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1.(non-recursive method) defines a function that finds a factor, returns a factorial of n, and calls the function to find the factorial, with the factorial of 0 and 1 both 1 2Recursively n!, note 0 and 1
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If you directly present the family who knows how to ask this question, you can directly bring in the formula of asking me, so that you can come up with it.
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How to ask for this to be cut into this, you should find a professional teacher or tell you about the content of this question.
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That is, how many classes are n? It's very simple to open up and look at this tutorial material, which talks about what it means.
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l Use recursion to achieve. 1. When n=0, n!=1;When n is not equal to 0, n!=n*(n-1)!
2. Define a function f(n) to implement recursion:
3. For example, find the factorial of 5, m= f(5), print(m), and the result is 120.
Find the factorial of n.
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An integerLadder Splitting God VerticalThat is, all those that are less than or equal to that numberPositive integerof the product. For example, the factorial of 3 is 3*2*1.
Computer: win10
Software: ISO
Software: Python
1. Use def** to create a large function, the name is func, and the parameter is ndef func(n):
2. Create a variable res, which is assigned as the parameter n of the function, ** as follows:
res = n。Blind ascension.
3. Then write the for range loop, as follows:
for i in range(1,n):
4. Next, in the for loop.
The calculation is performed and the res are returned, as follows:
res *=i
return res。
5. Printout the factorial of 3 with print, ** as follows:
print(func(3))。
6. The above ** realizes the factorial operation, and we can also use the recursive way. ** Below:
def func1(n):
if n==1:
return 1
else:return n *func1(n-1)print(func1(3))
The recursive way is that the function calls itself.
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def factorial(n):
result = n
for i in range(1,n):
result *= i
return result
def main():
print factorial(4)
if __name__ == '__main__':
main()
Introduction to Factorial:Keystone Carman.
Christian Kramp (1760-1826), an arithmetic notation invented in 1808, is a mathematical term.
A positive integer.
The factorial is the product of all positive integers less than and equal to the number, and the factorial of 0 is 1. Natural number.
The factorial of n is written as n!. In 1808, Keystone Carman introduced this notation.
That is, n!=1×2×3×..n-1)×n。The factorial can also be defined recursively: 0!=1,n!=(n-1)!×n。
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s = 1
for i in range(1, 10): In practice, the case that s,i is not 0 should also be considered in the actual program.
s*= i
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def recursion(n):
if n==1:
return 1
else:return n*recursion(n-1)list= Defines an empty list to append the factorial value generated by calling the recursive function to the list.
print("Write the factorial of 1-10 into the list and use the sum function to sum it") The display effect is obvious.
for i in range(1,11):
print(sum(list)) list.
sum_0=0
The display effect is obvious, center(80,"* Title placement.
print("The for loop directly calls the recursive function to sum".center(80,"*
for i in range(1,11):
sum_0 +=recursion(i)
print(sum_0)
This is the factorial sum of 1-10, you can refer to it!
The result of the output.
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That is, n!=1×2×3×..n-1)×n。The factorial can also be defined recursively: 0!=1,n!Number of drafts = (n-1)! ×n。
Create a function with def ** with the name func and the parameter is ndeffunc(n): create a variable res, and assign the value to the parameter n of the function, ** as follows: res=n.
Factorials are generally implemented recursively, as follows: Effect This function can only recognize integers, and even if you enter a trembling 0, it will report an error.
This problem requires writing a program to calculate the factorial of n. Input format: The input gives a positive integer n in a line.
Output format: Output the factorial value f in the format of "product=f" in one line, note that there is a space to the left and right of the equal sign. The problem ensures that the calculation result does not exceed the double precision range.
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python finds the factorial of n
Solution 1: Loop. The idea is relatively simple, that is, define a variable ns, give an initial value of 1, and then use the for loop to directly multiply to get the final result.
Solution 2: Recursive recursion is also a missing number is easier to understand, when n==2, return2*1;n==3,return3*(2*1);n==4,return4*(3*(2*1))。And so on, and then give the final result to the res to print it.
Both of these methods are relatively simple, but obviously they don't meet the requirements of the problem: "use an array a to represent a large integer a, a[0] represents the single digit of a, and a[1] represents the ten bits of a", so we need to find a way to use the array to get n!results.
Solution 3: Array.
First, define an array of ns to store n!For each digit or the value on Bishan, use the for loop to add 10000 0 values to ns, so that it is convenient to directly operate the array according to the index.
Then define length as the "length of the array" (a true value rather than an auto-added 0), i.e. n!The number of bits of the result. After that, you must also use the for loop for multiplication, but it is different from the direct multiplication shirt of solution 1, here the multiplier (i.e. i) is multiplied by the numbers on each bit respectively, if the result is greater than or equal to 10, then carry > 0 is the forward one digit value is carry, if the j cycle is over, carry > 0 means that you need to advance one place on the "length" of the current ns, so length+1 is the number of digits + 1, here carry plays the role of judging whether to carry, The length represents the number of bits of the result.
No, there should be in the New Year!
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