Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36450 Accepted Submission(s): 17456

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 ≤ 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

Sample Output

思路：

一个数的位数等于他对10取对数+1

源代码：

#include
    
     #include
     
      #include
      
       using namespace std;int main(){    int t;    int num;    double sum;    scanf("%d",&t);    while(t--)    {        sum=0;        scanf("%d",&num);        for(int i=1;i<=num;i++)        {            sum+=log10((double)i);        }        printf("%d\n",(int)sum+1);    }    return 0;}

上一篇：HDU1042---N!

下一篇：算法训练区间K大数

发表评论

关于作者

喝酒易醉，品茶养心，人生如梦，品茶悟道，何以解忧？唯有杜康！

-- 愿君每日到此一游！

Big Number

发表评论

最新留言

关于作者

推荐文章