| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 16331 | Accepted: 8203 |
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
Source
解题思路：
不说了，这题很水，就是用筛子把素数筛出来，然后从a开始，每次间隔d，在表里查，如果是素数，那么就把n--，直到最后n==0为止，输出p-d就好。。。
代码：
# include<iostream>
# include<cstdio>
# include<cstring>
# include<cmath>
using namespace std;
# define maxn 1000000
bool prime[maxn];
int main(void)
{
memset( prime, true, sizeof(prime) );
for ( int i = 3; i <= sqrt(maxn); i += 2)
{
for ( int j = 3; j <= maxn / i; j += 2 )
{
if ( prime[i] )
prime[i * j] = false;
}
}
for (int i = 4; i <= maxn; i += 2)
prime[i] = false;
prime[1] = prime[0] = false;
int a, d, n;
while (scanf("%d%d%d", &a, &d, &n) != EOF && !(a == 0 && d == 0 && n == 0))
{
int p;
for ( p = a; n > 0; p += d )
{
if ( prime[p] )
n--;
}
printf("%d\n", p - d);
}
return 0;
}