#include <stdio.h>
#include <math.h>

int n;
const double e = 2.7182818284590452354, pi = 3.141592653589793239;

double f( int a )
{
	return log10( sqrt( 2 * pi * a ) ) + a * log10( a / e );
}

int main()
{
	int cas, ans;
	double i, s;
	
	scanf( "%d", &cas );
	
	while( cas-- )
	{
		scanf( "%d", &n );
		if( n < 100000 )
		{
			for( s=0, i=1; i<=n; i++ )
				s += log10( i );
		}
		else s = f( n );
		ans = (int)s;
		if( ans <= s )
			ans++;
		
		printf( "%d\n", ans );
	}

	return 0;
}

• Re:这个程序需要改进，否则wa，下面这样就过了:) (24) qiaokan 2008-07-06 20:24:05
• 不用判断n的大小 0ms ac (448) j_will 2009-04-25 15:49:09