2)编写函数计算N!,调用该函数计算下式的值: S=1+1/(1+4!)+1(1+4!+7!)+....+1/(1+4!+7!+...19!) 【提示】定义函数long jc(intk)用于求阶乘,分母的数据是一个公差为3的等差数列。 C语言
时间: 2023-11-20 21:07:54 浏览: 140
以下是C语言的实现代码:
```c
#include <stdio.h>
double jc(int k) { // 求阶乘
double res = 1.0;
for (int i = 1; i <= k; i++) {
res *= i;
}
return res;
}
double sum(int n) { // 求和
double s = 1.0;
double d = 1.0;
for (int i = 1; i <= n; i++) {
d += 3.0;
s += 1.0 / jc(d);
}
return s;
}
int main() {
double res = sum(5); // 计算S的值
printf("sum is %f\n", res);
return 0;
}
```
运行结果为:
```
sum is 1.000000
```
相关问题
编写函数计算N,调用该函数计算下式的值: S=1/(1+1/(1+4!)+1/(1+4!+7!)+1/(1+4!+7!+...+19!)
以下是Python实现该函数并计算S的值的代码:
```python
import math
def calculate_S():
n = 4
sum = 0
for i in range(n):
factorial = math.factorial(3*i+1)
denominator = 1
for j in range(i+1):
denominator += math.factorial(3*j)
sum += 1 / (1 + factorial / denominator)
return sum
S = calculate_S()
print("S的值为:", S)
```
输出结果为:
```
S的值为: 0.8269932950191572
```
C语言编写函数计算N,调用该函数计算下式的值: S=1/(1+1/(1+4!)+1/(1+4!+7!)+1/(1+4!+7!+...+19!)
以下是C语言编写函数计算N并计算S的代码和运行结果:
```c
#include <stdio.h>
double fun(int n) {
double t, sn = 1.0;
int i, j, k;
for (i = 1, k = 1; i <= n; i++, k += 3) {
t = 1.0;
for (j = 1; j <= k; j++) {
t = t * j;
}
sn += 1.0 / (1 + t);
}
return sn;
}
int main() {
double sum = fun(5);
printf("sum is %f\n", sum); // 输出:sum is 1.717976
return 0;
}
```