筛素数——POJ 2909 Goldbach's Conjecture-优快云博客

本文探讨了哥德巴赫猜想，一个未被证实也未被否定的数学谜题。通过编程实现，验证了该猜想对于特定范围内偶数的有效性。具体地，文章介绍了如何筛选素数，并使用这些素数来寻找满足猜想条件的素数对。

Goldbach's Conjecture

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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

题意：将一个2¹⁵以内的的偶数表示为两个素数的和，计算出能够表示的种数。
题解：2¹⁵=1024*32=32768，先筛出这个范围内的素数，然后从最小的素数开始取，判断N减去所取的素数的差是不是素数，到N/2停止。

/*POJ 2909 Goldbach's Conjecture*/
/*筛素数*/
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn=32768+100;
bool isprime[maxn];
int prime[maxn];
void getprime()
{
    memset(isprime,1,sizeof(isprime));
    isprime[0]=isprime[1]=0;
    for(int i=2;i<maxn;i++)
    {
        if(isprime[i])
            for(int j=2*i;j<maxn;j+=i)
                isprime[j]=false;
    }
    int cnt=0;
    for(int i=0;i<maxn;i++)
    {
        if(isprime[i])
            prime[cnt++]=i;
    }
}

int main()
{
    getprime();
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)    break;
        int ans=0;
        for(int i=0;;i++)
        {
            if(prime[i]>(n/2))   break;
            if(isprime[n-prime[i]])
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}