可以计算到10000的阶乘的程序，很快

更快的算法涉及到二分区间，以haskell代码描述之
let facti n m = if (n >= m) then m else (facti n (n + (div (m - n) 2)) * (facti (n + 1 + (div (m - n) 2)) m))
let fact n = if (n <= 1) then 1 else (facti 1 n)
结果验证
Prelude> fact 63
1982608315404440064116146708361898137544773690227268628106279599612729753600000000000000
Prelude> product [1..63]
1982608315404440064116146708361898137544773690227268628106279599612729753600000000000000
C要实现得先实现O(nlogn)的大数乘法

我会写这个的时候我还不会算复杂度～

呆梦你怎么不放你的运行时间

#include <stdio.h>
#include <stdlib.h>
typedef struct _
{
struct _*p, *n;
unsigned short s[1000];
} num;
int main(void)
{
num *a, *b, *c, *d;
unsigned long e;
long i, j, n;
unsigned short *f, *g;
printf("Enter number:");
scanf("%d", &n);
a = (_*)malloc(sizeof(num));
b = (_*)malloc(sizeof(num));
c = b;
a->p = 0;
a->n = 0;
b->p = 0;
b->n = 0;
b->s[0] = 1;
b->s[1] = 0xFFFF;
for (i = 2; i <= n; i++)
{
c = a;
a = b;
d = a;
f = d->s;
b = c;
g = b->s;
e = 0;
j = e;
for (; d->p || 0xFFFF != *f;)
{
e += i * (*f);
*f++;
*g++ = e % 10000;
e /= 10000;
j++;
if (j > 1000)
{
if (!c->p)
{
c->p = (_*)malloc(sizeof(num));
c->p->p = 0;
c->p->n = c;
}
c = c->p;
g = c->s;
j = 0;
d = d->p;
f = d->s;
}
}
for (; e;)
{
*g = e % 10000;
*g++;
e /= 10000;
j++;
if (j >= 1000)
{
if (!c->p)
{
c->p = (_*)malloc(sizeof(num));
c->p->p = 0;
c->p->n = c;
}
c = c->p;
g = c->s;
j = 0;
}
}
*g = 0xFFFF;
}
for (n = j = 0; (n < 1000) && (c->s[n] != 0xFFFF); n++);
for (; c; n = 1000)
{
printf("%d", c->s[--n]);
j++;
for (n--; n >= 0; n--, j++)
printf("%04d", c->s[n]);
c = c->n;
}
return 0;
}

为啥你们都不喜欢用函数？