int n)
{	
	float n_sqrt;
    if(n == 1)  return 0;
	if(n == 2 || n == 3) return 1;
	if(n % 6 != 1 && n % 6 != 5) return 0;//不再6的倍数两侧的一定不是素数，在也不一定是素数

	n_sqrt = floor(sqrt((float)n));//开平方后，向下取整，取不大于要求值的最大的那个整数值
	
    for(int i = 5; i <= n_sqrt; i += 6)
	    if(n % i == 0 | n % (i + 2) == 0)
             return 0;

    return 1;
} 


int main()
{
    int a, d, n, cnt, ad;

    while(cin >> a >> d >> n)
    {
        if(a == 0 && d == 0 && n == 0)
            break;
        
        if(prime_num(a))    cnt = 1;
        else    cnt =0;
        
        ad = a;
        while(cnt != n)
        {
            ad += d;
            if(prime_num(ad))
                cnt++;
        }
        cout << ad << endl;
    }
}