#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;
}
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
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转载自blog.csdn.net/buxiangquaa/article/details/114988925
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