Big Number


Time Limit: 1000MS

 

Memory Limit: 65536K

Total Submissions: 27526

 

Accepted: 8784


Description


In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number.


Input


Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 <= m <= 10^7 on each line.


Output


The output contains the number of digits in the factorial of the integers appearing in the input.


Sample Input


2 10 20


Sample Output


7 19


//题意：


给你一个n，让你算出 n！的数的位数。


思路：


hdoj1018和这个题一样，直接对每个n取对数就行了，但这个题交上去竟然超时了。。。所以就转换成打表了，直接先将所有的（1--10000000）的值都打到数组a中，感觉挺快的，但是交上去超内存了。。。最后用打表和暴力结合，先打出（1--8000000）的值存到数组a中，如果n>8000000,那么在暴力求出从8000000--n之间的位数加上a[8000000],就行了。。。


#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define ll long long
#define ull unsigned long long
#define IN __int64
#define N 8000010
#define M 1000000007
using namespace std;
double a[N];
int init()
{
	a[0]=0;
	for(int i=1;i<N;i++)
	{
		a[i]=a[i-1]+log10(i);
	}
}
int count(int n)
{
	double m=a[N-1];
	for(int i=N;i<=n;i++)
		m+=log10(i);
	return (int)m+1;
}
int main()
{
	int t;
	int n,i;
	init();
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		if(n<N)
			printf("%d\n",(int)a[n]+1);
		else
			printf("%d\n",count(n));
	}
	return 0;
}