poj2909

Goldbach's Conjecture

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

610120

Sample Output

121

Source

Svenskt Mästerskap i Programmering/Norgesmesterskapet 2002

求对于给定的数字，有多少组满足哥德巴赫猜想的解。思路比较清楚，暴力枚举即可。

代码如下：

#include<stdio.h>int is_prime(int x) {//判断一个数是否是素数    if (x == 2)//2是素数，直接返回1        return 1;    int flag = 0;    for (int i = 2; i * i <= x; i++) {        if (!(x % i)) {//如果找到了该数的一个因子，那么这个数不是素数，标记后中断循环            flag = 1;            break;        }    }    if (flag)//如果找到了一个因子，说明这个数不是素数，返回0        return 0;    return 1;//没有找到因子，这个数是素数，返回1}int main(void) {    int n;    while (scanf("%d", &n), n) {        int count = 0;        for (int i = 2; i <= n / 2; i++) {            if (is_prime(i) && is_prime(n - i))//如果i和n-i都是素数，找到了一组满足哥德巴赫猜想的解，count的数量+1                count++;        }        printf("%d\n", count);//输出结果    }    return 0;}