求1到5阶乘之和 求1到5阶乘之和1⽅法⼀：2 s = 1 3 sum = 0 4for i in range(1,6):5 s = s*i 6 sum += s 7print(s)8print(sum)1⽅法⼆：效率低，每次都要重新算k 2 3 n = 0 4for i in range(1,6):5 k = 1 6for j in range(1, i + 1):7 k =...
求1到5阶乘之和 1方法一:2s = 13sum =04foriinrange(1,6):5s = s*i6sum +=s7print(s)8print(sum) 1方法二:效率低,每次都要重新算k23n =04foriinrange(1,6):5k = 16forjinrange(1, i + 1):7k = k *j8n = n +k9print(n) 为什么要坚持,想一想当初!
include <stdio.h>void main(){int i,k=1,s=0;for(i=1;i<=5;i++){k*=i;s+=k;}printf("1!+2!+3!+4!+5!=%d",s);}
#求1到5阶乘之和 # a = 1 sum = 0 for i in range(1,6): a = i*a sum = sum+a print(sum)
Private Sub Command1_Click()Dim s As Double, ss As String, i As Integer For i = 1 To 5 s = s + jc(i)ss = ss & i & "!+"Next i Print Mid(ss, 1, Len(ss) - 1) & "=" & s End Sub Function jc(n As Integer)jc = 1 For i = 1 To n jc = jc * i ...
通过C语言编程，我们可以求解奇数阶乘之和，比如1！，3！，5！一直到19！的和。首先，我们需要定义一个主函数，函数的实现代码如下：在主函数中，首先声明三个整型变量，n表示阶乘的终止值，i表示内层循环的计数变量，sum用于存储奇数阶乘之和。接下来，使用一个for循环，循环条件为n从1递增到19，每次...
编写函数求1~n的阶乘之和。例如,若n值为5,则结果为153 相关知识点: 试题来源: 解析 # include int fun(int n) { int i,f=1,s=0; for(i=1;i<=n;i++) {f=f*i; s=s+f;} return s; } void main() { int n; scanf("%d",&n); printf("%d ",fun(n)); }...
不需要用fact函数，用 fact函数这题就没有意义：Sub xxx() For i = 1 To 5 '变量 i从1到5 s = 1 '初始值1,从1开始乘 For j = 1 To i '变量 j 从1 到 i (i分别=1,2,3,4,5) s = s * j '阶乘结果 1*...*i (1*1,1*2,1*2*3 Next jc =...
“从键盘输入n,求1+2!+3!+...+n!的和” 对于此题,我们可以用定义一个函数来解决,接着用一个for循环语句来设置从1到n,接下来一起来编写这个代码吧。 解决方案 假定这个函数名称为f def f(x): f = 1 for i in range(1,x+1): f *= i return f n = i...
'先写一个求阶乘的函数 Private Function factorial(ByVal n As Integer) As Long Dim i As Integer, l As Long l = 1 For i = 1 To n l = l * i Next factorial = l End Function '添加一个按钮控件 Private Sub Command1_Click()Dim result As Long, i As Integer result = 0 ...