nyoj 28 nyoj 69-大数阶乘

http://acm.nyist.net/JudgeOnline/problem.php?pid=28 大数水题~~

#include<stdio.h>
int a[20000];
void ch(int *a, int n)
{
int i, k;
for(i = 19999; i >= 0; i--)
{
if(a[i])
break;
}
k = i;
for(i = 0; i <= k; i++)
{
a[i] = a[i] * n;
}
for(i = 0; i < 20000; i++)
{
if(a[i] > 9)
{
a[i+1] += a[i] / 10;
a[i] = a[i] % 10;
}
}
}
int main()
{
int n, i;
a[0] = 1;
scanf("%d", &n);
for(i = 1; i <= n; i++)
ch(a, i);
for(i = 19999; i >= 0; i--)
{
if(a[i])
break;
}
for(; i >= 0; i--)
printf("%d", a[i]);
putchar(10);
return 0;
}

这是具体计算阶乘的~

如果只需要计算大数阶乘的位数，像http://acm.nyist.net/JudgeOnline/problem.php?pid=69，有公式可以套~~~ n!的位数 = log10(2*PI*n)/2+n*log10(n/e);

代码如下：

#include<iostream>
#include<math.h>
using namespace std;
int main()
{
int tcases, n;
cin >> tcases;
while(tcases--)
{
cin >> n;
if(n == 1)
cout << 1 << endl;
else
{
int result = (int) (log10(4.0 * acos(0.0) * n)/2 + 1.0 * n * log10(1.0 *n / exp(1.0))) + 1;  //arccos（0）= PI/2
cout << result << endl;
}
}
return 0;
}