Goldbach`s Conjecture

最新推荐文章于 2021-07-21 16:18:11 发布

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本文介绍了一个程序设计任务：验证哥德巴赫猜想对于不超过10^7的偶数是否成立。通过预处理所有素数并使用这些素数来检查每个测试案例，程序能够找出表达为两个素数之和的所有可能方式。

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Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states:

Every even integer, greater than 2, can be expressed as the sum of two primes [1].

Now your task is to check whether this conjecture holds for integers up to 10⁷.

Input

Input starts with an integer T (≤ 300), denoting the number of test cases.

Each case starts with a line containing an integer n (4 ≤ n ≤ 10⁷, n is even).

Output

For each case, print the case number and the number of ways you can express n as sum of two primes. To be more specific, we want to find the number of(a, b) where

1) Both a and b are prime

2) a + b = n

3) a ≤ b

Sample Input

Sample Output

Case 1: 1

Case 2: 1

Hint

1. An integer is said to be prime, if it is divisible by exactly two different integers. First few primes are 2, 3, 5, 7, 11, 13, ...

求一个偶数由多少对素数之和构成

#include <bits/stdc++.h>

using namespace std;

bool prime[10000010];
int primes[700005];
int sum;
void Init()
{
    sum = 0;
    memset(prime, 1, sizeof(prime));
    memset(primes, 0, sizeof(primes));
    primes[sum++] = 2;
    prime[0] = prime[1] = 0;
    for(int i = 4; i < 10000010; i += 2)
        prime[i] = 0;
    for(long long i = 3; i < 10000010; i += 2)
    {
        if(prime[i])
        {
            primes[sum++] = i;
            for(long long j = i * i; j < 10000010; j += 2 * i)
                prime[j] = 0;
        }
    }
}
int main()
{
    Init();
    int t;
    long long x, ans;
    scanf("%d", &t);
    for (int cases = 1; cases <= t; ++cases)
    {
        scanf("%lld", &x);
        ans = 0;
        for(int i = 0; primes[i] <= x/2; i++)   //预处理出所有的素数，直接比较素数
            if(prime[x - primes[i]])           
                ans++;
        printf("Case %d: %lld\n", cases, ans);
    }
    return 0;
}