编程求e=1+1/1!+1/2!+1/3!+…+1/n!,直到最后一项小于10的-5次方

#include<stdio.h>
#include<math.h>
int main()
{
    
    
	int i=1;
	double e=1,sum=1,t=1;
	do{
    
    
		sum=sum*i;
		t=1.0/sum;
		e=e+t;
		i++;
	}while(fabs(t)>=1e-5);
	printf("%0.2lf",e);
	return 0;
}