poj 1423 Big Number

最新推荐文章于 2019-07-24 00:23:00 发布

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Big Number

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24637		Accepted: 7895

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

Source

Dhaka 2002

思路：log10(n!)=log10(1*2*3…*n)=log10(1)+log10(2)+…+log10(n)

方法一：用空间换时间，开个数组把结果存起来一部分！

#include"stdio.h"
#include"math.h"
#include"string.h"
#define N 8000000
double a[N];
void Inti()
{
	int i;
	a[0]=0;
	for(i=1;i<N;i++)
		a[i]=a[i-1]+log10(i);
}
int main()
{
	int t,n,i;
	double s;
	Inti();
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		if(n<N)
		{
			printf("%d\n",(int)a[n]+1);
			continue;
		}
		s=a[N-1];
		for(i=N;i<=n;i++)
			s+=log10(i);
		printf("%d\n",(int)ceil(s));
	}
	return 0;
}

方法二：

n! = sqrt(2*π*n) * ((n/e)^n) * (1 + 1/(12*n) + 1/(288*n*n) + O(1/n^3)) π = acos(-1) e = exp(1) 两边对10取对数
忽略log10(1 + 1/(12*n) + 1/(288*n*n) + O(1/n^3)) ≈ log10(1) = 0 得到公式
log10(n!) = log10(sqrt(2 * pi * n)) + n * log10(n / e)。

#include"stdio.h"
#include"math.h"
#include"string.h"
#define N 80000
int main()
{
int t,n,i;
double s,pi,e;
pi=acos(-1);
e=exp(1);
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
s=log10(sqrt(2*pi*n))+n*log10(n/e);
if(n==1)
s=1;
printf("%d\n",(int)ceil(s));
}
return 0;
}