HDU 1018 Big Number

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38814    Accepted Submission(s): 18850

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   |   We have carefully selected several similar problems for you:  1013 1017 1016 1071 1019

 

分析：  对于一个数n  它的位数等于 log10(n)+1;

            那么n! 的位数为 log10(n*(n-1).....*1) +1

          公式展开  log10(n)+log10(n-1)+....log10(1)+1;

代码如下：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <map>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int MAXN=40000;
int ans[MAXN];
int flag;
int main()
{
    double sum;
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        sum=1;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
          sum+=log10(i);
        }
        printf("%d\n",(int)sum);
    }
    return 0;
}