求用while和for循环写的求1！+2！+3！+。。。+n！的代码，急 急 急 急 急

Answer 1

int n,m=0;

cin>>n;

for（int i=1;i<=n;i++)

m+=i;
cout<<m;
后面自己写，也就是不从完整，就那几条语句
int n,m=0,i=1;
while(i>n)
{
m+=i;
i++;
}
cout<<m;

追问

用IDL语言写出来会吗？要求使用IDL写 谢谢了 急啊

追答

刚刚收出个类似的 希望对你有用
sum = 0L
i = 1
WHILE i LE 1000DO BEGIN
sum = sum + i
i++
ENDWHILE
PRINT, ' 1 + 2 + ..+1000 = ', sum
END
for的没收出来

Answer 2

while循环

# include <stdio.h>
double Factorial (double n);
void main()
{
double i;
double n;
printf ("请输入n\n");
scanf ("%lf",&n);
double summent;
double sum;
i = 1.0;
summent = 0;
while (i <= n)
{
sum = Factorial (i);
summent = summent + sum;
i++;
}
printf ("1到n的阶乘和为：");
printf ("%lf\n",summent);
}
double Factorial (double n)
{
int i;
double sum;
sum = 1;
for (i = 1;i <= n;i++)
{
sum = sum * i;
}
return sum;
}

for循环

# include <stdio.h>
double Factorial (double n);
void main()
{
double i;
double n;
printf ("请输入n\n");
scanf ("%lf",&n);
double summent;
double sum;
summent = 0;
for (i = 1;i <= n;i++)
{
sum = Factorial (i);
summent = summent + sum;
}
printf ("1到n的阶乘和为：");
printf ("%lf\n",summent);
}
double Factorial (double n)
{
int i;
double sum;
sum = 1;
for (i = 1;i <= n;i++)
{
sum = sum * i;
}
return sum;
}

Answer 3

#include<stdio.h>
int main()
{
int n,num=0,i,x=1;
while(1){
printf("input n:");
scanf("%d",&n);
x=1;
num=0;
if(n<=0||n>12)
break;
for(i=1;i<=n;i++){
x*=i;
num+=x;
}
printf("阶乘为：%d\n",num);
}
return 0;
}

由于int变量只能算到12，所以我设置大于12直接退出

Answer 4

#include<stdio.h>
main()
{
int n,k,c;
long int sum=0;
printf("输入你需要的阶层:");
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
c=1;
for(k=1;k<=i;k++)
c=c*k;
sum+=c;
}
printf("阶层的结果=%d",sum);
}
这是for的部分
#include<stdio.h>
main()
{
int n,k,c,i=1;
long int sum=0;
printf("输入你需要的阶层:");
scanf("%d",&n);
while(i<=n)
{

c=1;
for(k=1;k<=i;k++)
c=c*k;
sum+=c;
i++;
}
printf("阶层的结果=%d",sum);
}
这是while的部分

Answer 5

#include<stdio.h>
int jc(int n)
{
int i,s=1;
for(i=2;i<=n;i++)
{
s=s*i;
}
return s;
}
int main()
{
int i,n,s=0;
printf("n=");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
s=s+jc(i);
}
if(n==1)
{
printf("1!=%d\n",s);
}
else if(n==2)
{
printf("1!+2!=%d\n",s);
}
else
printf("1!+2!+...+%d!=%d\n",n,s);
}