Big Number

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

21020

 

 

Sample Output

719

 

 

Source

Asia 2002, Dhaka (Bengal)

 

 

Recommend

JGShining

答案：

#include <stdio.h>
#include <cmath>

int main()
{
 int n, t, i;
 double sum;
 if (scanf("%d", &t) && t)
 {
  while (t--)
  {
   scanf("%d", &n);
   sum = 0;
   for (i = 1; i <= n; ++i)
   {
    sum += log10((double) i);
   }
   printf("%d/n", (long)ceil(sum));
  }
 }
 return 0;
}

下面是另一种算法，不过超时了，并且处理位数有限，但是可以求出相关的阶乘数：

#include <string>

int main()
{
 long n, s, tmp;
 long v;
 char sum[100000000];
 int  j, w;
 
 scanf("%d", &n);
 while (n--)
 {
  scanf("%d", &v);
  memset(sum, 0, sizeof(sum));
  sum[0] = '1';
  for (j = 2; j <= v; ++j)
  {
   s = 0;
   for (w = 0; w < strlen(sum); ++w)
   {
    tmp = j*(sum[w]-'0')+s;
    s = tmp/10;
    sum[w] = tmp%10+'0';
   }
   if (s > 0)
   {
    do
    { 
     sum[w++] = s%10+'0';
     s = s/10;
    }while (s > 0);
   }
  }
  printf("%d/n", strlen(sum));
 }

 return 0;
}