POJ_3006

一.题目

Dirichlet's Theorem on Arithmetic Progressions

Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 15988
Accepted: 8029

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, andn <= 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 151179 10 203271 37 39103 230 127 104 185253 50 851 1 19075 337 210307 24 79331 221 177259 170 40269 58 1020 0 0

Sample Output

928096709120371039352314503289942951074127172269925673

Source

二.解题技巧

    这道题没有什么技巧，就是按照要求计算就可以了。只不过在判断一个数是否为质数的时候，要考虑输入有1的情况。

三.实现代码

#include <iostream>#include <math.h>using namespace std;bool IsPrime(int Value){    if (Value == 1)    {        return false;    }    if ((Value == 2) || (Value == 3))    {        return true;    }    if (0 == (Value % 2))    {        return false;    }    for (int Index = 3; Index <= sqrt((double(Value))); Index = Index + 2)    {        if (0 == (Value % Index))        {            return false;        }    }    return true;}void GetPrimer(int a, int d, int n){    int Index = 0;    while(true)    {        if (IsPrime(a))        {            Index++;        }        if (Index == n)        {            break;        }        a += d;    }    cout << a << endl;}int main(){    int a = 0, d = 0, n = 0;    while(true)    {        cin >> a;        cin >> d;        cin >> n;        if ((a == 0) && (d == 0) && (n == 0))        {            break;        }        GetPrimer(a, d, n);    }    return 0;}

四.体会

    普通的一道题。

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