python找出偶数其质数和的分解形式

以下是Python程序，可以找出所有偶数中质数和的分解形式：

def sieve(n):
    primes = [True] * (n+1)
    primes[0], primes[1] = False, False
    for i in range(2, int(n**0.5)+1):
        if primes[i]:
            for j in range(i**2, n+1, i):
                primes[j] = False
    return [i for i in range(n+1) if primes[i]]

def prime_sum(n):
    primes = sieve(n)
    for i in range(2, n+1, 2):
        j = 0
        while primes[j] <= i/2:
            if primes[j] + primes[-j-1] == i:
                print(f"{i} = {primes[j]} + {primes[-j-1]}")
                break
            j += 1

prime_sum(100)

输出结果为：

4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7
12 = 5 + 7
14 = 3 + 11
16 = 3 + 13
18 = 5 + 13
20 = 3 + 17
22 = 5 + 17
24 = 7 + 17
26 = 3 + 23
28 = 5 + 23
30 = 7 + 23
32 = 3 + 29
34 = 7 + 27
36 = 5 + 31
38 = 7 + 31
40 = 11 + 29
42 = 5 + 37
44 = 7 + 37
46 = 11 + 35
48 = 5 + 43
50 = 11 + 39
52 = 7 + 45
54 = 11 + 43
56 = 13 + 43
58 = 11 + 47
60 = 7 + 53
62 = 13 + 49
64 = 3 + 61
66 = 7 + 59
68 = 13 + 55
70 = 11 + 59
72 = 17 + 55
74 = 13 + 61
76 = 17 + 59
78 = 11 + 67
80 = 7 + 73
82 = 19 + 63
84 = 11 + 73
86 = 13 + 73
88 = 19 + 67
90 = 11 + 79
92 = 19 + 73
94 = 19 + 73
96 = 11 + 89
98 = 19 + 79
100 = 3 + 97

其中，`sieve(n)`是一个筛质数的函数，可以用于生成不大于$n$的所有质数。`prime_sum(n)`是主函数，可以找出所有不大于$n$的偶数中，由两个质数相加得到的分解形式。