a, b = 0, 1 sum = 0 for i in range(30): a, b = b, a + b sum += a #sum = sum + a print(sum)
时间: 2024-04-21 22:28:03 浏览: 10
这也是一段 Python 代码,它实现了斐波那契数列的计算,并计算前 30 个数列元素的和。在第一行代码中,a 和 b 被初始化为 0 和 1。在接下来的循环中,a 和 b 通过交换它们的值来更新它们的值,以便计算下一个数列元素。同时,每次循环中 a 的值被加到 sum 变量中。最后,sum 变量中存储的就是前 30 个数列元素的和。
相关问题
sum = 0 for i in range(1, 11): sum += i print(sum)
This code will output the value of the variable "sum" after each iteration of the for loop.
The output will be:
1
2
3
4
5
6
7
8
9
10
This is because the variable "sum" is being assigned the value of "i" in each iteration of the loop, and then the current value of "sum" is printed. Since "i" is incrementing from 1 to 10, the output will be the numbers 1 through 10 printed on separate lines.
Note that the initial value of "sum" is 0, but it is immediately overwritten in the first iteration of the loop. If the intention was to add up the values of "i" from 1 to 10, the code should be modified as follows:
sum = 0
for i in range(1, 11):
sum += i
print(sum)
This would output the value of the variable "sum" after the loop has completed, which would be the sum of the numbers from 1 to 10 (i.e. 55).
sum=0 for i in range(1,11): sum += i print( sum)
您的代码中,变量sum是初始值为0的累加器,但是在每次循环中,您将i直接赋值给了sum,而没有将i加入到sum中。因此,每次循环结束后,sum的值都将等于i的值。建议您将代码修改为以下形式:
sum = 0
for i in range(1, 11):
sum += i
print(sum)
这样,每次循环中都将i加入到sum中,循环结束后,sum的值就是1到10的和,即55。