HDU - 1397 Goldbach's Conjecture（哥德巴赫猜想）

题目：

Description

Goldbach's Conjecture: 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. 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 2^15. 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

610120

Sample Output

121

这个题目是问1个数n可以有多少种方法表示成2个素数的和，n=p1+p2且p1<=p2

除了n=4的情况，p1和p2肯定是2个奇素数。

代码：

#include<iostream>using namespace std;int prime[32768];int num[32768];void get_prime(){for (int i = 0; i < 32768; i++)prime[i] = 1;for (int i = 2; i < 32768; i++)if(prime[i])for (int j = i + i; j < 32768; j += i)prime[j] = 0;}int main(){get_prime();for (int n = 4; n < 32768; n+=2){int sum = (n == 4);for (int i = 3; i <= n / 2; i += 2)if (prime[i] && prime[n - i])sum++;num[n] = sum;}int n;while (scanf("%d",&n)){if (n == 0)break;printf("%d\n", num[n]);}return 0;}

prime存1或者0，prime[i]=(isprime(i));

get_prime是利用筛法给prime进行初始化。

筛法的效率很高，接近线性时间。

上面的代码是436ms AC，我觉得还得优化。

主要就是num的初始化是θ（n^2）的，所以比较慢。

于是我发现，num的初始化也可以用类似筛法的方法！

改进之后只花了78ms

代码：

#include<iostream>using namespace std;int prime[32768];int num[32768];void get_prime(){for (int i = 0; i < 32768; i++)prime[i] = 1;for (int i = 2; i < 32768; i++)if(prime[i])for (int j = i + i; j < 32768; j += i)prime[j] = 0;}int main(){get_prime();for (int i = 4; i < 32768; i += 2)num[i] = 0;for (int i = 3; i < 32768; i+=2){if (prime[i] == 0)continue;for (int j = i; i + j < 32768; j += 2)if (prime[j])num[i + j]++;}int n;while (scanf("%d",&n)){if (n == 0)break;printf("%d\n", num[n]);}return 0;}