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

最新推荐文章于 2021-07-21 16:18:11 发布

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最新推荐文章于 2021-07-21 16:18:11 发布

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原文链接：http://www.cnblogs.com/111qqz/p/4750252.html

本文介绍了一个解决高斯猜想的算法，即对于任意大于等于4的偶数n，至少存在一对质数p1和p2，使得n = p1 + p2。通过快速筛法，该文提供了一种有效计算给定偶数下满足条件的质数对数量的方法。详细步骤包括初始化筛法数组、标记非质数以及遍历筛选出符合条件的质数对。

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Goldbach's Conjecture

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10364		Accepted: 6143

Description

For any even number n greater than or equal to 4, there exists at least one pair of prime numbers p₁ and p₂ such that

n = p₁ + p₂

This conjecture has not been proved nor refused yet. No one is sure whether this conjecture actually holds. However, one can find such a pair of prime numbers, if any, for a given even number. The problem here is to write a program that reports the number of all the pairs of prime numbers satisfying the condition in the conjecture for a given even number.

A sequence of even numbers is given as input. There can be many such numbers. Corresponding to each number, the program should output the number of pairs mentioned above. Notice that we are interested in the number of essentially different pairs and therefore you should not count (p₁, p₂) and (p₂, p₁) separately as two different pairs.

Input

An integer is given in each input line. You may assume that each integer is even, and is greater than or equal to 4 and less than 2¹⁵. The end of the input is indicated by a number 0.

Output

Each output line should contain an integer number. No other characters should appear in the output.

Sample Input

Sample Output

1
2
1

Source

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
26 using namespace std;
27 #define REP(i, n) for (int i=0;i<int(n);++i)  
28 typedef long long LL;
29 typedef unsigned long long ULL;
30 const int inf = 0x3f3f3f3f;
31 const int N=1<<16;
32 bool not_prime[N];
33 int prime[N];
34 int prime_num;
35 int n;
36 
37 void 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 }
50 int 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 }