1! + 3! + 5! + …… + n! 的和。 n=eval(input()) s=0
时间: 2023-09-26 09:09:12 浏览: 42
那么可以按照之前给出的方法,在循环中累加每个 i 的阶乘。代码如下:
```python
n = eval(input())
s = 0
factorial = 1
for i in range(1, n+1, 2):
factorial *= i
s += factorial
print(s)
```
其中,n 是输入的数值,s 是累加结果,factorial 是阶乘。在循环中,每次更新 factorial 和 s 的值,最后输出累加结果即可。
相关问题
for c in n: s += eval(c)**2
I'm sorry, as an AI language model, I cannot provide a complete answer as there is no context to this code snippet. However, I can explain what this code does:
This code iterates over each character in the string "n" and evaluates the square of the integer represented by that character using the "eval()" function. The result of each evaluation is then assigned to the variable "s".
For example, if "n" is the string "123", this code would evaluate the squares of the integers 1, 2, and 3, and assign the results to "s". The final value of "s" would depend on the value of "n".
n=eval(input()) sum1=0 while sum1+=n%10 print(sum1)
这段代码中的 while 循环没有条件,会导致无限循环。我猜测代码的意图是对输入的整数 n 进行各个位上数字的求和,可以改成如下代码:
```
n = eval(input())
sum1 = 0
while n > 0:
sum1 += n % 10
n //= 10
print(sum1)
```
这里使用了取余和整除操作,将 n 的个位数字加入 sum1 ,然后将 n 的值除以 10 取整,相当于将 n 的个位数字去掉。当 n 变成 0 时,所有数字都已经被求和完成,输出 sum1 即可。