您的 d 变量在外循环的每次迭代中都会被重置.将初始化移出该循环.

Your d variable is being reset during each iteration of your outer loop. Move the initialization out of that loop.

此外，a == 2 检查应该在外循环的每次迭代中只发生一次.将其移出内循环.

Additionally, the a == 2 check should only occur once per iteration of the outer loop. Move it out of the inner loop.

b=1
d = 0
#generates a list of numbers.
while b<100:
    b=b+1
    x = 0.0
    a = 0
    #generates a list of numbers less than b. 
    while x<b:
        x=x+1
        #this will check for divisors. 
        if (b/x)-int(b/x) == 0.0:
            a=a+1
    if a==2:
        #if it finds a prime it will add it.
        d=d+b
print d 

结果:

1060

在此期间，让我们尝试清理代码，使其更易于理解.你可以把内循环移到它自己的函数中，让读者更清楚地了解它的用途:

While we're at it, let's try cleaning up the code so it's more comprehensible. You can move the inner loop into its own function, so readers can more clearly understand its purpose:

def is_prime(b):
    x = 0.0
    a = 0
    while x<b:
        x=x+1
        #this will check for divisors. 
        if (b/x)-int(b/x) == 0.0:
            a=a+1
    if a==2:
        return True
    else:
        return False

b=1
d=0
#generates a list of numbers.
while b<100:
    b=b+1
    if is_prime(b):
        d=d+b
print d

使用变量名称来描述它们所代表的内容也很有用:

It's also useful to use variable names that describe what they represent:

def is_prime(number):
    candidate_factor = 0
    amount_of_factors = 0
    while candidate_factor<number:
        #A += B is equivalent to A = A + B
        candidate_factor += 1
        #A little easier way of testing whether one number divides another evenly
        if number % candidate_factor == 0:
            amount_of_factors += 1
    if amount_of_factors == 2:
        return True
    else:
        return False

number=1
prime_total=0
#generates a list of numbers.
while number<100:
    number += 1
    if is_prime(number):
        prime_total += number
print prime_total

for 循环比增加计数器的 while 循环更具有惯用性:

for loops are more idomatic than while loops that increment a counter:

def is_prime(number):
    amount_of_factors = 0
    for candidate_factor in range(1, number+1):
        if number % candidate_factor == 0:
            amount_of_factors += 1
    if amount_of_factors == 2:
        return True
    else:
        return False

prime_total=0
#generates a list of numbers.
for number in range(2, 101):
    if is_prime(number):
        prime_total += number
print prime_total

如果您感觉很大胆，可以使用列表推导式来减少使用的循环次数:

If you're feeling bold, you can use list comprehensions to cut down on the number of loops you use:

def is_prime(number):
    factors = [candidate_factor for candidate_factor in range(1, number+1) if number % candidate_factor == 0]
    return len(factors) == 2

#generates a list of numbers.
primes = [number for number in range(2, 101) if is_prime(number)]
prime_total = sum(primes)
print prime_total