After learning the loop, I came across the problem of writing factorial of N, which caused a series of problems, including some common pitfalls for beginners, and how to simplify the code

1. Calculation n The factorial
2. Calculation 1！+2！+3！…+n！

1 Calculation n The factorial

#include <stdio.h>

int main()
{
	// Input 
	int n = 0;
	scanf("%d", &n);
	// produce 1~n The number of 
	int i = 0;
	int ret = 1;
	for (i = 1; i <= n; i++)
	{
		ret = ret * i;
	}
	printf("%d\n", ret);
	return 0;
}

This is very simple , Things that can be solved in one cycle . Let's look at the train of thought ,n The factorial （ Here 3 give an example ）3 The factorial of is 123 , Let's give a cycle , produce 1 To 3 Number of numbers , Then multiply it by the variable you define ret in （ Note that there ret The initial value of can only be set to 1, Initial value if 0, Then any number multiplied by the past is 0 了 , You know what ）

2 Calculation 1！+2！+3！…+n！

2.1 Look at the code that hides the pit

#include <stdio.h>

int main()
{
	// Input 
	int n = 0;
	scanf("%d", &n);
	
	int i = 0;
	int ret = 1;
	int sum = 0;
	int j = 0;
	for (j = 1; j <= n; j++)
	{
		for (i = 1; i <= j; i++)
		{
			ret = ret * i;
		}
		sum = sum + ret;
	}

	printf("%d\n", sum);
	return 0;
}

Look at the results

Here it comes out 1 To 3 The sum of the factorials of is 15,, But actually it should be 9. What's the reason ？ Many people are the result of the first question above , Simply and roughly put a loop on the outside and add them up . But the problem is , Namely When the inner circulation is carried out once , When entering ret Is initialized to 1, Lead to calculation 3 When we take the factorial of the ,ret The initial value is 2, So it leads to mistakes

2.2 Point out the pit , And give the correct code

#include <stdio.h>

int main()
{
	// Input 
	int n = 0;
	scanf("%d", &n);

	int i = 0;
	int ret = 1;
	int sum = 0;
	int j = 0;
	for (j = 1; j <= n; j++)
	{
		ret = 1;// The solution is to add this sentence ！！！
		for (i = 1; i <= j; i++)
		{
			ret = ret * i;
		}
		sum = sum + ret;
	}

	printf("%d\n", sum);
	return 0;
}

2.3 Code simplification

First, simplify the code

#include <stdio.h>

int main()
{
	// Input 
	int n = 0;
	scanf("%d", &n);

	int ret = 1;
	int sum = 0;
	int j = 0;
	for (j = 1; j <= n; j++)
	{
		ret = ret * j;
		sum = sum + ret;
	}

	printf("%d\n", sum);
	return 0;
}

Observe carefully 2.2 Code for , You will find that in these two cycles , A lot of things have been calculated repeatedly , So in simplified code , It can be adjusted .
Here are some examples to help understand , At the end of the calculation 1 After the factorial of ,2 The factorial of only needs to be in 1 On the basis of factorial of 2 Just go , And so on ,3 The factorial of is 2 The result of the factorial of is multiplied by 3. So after giving up a cycle , Code efficiency has greatly improved

3 Summary

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