本题要求实现一个计算非负整数阶乘的简单函数,并利用该函数求 1!+2!+3!+...+n! 的值。
函数接口定义:
double fact( int n );
double factsum( int n );
函数fact
应返回n
的阶乘,建议用递归实现。函数factsum
应返回 1!+2!+...+n
! 的值。题目保证输入输出在双精度范围内。
裁判测试程序样例:
#include <stdio.h>
double fact( int n );
double factsum( int n );
int main()
{
int n;
scanf("%d",&n);
printf("fact(%d) = %.0f\n", n, fact(n));
printf("sum = %.0f\n", factsum(n));
return 0;
}
/* 你的代码将被嵌在这里 */
输入样例1:
10
输出样例1:
fact(10) = 3628800
sum = 4037913
输入样例2:
0
输出样例2:
fact(0) = 1
sum = 0
1 #include <stdio.h> 2 double fact(int n); 3 double factsum(int n); 4 5 int main() { 6 int n; 7 8 scanf("%d", &n); 9 printf("fact(%d) = %.0f\n", n, fact(n)); 10 printf("sum = %.0f\n", factsum(n)); 11 12 return 0; 13 } 14 double fact(int n) { 15 double res, sum; 16 if (n == 1 || n == 0) { 17 res = 1; 18 } else { 19 res = n * fact(n - 1); 20 } 21 return res; 22 } 23 double factsum(int n) { 24 double sum = 0; 25 if (n == 1) 26 sum = 1; 27 else if (n == 0) 28 sum = 2; 29 else { 30 sum = fact(n) + factsum(n - 1); 31 } 32 33 return sum; 34 }