题目1440：Goldbach's Conjecture

最新推荐文章于 2021-07-21 16:18:11 发布

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本文探讨了高斯猜想，即对于任意大于等于4的偶数，至少存在一对质数使得该偶数等于这两质数之和。通过编程实现，本文展示了如何为给定的偶数找到满足条件的质数对数量，并处理了输入输出细节。

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题目描述：

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.

输入：

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.

输出：

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

样例输入：

样例输出：

1
2
1

代码：

#include <stdio.h>
int prime[100000];
int primeSize;
bool num[100000];
void init() {
    int i;
    for(i=0;i<100000;i++)
        num[i] = false;
    primeSize = 0;
    for(i=2;i<100000;i++) {
        if(num[i]) continue;
        prime[primeSize++] = i;
        int j;
        for(j=2*i;j<100000;j+=i)
            num[j] = true;
    }
}

int main() {
    int n;
    init();
    while(scanf("%d",&n)!=EOF) {
        if(n == 0) break;
        int i,j,count=0;
        for(i=0;i<primeSize&&prime[i]<n;i++) 
            for(j=i;j<primeSize&&prime[j]<n;j++) {
                if(prime[i]+prime[j] == n)
                    count++;
            }
        printf("%d\n",count);
    }
    return 0;
}