问题
how to get the value of an integer x, indicated by x!, it is the product of the numbers 1 to x.
Example: 5! 1x2x3x4x5 = 120.
int a , b = 1, c = 1, d = 1;
printf("geheel getal x = ");
scanf("%d", &a);
printf("%d! = ", a);
for(b = 1; b <= a; b++)
{
printf("%d x ", c);
c++;
d = d*a;
}
printf(" = %d", d);
回答1:
how to get the som of an integer x, indicated by x!, is the product of the numbers 1 to x.
Did you mean factorial of x ?
Change d = d*a; to d = d*b inside the loop
回答2:
You can simply do:
for(b = 1; b <= a; b++) {
d *= b;
}
// d now has a!
回答3:
This is the optimal implementation in size and speed:
int factorial(int x)
{
static const int f[13] = { 1, 1, 2, 6, 24, 120, /* ... */ };
if ((unsigned)x < (sizeof f/sizeof f[0])) return f[x];
else return INT_MAX+1; /* or your favorite undefined behavior */
}
Hint: x! (x factorial) does not fit in an int except for very very small values of x.
回答4:
Try
d = d * b;
instead of
d = d * a
and it should work fine
回答5:
You actually have a lot of redundant code there, that might be why you did not spot the error yourself.
To calculate the factorial, you only need the accumulator (d in the above code) and the input (a). Why?
回答6:
My code is not good as other but it works for me:
#include <iostream>
using namespace std;
unsigned int fattoriale (int n){
if (n == 1){
return 1;
}
else {
return n * fattoriale(n-1);
}
}
int main() {
int tmp, num;
cin >> num;
tmp = fattoriale(num);
cout << "Stampo il fattoriale del numero inserito: " << tmp << endl;
}
回答7:
int factorial(int x)
{
int f;
if (x == 0)
{
f = 1;
}
else if (x > 0)
{
f = x*factorial(x-1);
}
return f;
}
int main()
{
int n = 0;
cout << factorial(n);
return 0;
}
来源:https://stackoverflow.com/questions/3786207/howto-compute-the-factorial-of-x