POJ 3006 Dirichlet’s Theorem on Arithmetic Progressions(素數表)

//簡單題。篩選法打素數表。<br />#include<iostream><br />using namespace std;<br />const int MAXP = 1100000;<br />bool isPrime[MAXP];<br />int prime[MAXP];<br />void primeList()<br />{<br />memset(isPrime,true,sizeof(isPrime));<br />isPrime[1] = false;<br />for(int i = 2;i <= MAXP;++i)<br />{<br />if(isPrime[i])prime[++prime[0]] = i;<br />for(int j = 1,k;(j <= prime[0]) && (k = i * prime[j]) <= MAXP;++j)<br />{<br />isPrime[k] = false;<br />if(i % prime[j] == 0)break;<br />}<br />}<br />}<br />int main()<br />{<br />//freopen("in.txt","r",stdin);<br />primeList();<br />int a,d,n,ans;<br />while(scanf("%d%d%d",&a,&d,&n) && a != 0)<br />{<br />for(int i = a;n > 0;i += d)<br />{<br />if(isPrime[i])--n;<br />if(n == 0)ans = i;<br />}<br />printf("%d/n",ans);<br />}<br />return 0;<br />}