Question

求 e=1+1/1!+1/2!+1/3!+…… +1/n! 之值，设N=8，请编程。

Answer 1

只给出主要部分
int e=1;
for (int n=1 ; n<=8 ; n++)
{
int a=1
for (int m=1 ; m<=n ; m++)
{
a = a * m;
}
e = e + 1 / a;
}

Answer 2

Private Sub Form_Click()
s = 1
For i = 1 To 8
s = s * i
If i Mod 2 = 0 Then
sum1 = sum1 - 1 / s
Else
sum2 = sum2 + 1 / s
End If
Next i
Sum = 1 + sum1 + sum2
Print Sum
End Sub

Answer 3

main()
{float s=0,t=1;
int n;
for(n=1;n<=8;n++)
{
t=t*n; /*求n！*/
s=s+1/t; /*将各项累加*/
}
printf("s=%e\n",s);
}

Answer 4

应该是考的 递归 的思想吧
很简单啊
#include <iostream.h>
const int N=8;
int jiecheng(int n);
int main()
{
float e=0;
for(int i=1; i<=N; i++)
{
e+=(float)1/jiecheng(i);
}

cout<<e<<endl;
return 0;
}

int jiecheng(int n)
{
if(n==0)
return 1;
return n*jiecheng(n-1);
}

Answer 5

input \"pleaase enter n;\"n
e=1:t=1/i:i=1
while s=s+t
i=i+i
t=t*1/1
wend
print \"s=\";s ;\"i=\";i