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快速筛

2015

poj 2909 Goldbach's Conjecture (哥德巴赫猜想)

·1 min
 水题 写一遍的目的是。。。复习一下快速筛的写法 喵呜 1/************************************************************************* 2 > File Name: code/poj/2909.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年08月22日 星期六 14时25分34秒 6 ************************************************************************/ 7 8#include<iostream> 9#include<iomanip> 10#include<cstdio> 11#include<algorithm> 12#include<cmath> 13#include<cstring> 14#include<string> 15#include<map> 16#include<set> 17#include<queue> 18#include<vector> 19#include<stack> 20#define y0 abc111qqz 21#define y1 hust111qqz 22#define yn hez111qqz 23#define j1 cute111qqz 24#define tm crazy111qqz 25#define lr dying111qqz 26using namespace std; 27#define REP(i, n) for (int i=0;i<int(n);++i) 28typedef long long LL; 29typedef unsigned long long ULL; 30const int inf = 0x3f3f3f3f; 31const int N=1<<16; 32bool not_prime[N]; 33int prime[N]; 34int prime_num; 35int n; 36 37void init(){ 38 not_prime[0] = true; 39 not_prime[1] = true; 40 for ( int i =2 ; i < N ; i++){ 41 if (!not_prime[i]){ 42 prime[++prime_num] = i; 43 } 44 for ( int j = 1 ; j <= prime_num&&i*prime[j]<N ; j++){ 45 not_prime[i*prime[j]] = true; 46 if (i%prime[j]==0) break; 47 } 48 } 49} 50int main() 51{ 52 init(); 53 int n ; 54 while (scanf("%d",&n)&&n){ 55 int ans = 0 ; 56 for ( int i = 2 ; i <= n /2 ; i++){ 57 if (!not_prime[i]&&!not_prime[n-i]){ 58 ans++; 59 } 60 } 61 printf("%d\n",ans); 62 } 63 64 return 0; 65}