HDU 1108

Posted 2020-10-01 ISGuXing

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38440    Accepted Submission(s): 18627

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

题目大意：问n的阶乘有多少位。

思路：数学！我们发现阶乘的位数可以这么计算：

t+=log10(i*1.00);

sum=int(t)+1;

这样就很好的求出位数了。
代码：

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<string.h>
 4 #include<cmath>
 5 #include<math.h>
 6 using namespace std;
 7 int main(){
 8     int T;
 9     cin>>T;
10     while(T--){
11         int n;
12         scanf("%d",&n);
13         int sum=0;
14         double t=0.0;
15         for(int i=1;i<=n;i++){
16             t+=log10(i*1.00);
17         }
18         sum=int(t)+1;
19         cout<<sum<<endl;
20     }
21     return 0;
22 }