C language introduction practice (12): find the value of natural constant e

This is a 《C Language introduction practice 》 Series No 12 piece .
Last one ：C Language introduction practice (11)： Enter a set of positive integers , Find the sum of inverse numbers

---

subject

Using the formula ：

$$e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}+...$$

Find the natural constant e, The control accuracy is required to be 0.000001, Please program to solve this problem .

requirement

There is no need to enter , Results output only e The value of the can .

Reference code

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
    
	float e = 1;// The first 1 term ,1/0!
	float d = 1;
	
	int i;
	int pro = 1;// The denominator  
	for(i=1;d>=1e-6;i++) {
    
		pro *= i;
		d = (float)1/pro;
		e += d;
	}
	printf("%f", e);//2.718282
	return 0;
}

Code duplicate

At first ,e Initialize to the value of the first item , And then use it n! = n * (n-1)! The nature of ,for In the cycle, calculate the denominator of each item starting from the second item .