| 500! |
In these days you can more and more often happen to see programs which perform some useful calculations being executed rather then trivial screen savers. Some of them check the system message queue and in case of finding it empty (for examples somebody is editing a file and stays idle for some time) execute its own algorithm.
As an examples we can give programs which calculate primary numbers.
One can also imagine a program which calculates a factorial of given numbers. In this case it is the time complexity of order O(n) which makes troubles, but the memory requirements. Considering the fact that 500! gives 1135-digit number no standard, neither integer nor floating, data type is applicable here.
Your task is to write a programs which calculates a factorial of a given number.
Assumptions: Value of a number ``n" which factorial should be calculated of does not exceed 1000 (although 500! is the name of the problem, 500! is a small limit).
Input
Any number of lines, each containing value ``n" for which you should provide value of n!
Output
2 lines for each input case. First should contain value ``n" followed by character `!‘. The second should contain calculated value n!.
Sample Input
10 30 50 100
Sample Output
10! 3628800 30! 265252859812191058636308480000000 50! 30414093201713378043612608166064768844377641568960512000000000000 100! 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
算法竞赛入门经典上的解法如下：会超时。
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 3000;
int a[maxn];
int main()
{
int n;
int i, j;
int c, s;
while (scanf("%d", &n) == 1)
{
memset(a,0,sizeof(a));
a[0] = 1;
for (i = 2; i <= n; i++)
{
c = 0;
for (j = 0; j < maxn; j++)
{
s = a[j] * i + c;
a[j] = s % 10;
c = s/10;
}
}
for (i = maxn; i >= 0;i--)
if (a[i])
break;
printf("%d!\n",n);
for (j = i; j >= 0; j--)
printf("%d",a[j]);
printf("\n");
}
return 0;
}
修改后如下：
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 500;
int a[maxn];
int main()
{
int n;
int i, j;
int len;
int c, s;
while (scanf("%d", &n) == 1)
{
memset(a, 0, sizeof(a));
a[0] = 1;
len = 1;
for (i = 2; i <= n; i++)
{
c = 0;
for (j = 0; j < len; j++)
{
s = a[j] * i + c;
a[j] = s % 1000000;
c = s / 1000000;
}
if (c)
a[len++] = c;
}
printf("%d!\n", n);
printf("%d",a[len-1]);
for (j = len-2; j >= 0; j--)
printf("%06d", a[j]);
printf("\n");
}
return 0;
}