有关求e的近似值的一道题求解和讲解一下看不懂
求解和讲解一下 代码下示,有疑问请追贴,谢谢#include <stdio.h>
#include <stdlib.h>
double f(int x); //定义函数以求阶乘的倒数
int main(void)
{
int n;
double s = 1;
scanf("%d", &n);
while (n) //迭代相加
{
s += f(n);
n--;
}
printf("%.8lf", s);
system("Pause");
return 0;
}
double f(int x)
{
double m, n = 1;
while (x) //m=1/n,n=x!
{
n *= x;
x--;
}
m = 1 / n;
return m;
} #求e的近似值
s = 0
def my_e(n):
global s
if n==0:
s = s + 1
else:
s = s+jc(n)
my_e(n-1)
def jc(n):
temp = 1
for i in range(1,n+1):
temp *= i
return 1/temp
my_e(10)
print(s) 本帖最后由 听风夜雨 于 2020-3-29 17:07 编辑
def g(y):
def f(x): #定义阶乘,零基础学习python递归章节
if x==1:
return 1
else:
return 1/x*f(x-1)
b=1
for x in range(1,y+1):
b+=f(x)
print('%9.8f'%b)
g(10)
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