POJ 2262 Goldbach's Conjecture(素数表)

//验证哥德巴赫猜想，结论必然正确……故不必要检测错误情况//用筛选法打表//如果i是素数，n-i也是素数，则这两个数就是分解的结果//复杂度必须O(n)才能过，O(n^2)必定TLE#include<iostream>#include<cstring>using namespace std;const int MAXP = 10000010;bool isPrime[MAXP];int prime[MAXP];void primeList(){memset(isPrime,true,sizeof(isPrime));for(int i = 2;i <= MAXP;++i){if(isPrime[i])prime[++prime[0]] = i;for(int j = 1,k;(j <= prime[0]) && (k = i * prime[j]) <= MAXP;++j){isPrime[k] = false;if(i % prime[j] == 0)break;}}}int main(){primeList();int n,a,b;while(scanf("%d",&n) && n != 0){for(int i = 1;i <= prime[0];++i){if(isPrime[n - prime[i]]){a = prime[i];b = n - prime[i];break;}}printf("%d = %d + %d/n",n,a,b);}return 0;}