hdu Goldbach's Conjecture

Problem 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
------------------------------------------------------------
题意大致为：
给一个大于等于4的偶数，将它分解为 两个素数的和，(p1,p2)和(p2,p1)是算做是一对，
求这样的素数对有多少个；
思路：
这样的素数对，一个比n/2大一个比n/2小，
所以，可以遍历1到n/2(区间左开，右闭)，并且两者都是素数即可；
#include<stdio.h>#include<math.h>int nprime[100010];int main(){int isprime(int n);int n,count,i,j;for(i=2;i*i<100010;i++){if(!nprime[2]){for(j=i*i;j<100010;j+=i)nprime[j]=1;}}while(scanf("%d",&n)&&n){count=0;for(i=2;i<=n/2;i++){if(!nprime[i]&&!nprime[n-i])count++;}printf("%d\n",count);}return 0;}
ps : 每次都判断会超时；所以用打表的方法；/*int isprime(int n){int i;if(n==1)return 0;for(i=2;i<=sqrt(n);i++){if(n%i==0)return 0;}return 1;}*/