(Prime number) I-Dirichlet's Theorem on arithmetic progressions (1.5.5)

Time limit:1000 ms Memory limit:65536kb 64bit Io format:% I64d & % i64usubmit status

Description

IfAAndDAre relatively prime positive integers, the arithmetic sequence beginningAAnd increasingD, I. e .,A,A+D,A+ 2D,A+ 3D,A+ 4D,..., Contains infinitely implements 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 invalid prime numbers

2, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89 ,....

Your mission, shocould you decide to accept it, is to write a program to findNTh prime number in this arithmetic sequence for given positive integersA,D, AndN.

Input

The input is a sequence of datasets. a dataset is a line containing three positive integersA,D, AndNSeparated by a space.AAndDAre relatively prime. You may assumeA<= 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 shoshould be composed of as many lines as the number of the input datasets. Each line shoshould contain a single integer and shoshould never contain extra characters.

The output integer corresponding to a datasetA,D,NShocould beNTh prime number among those contained in the arithmetic sequence beginningAAnd increasingD.

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

#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;}