hdoj--1018--Big Number(简单数学)

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 32410    Accepted Submission(s): 15175

Problem 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 ≤ n ≤ 10⁷ 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

 

Source

Asia 2002, Dhaka (Bengal)

 

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刚看题的时候没看数据范围，傻乎乎的用大数阶乘，后来才发现有规律啊！！

附上大数的，但是不会过

//#include<stdio.h>
//#include<string.h>
//int main()
//{
//	int t;
//	scanf("%d",&t);
//	int n,num[200];
//	while(t--)
//	{
//		memset(num,0,sizeof(num));
//		num[0]=1;
//		scanf("%d",&n);
//		int j;
//		int c=0;
//		for(int i=1;i<=n;i++)
//		{
//			for(j=0;j<=199;j++)
//			{
//				num[j]=num[j]*i+c;
//				if(num[j]>=10)
//				{
//					c=num[j]/10;
//					num[j]-=c*10;
//				}
//				else c=0;
//			}
//			
//		}
//		for(j=199;j>=0;j--)
//		if(num[j]!=0) 
//		break;
//		printf("%d
",j+1);
//	}
//	return 0;
//} 
#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		double ans=0;
		for(int i=1;i<=n;i++)
		ans+=log10(i);
		printf("%d
",(int)ans+1);
	}
	return 0;
}