杭电 ACM 1018

Big Number


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17625    Accepted Submission(s): 7897




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)
 


Recommend
JGShining

这个题一般，但是想过缩短时间就难了，下面是拷的别人的方法；可以看一下，后一个方法0ms,第一个900ms还多，

朴素公式为：

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

求位数：log10(n!)=log10(1*2*3…*n)=log10(1)+log10(2)+…+log10(n)，对log10(n!)的值取整加1就是n!的位数。

《计算机程序设计艺术》中给出了另一个公式

    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)

代码：

方法1：

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


int main(void)
{
int n;
int a;
int sum;
double multi;
int i, j;
scanf("%d", &n);
for(i = 0;i < n;i++)
{
scanf("%d", &a);
multi = 0.0;
for(j = 1;j <= a;j++)
{
multi += log10((double)j);
}
sum = (int)multi;
printf("%d\n", sum + 1);
}
return 0;
}

方法2：

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


int main(void)
{
int n, a;
int i, sum;
double pi, e;
pi = acos(-1.0);
e = exp(1.0);
scanf("%d", &n);
for(i = 0;i < n;i++)
{
sum = 0;
scanf("%d", &a);
sum = (int)(log10(sqrtl(2*pi*a))+a*log10(a/e)+1);;
if(a == 1)
sum = 1;
printf("%d\n", sum);
}
return 0;
}