利用求阶乘函数Fact(),编程计算并输出1! + 2! + …+ n!的值。 **输入格式要求:"%u" 提示信息:"Input n(n>0):" **输出格式要求:"sum = %lu\n" 程序运行示例: Input n(n>0):5 sum = 153
时间: 2024-03-03 22:51:32 浏览: 48
下面是实现该功能的代码:
```c
#include <stdio.h>
unsigned long Fact(unsigned int n);
unsigned long Sum(unsigned int n);
int main()
{
unsigned int n;
printf("Input n(n>0):");
scanf("%u", &n);
printf("sum = %lu\n", Sum(n));
return 0;
}
unsigned long Fact(unsigned int n)
{
unsigned long result = 1;
for(unsigned int i = 1; i <= n; i++)
{
result *= i;
}
return result;
}
unsigned long Sum(unsigned int n)
{
unsigned long result = 0;
for(unsigned int i = 1; i <= n; i++)
{
result += Fact(i);
}
return result;
}
```
在上述代码中,我们定义了两个函数:`Fact()`和`Sum()`,其中`Fact()`用于求n的阶乘,而`Sum()`则用于计算1! + 2! + …+ n!。在主函数中,我们先读入n的值,然后调用`Sum()`函数计算结果并输出。
相关问题
C语言,利用求阶乘函数fact,编程计算并输出1!+2!+...+n!的值
以下是C语言实现计算1!+2!+3!+……+n!的代码,其中利用了求阶乘函数fact(n):
```c
#include <stdio.h>
double fact(int n) {
double res = 1;
for (int i = 1; i <= n; i++) {
res *= i;
}
return res;
}
double fun(int n) {
double res = 0;
for (int i = 1; i <= n; i++) {
res += fact(i);
}
return res;
}
int main() {
int n;
printf("Enter n: ");
scanf("%d", &n);
printf("%.0f\n", fun(n));
return 0;
}
```
利用求阶乘函数Fact(),编程计算并输出1!+2!+……+n!的值
以下是利用求阶乘函数Fact(),编程计算并输出1!+2!+……+n!的值的C语言代码:
```c
#include <stdio.h>
int Fact(int n) {
if (n == 0 || n == 1) {
return 1;
} else {
return n * Fact(n - 1);
}
}
int main() {
int n, sum = 0;
printf("请输入一个正整数n:");
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
sum += Fact(i);
}
printf("1!+2!+...+%d!的值为:%d\n", n, sum);
return 0;
}
```
运行结果如下:
```
请输入一个正整数n:5
1!+2!+...+5!的值为:153
```
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