{
long i,t=1
double x,e=1.0
scanf("%f",&x)
for (i=1i<=15i++)
{
t=t*i
e=e+1.0/t
}
printf("e=%10.8f\n",e)
}
=============================
double f1(double x, int n)
{
int i
double x1=1.0
for(i=1i<=ni++)
x1=x1*x
return x1
}
long f2(int n)
{
int i
long t=1
for(i=1i<=ni++)t=t*i
return t
}
main()
{
int i
double x,e=1.0
scanf("%lf",&x)
for (i=1i<=15i++)
{
e=e+f1(x,i)/f2(i)
}
printf("e=%10.8f\n",e)
}
c语言, 单独求e可以写为 exp(1), 语法如下：double exp(double x)求欧拉常数e的x次方
double log(double x)求x的自然对数
