哥德巴赫猜想(使用Python从0开始实现)

1. 首先验证是否为素数.(是素数返回True; 否则返回False)

def prime_number(N):
    count = 0
    for i in range(2, N):
        if N % i == 0:
            count += 1
    if count == 0:
        return True
    else:
        return False

2. 当作练手, 传入你想知道的0到N的范围内有几个素数, 且都是哪些素数.

    def All_prime_number(N):
        count = 0
        for i in range(2, N + 1):
            if prime_number(i) == True:
                count += 1
                print(i, end="t")
        print()
        print(f"一共有{count}个素数")

3. 哥德巴赫猜想

    def Goldbach_conjecture(N):
        if N % 2 == 0 and N > 2:
            for i in range(2, N + 1):
                for j in range(i, N + 1):
                    if prime_number(i) == True and prime_number(j) == True and i + j == N:
                        print(f"{i} + {j} = {N}")
        else:
            print("请输入大于2的偶数！")

4. 方便读者获取代码

# 哥德巴赫猜想：任一大于2的偶数都可写成两个质数之和
# 质数：质数是指在大于1的自然数中，除了1和它本身以外不再有其他因数的自然数。
    def prime_number(N):
        count = 0
        for i in range(2, N):
            if N % i == 0:
                count += 1
        if count == 0:
            return True
        else:
            return False

    def All_prime_number(N):
        count = 0
        for i in range(2, N + 1):
            if prime_number(i) == True:
                count += 1
                print(i, end="t")
        print()
        print(f"一共有{count}个素数")

    def Goldbach_conjecture(N):
        if N % 2 == 0 and N > 2:
            for i in range(2, N + 1):
                for j in range(i, N + 1):
                    if prime_number(i) == True and prime_number(j) == True and i + j == N:
                        print(f"{i} + {j} = {N}")
        else:
            print("请输入大于2的偶数！")

    N = int(input("请输入大于2的偶数:"))
    Goldbach_conjecture(N)

当输入N为100时候