POJ 2909 Goldbach’s Conjecture(素數表)

//這次是驗證哥德巴哈猜想的解有多少個，同樣核心是打素數表<br />//思路和POJ 2262差不多，篩法打素表<br />#include<iostream><br />using namespace std;<br />const int MAXP = 400000;<br />bool isPrime[MAXP];<br />int prime[MAXP];<br />void primeList()<br />{<br />memset(isPrime,true,sizeof(isPrime));<br />for(int i = 2;i <= MAXP;++i)<br />{<br />if(isPrime[i])prime[++prime[0]] = i;<br />for(int j = 1,k;(k = i * prime[j]) <= MAXP && j <= MAXP;++j)<br />{<br />isPrime[k] = false;<br />if(i % prime[j] == 0)break;<br />}<br />}<br />}<br />int main()<br />{<br />primeList();<br />int n,ans;<br />while(scanf("%d",&n) && n != 0)<br />{<br />ans = 0;<br />for(int i = 1;prime[i] <= n/2;++i)<br />{<br />if(isPrime[n - prime[i]])<br />++ans;<br />}<br />printf("%d/n",ans);<br />}<br />return 0;<br />}<br />