原題網址：http://poj.org/problem?id=3006
題目描述：
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 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
92809 6709 12037 103 93523 14503 2 899429 5107 412717 22699 25673
大致題意：
給定一個素數a和d，則a+d，a+2d.....a+nd中存在無窮多的素數，輸入a，d和m，輸出a+nd數列中的第m個素數。大致看下輸入輸出樣例就可以明白了。
題解思路：
這個題如果一個一個判斷的話絕對會超時2333，所以就要找別的辦法了。看題目中給的很特殊的數據範圍，還有一句話“輸出結果一定不大於10^6”，這意圖就很明顯了，打表哇！先生成素數表，然後拿着a+nd和素數表比較來判斷是否是素數，這樣就絕對不會超時了。
下面是代碼：
#include<iostream>
#include<cstring>
using namespace std;
#define maxn 1000000
const int n = 1000000;
bool isprime[n + 1];
int primenum[maxn], cnt;
void Euler() //歐拉篩，生成素數表
{
memset(isprime, true, sizeof(isprime));
isprime[1] = false;
for (int i = 2; i <= n; i++)
{
if (isprime[i])primenum[++cnt] = i;
for (int j = 1; j <= cnt && i*primenum[j] <= n; j++)
{
isprime[i*primenum[j]] = false;
if (i%primenum[j] == 0)break;
}
}
}
int main()
{
Euler();
int a, d, n;
while (cin >> a >> d >> n)
{
if (a == 0 && d == 0 && n == 0)
{
break;
}
int flag = 0;
int count = 0;
while (count != n)
{
while (primenum[flag] < a)
{
flag++;
}
if (a == primenum[flag])
{
count++;
}
a += d;
}
cout << primenum[flag] << endl;
}
return 0;
}