POJ 2262 Goldbach’s Conjecture(素數表)

//驗證哥德巴哈猜想，結論必然正確……故不必要檢測錯誤情況<br />//用篩選法打表<br />//如果i是素數，n-i也是素數，則這兩個數就是分解的結果<br />//複雜度必須O(n)才能過，O(n^2)必定TLE<br />#include<iostream><br />#include<cstring><br />using namespace std;<br />const int MAXP = 10000010;<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;(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 />primeList();<br />int n,a,b;<br />while(scanf("%d",&n) && n != 0)<br />{<br />for(int i = 1;i <= prime[0];++i)<br />{<br />if(isPrime[n - prime[i]])<br />{<br />a = prime[i];<br />b = n - prime[i];<br />break;<br />}<br />}<br />printf("%d = %d + %d/n",n,a,b);<br />}<br />return 0;<br />}<br />