为给定测试用例查找阶乘的程序

Question

Input
An integer t, 1<=t<=100, denoting the number of test cases, followed by t lines, each containing a single integer n, 1<=n<=100.

Output
For each integer n given at input, display a line with the value of n!

I have typed the code below and it doesn't get accepted on codechef.
I don't know why.

#include<stdio.h>
double fact(int n);
void disp(double t);

int main() {
  int a, i, c[100];
  scanf("%d", &a);
  for (i = 0; i < a; i++) {
    scanf("%d", &c[i]);
  }
  for (i = 0; i < a; i++) {
    disp(fact(c[i]));
  }
  return 0;
}

double fact(int n) {
  double f;
  if (n == 1 || n == 0)
    f = (double) 1;
  else
    f = (double) n * fact(n - 1);
  return f;
}

void disp(double t) {
  printf("%f\n", t);
}

Answer 1

I have typed the code below and it doesn't get accepted on codechef.
I don't know why.

OP's code fails as double lacks precision for this task. Given "containing a single integer n, 1<=n<=100." and fact(100) , the printed result needs far more precision than afforded by double .

---

100! is exactly

93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

double (precision) and uint64_t (range) are both insufficient for this task. Need new approach. Something that extends exact integer multiplication.

Perhaps create a function that multiplies a string ? mult_str(char *s, int n) ? "1" * 2 --> "2" .

Do-able in about 30 lines of C code without special libraries.

Answer 2

This seemed like an interesting short program so I wrote it up and submitted it to CodeChef. It was accepted on the first attempt -- not bad for BOTE (Back of the Envelope) code.

/* FCTRL2.c

   CodeChef Problem Code: FCTRL2 (Small Factorials)
*/

#include <stdio.h>

#define LEN         158

int nextFactorial (const int n, char fact[], int limit)
{
    int carry = 0;
    for ( int index = LEN - 1; index >= limit; --index ) {
        int product = fact[index] * n + carry;
        fact[index] = product % 10;
        carry = 0;
        if ( product > 9 ) {
            carry = product / 10;
            if ( index == limit )
                --limit;
        }
    }

    return limit;
}

void displayFactorial (const char fact[], const int limit)
{
    for ( int index = limit; index < LEN; ++index )
        printf ("%c", fact[index] + '0');
    printf ("\n");
}

int main (void)
{
    int count;
    scanf ("%i", &count);

    int n[count];
    for ( int i = 0; i < count; ++i )
        scanf ("%i", &n[i]);

    for ( int i = 0; i < count; ++i ) {
        char fact[LEN] = { [LEN - 1] = 1 }; 
        int limit = LEN - 1;
        for ( int j = 2; j <= n[i]; ++j )
            limit = nextFactorial (j, fact, limit);
        displayFactorial (fact, limit);
    }

    return 0;
}