C语言求1+2！+3！+...+20!的和。使用while循环语句以及递推算法

Answer 1

#include <stdio.h>
void main()
{
int i=1,j=1,k=1,sum=0;
while(i<=20)
{ while(j<=i)
{
k*=j;
j++;
sum+=k;
}
i++;
}
printf("%d\n",sum);
}

Answer 2

这个是你需要的程序，我在vc6下调试运行正确了，其中用递归计算n！，用while来计算总的和
#include<stdio.h>
void main()
{
long int sum=0;
long int count(int n);
int i=1;
while(i<=20)
{
sum+=count(i);
i++;
}
printf("result=%ld",sum);
}
long int count(int n)
{
if(n>1)
return count(n-1)*n;
else
return 1;
}

Answer 3

1+2！+3！+...+20!

long sum = 0;
int i = 1;
long fac = 1;
while(i <= 20){
fac = fac * i;
sum = sum + fac;
i++;
}

printf(sum);

---------------------------------------------------递归
long sum =0;
int i = 1;
while(i <= 20){
sum = sum + calcFac(i);
i++;
}

long calcFac(int num){
if(num == 1) {
return 1;
else{
return calcFac(num-1) * num;
}

}

Answer 4

#include <stdio.h>
void main()
{
int i=1,j=1,k=1,sum=0;
while(i<=20)
{ while(j<=i)
{
k*=j;
j++;
sum+=k;
}
i++;
}
printf("%d\n",sum);
}
vc6.0调试过正确

Answer 5

<stdio.h>
#include <conio.h>
void main()
{
float n,s=0,t=1;
for(n=1;n<=20;n++)
{
t*=n;
s+=t;
}
printf("1+2!+3!...+20!=%e\n",s);
getch();
}