POJ3006:Dirichlet’s Theorem on Arithmetic Progressions

Description

If a and d are relatively prime positive integers, the arithmetic sequence beginning witha 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 thenth prime number in this arithmetic sequence for given positive integersa,
d, and n.

Input

The input is a sequence of datasets. A dataset is a line containing three positive integersa,
d, and n separated by a space. a and d are relatively prime. You may assumea <= 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 thenth prime number among those contained in the arithmetic sequence beginning witha 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

//題意很簡單，找出以a開始,以b遞增的第n個素數，果斷打表

#include <iostream>#include <string.h>using namespace std;const int maxn = 1000005;int num[maxn],i,j;int main(){    int a,d,n;    memset(num,1,sizeof(num));    num[1] = num[0] = 0;    for(i = 2; i<=maxn/2; i++)    {        if(num[i])        {            for(j = i+i; j<maxn; j+=i)            {                num[j] = 0;            }        }    }    while(cin >> a >> d >> n)    {        if(!a && !d && !n)        {            break;        }        for(i = a;n;i+=d)        {            if(num[i])            {                n--;            }        }        cout << i-d << endl;    }    return 0;}